An introduction to group actions

October 1, 2014

Reproduced below is a set of lecture notes I wrote for an undergraduate course on abstract algebra.

The below notes assume that the reader is familiar with basic group theory, up to the first isomorphism theorem. The reader is also expected to have a nodding acquaintance with the language of generators and relations.

1. Groups of Transformations and Permutations

$GL(2,\mathbb{R})$, the set of invertible 2-by-2 matrices with real entries, can be made into a group by considering the matrix multiplication operation as the binary operation on $GL(2,\mathbb{R})$. Matrix multiplication is associative; the identity matrix $$I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ serves as the group identity; every invertible matrix is, well, invertible. In this manner, $GL(2,\mathbb{R})$ is a group of matrices.

Recall, however, that every matrix defines a linear transformation. In the case of $GL(2,\mathbb{R})$, every $M \in GL(2,\mathbb{R})$ defines a linear transformation $L_M:\mathbb{R}^2 \to \mathbb{R}^2$ given by the formula $L_M(v) = M \cdot v$, where $\cdot$ represents the matrix-vector multiplication. In other words, every element of the group $GL(2,\mathbb{R})$ transforms 2-vectors to some other 2-vectors. From this angle, $GL(2,\mathbb{R})$ is a group of transformations on $\mathbb{R}^2$.

Similarly, then, we can define the group $GL(n,\mathbb{F})$ of invertible $n$-by-$n$ matrices with entries in a field $\mathbb{F}$ and consider it a group of transformations on $\mathbb{F}^n$, the space of $n$-vectors with entries in $\mathbb{F}$. Generalizing even further, we define the general linear group on a vector space $V$ over a field $\mathbb{F}$, denoted by $GL(V)$, to be the group of all invertible linear transformations $T:V \to V$ with function composition as the group operation. Each element of $GL(V)$ takes vectors in $V$ and transforms them into other vectors in $V$. In other words, $GL(V)$ is a group of transformations on $V$.

But what is a “group of transformations”, really? Every group we have discussed above has one thing in common: each group $G$ comes with some set $X$ on which each element of $G$ can be thought of as a function $f:X \to X$. The most general group of this kind is the symmetric group on a nonempty set $X$, which is the collection $S_X$ of all bijections $f:X \to X$. We require the functions to be bijections, so that $S_X$ with respect to the function composition operation is a bona fide group. Note that the general linear group $GL(V)$ is a subgroup of $S_V$, the symmetric group on the vector space $V$.

Being a bijection, an arbitrary element $f$ of the symmetric group $S_X$ transforms each element of $X$ in a way that preserves $X$. Indeed, if we let $f$ act on the elements of $X$, then we end up with $X$, relabeled. In this sense, the elements of $S_X$ are called permutations on $X$, and subgroups of $S_X$ permutation groups on $X$.

2. Cayley’s Theorem; Basic Definitions and Examples

Let us now consider an abstract group $G$. Unlike $GL(V)$ or $S_X$, the group $G$ does not come equipped with an obvious set on which the elements of $G$ can act as permutations. What we do have is a binary operation on $G$, which takes two elements of $G$ and outputs one element of $G$. This is to say that, by fixing one of the two input parameters, the group operation can be thought of as a function on $G$. We formalize this observation as follows:

Theorem 2.1 (Cayley). Every group is isomorphic to a permutation group on $G$.

Proof. Given a $g \in G$, we define the left regular representation $L_g:G \to G$ by setting $L_g(x) = gx$ for each $x \in G$. We note that $L_g$ is a bijection, as $L_g^{-1}$ is the inverse of $L_g$.

Let us now define a left-multiplication map $L:G \to S_G$ by declaring $L(g) = L_g$ for each $g \in G$. We fix $g,h \in G$ and observe that

$$\begin{align*} L(gh)(x) &= L_{gh}(x) = ghx \\ &= g(hx) = L_g(L_h(x)) \\ &= (L_g \circ L_h)(x) = (L(g) \circ L(h))(x) \end{align*}$$

for all $x \in G$. From this, it follows that $L(gh) = L(g)L(h)$. Since $g$ and $h$ were arbitrary, we conclude that $L$ is a group homomorphism.

It remains to show that $L$ is injective, whence $G$ is isomorphic to $\operatorname{im} L$, a subgroup of $S_G$. To this end, we fix $g,h \in G$ and suppose that $L(g) = L(h)$. This, in particular, implies that

$$1_G = L_g(g^{-1}) = L(g)(g^{-1}) = L(h)(g^{-1}) = L_h(g^{-1}) = hg^{-1}.$$

Since the inverse element of a fixed group element is unique, we conclude that $g = h$. This completes the proof of Cayley’s theorem. $\square$

We glean from Cayley’s theorem an important idea: we can make an abstract group $G$ act on a set $X$ by identifying $G$ with a permutation group on $X$.

Definition 2.2. Let $G$ be a group. A permutation representation of $G$ is a group homomorphism $\varphi:G \to S_X$ into the symmetric group $S_X$ on some set $X$. Given a nonempty set $X$, we say that $G$ acts on $X$ if there exists a permutation representation $\varphi:G \to S_X$. In this case, $X$ is referred to as a $G$-set. If, in addition, $\varphi:G \to S_X$ is injective, then we say that $G$ acts faithfully on $X$.

Given a permutation representation $\varphi:G \to S_X$, we often write

$$g \cdot x \, \, \, \, \, \, \, \, \, \, \, \text{ or } \, \, \, \, \, \, \, \, \, \, \, \, gx$$

to denote $\varphi(g)(x)$. This is the left action notation, which we shall use throughout this post. Also prevalent in the literature is the right action notation, which employs

$$x \cdot g \, \, \, \, \, \, \, \, \, \, \, \text{ or } \, \, \, \, \, \, \, \, \, \, \, \, xg$$

to denote $\varphi(g)(x)$. We will not use the right action notation in this post.

Here is a simple exercise to get you used to the left action notation.

Exercise 2.3. If $G$ acts on $X$, then the following holds:
$1_G \cdot x = x$ for all $x \in X$.
$g \cdot (h \cdot x) = (gh) \cdot x$ for all $g,h \in G$ and $x \in X$.
Conversely, if (1) and (2) hold, then the mapping $\varphi:G \to S_X$ given by the formula $\varphi(g)(x) = g \cdot x$ is a permutation representation.

We shall therefore use the left action notation and the permutation representation notation interchangeably. In construcitn a group action, we often define a map and check that it is a permutation representation, or define a dot operation and check that properties (1) and (2) hold. If a group action arises from restricting an already-extant functions $f$ on a smaller domain $D$, then we must also check that the range of $f\vert_D$ is a subset of $D$: see Example 2.10.

Let us now study several examples of group actions. There is no need to peruse all of these on the first read: take a look at a few, move on to Section 3, come back whenever certain examples are cited.

Example 2.4. Let $X$ be a nonempty set. The symmetric group $S_X$ acts faithfully on $X$ via the identity representation $\operatorname{id}:S_X \to S_X$ that sends each $\sigma \in \S_X$ to itself. If $G$ is a subgroup of $S_X$, then the canonical injection map $\iota_G:G \to S_X$ that sends each $g \in G$ to itself is an injective permutation representation of $G$ on $X$. Therefore, $G$ acts faithfully on $X$.

Now, if $Y$ is a set that contains $X$ as a subset, then we can define an injective group homomorphism $\varphi:S_X \to S_Y$ by setting

$$\varphi(\sigma)(x) = \begin{cases} \sigma(y) & \text{ if } y \in X; \\ y & \text{ if } y \in Y \smallsetminus X. \end{cases}$$

In other words, we identify $S_X$ with the subgroup of permutations on $Y$ that permutes $X$ and fixes $Y \smallsetminus X$. It now follows that $S_X$ acts faithfully on $Y$. Moreover, $\varphi \circ \iota_G:G \to S_Y$ is an injective group homomorphism, whence $G$ acts faithfully on $Y$.

Example 2.5. In general, if $G$ acts on $X$, then every subgroup $H$ of $G$ acts on $X$. To see this, we take a permutation representaiton $\varphi:G \to X$ and precompose it with the canonical injection map $\iota_H:H \to G$, obtaining another permutation representation $\varphi \circ \iota_H:H \to S_X$. Note that if the action of $G$ on $X$ is faithful, then so is induced action of $H$ on $X$.

Even more generally, if $G$ acts on $X$, and if $\varphi:H \to G$ is a group homomorphism from another group $H$ into $G$, then $H$ acts on $X$. $\square$

Example 2.6. The general linear group $GL(V)$ is a subgroup of the symmetric group $V$, and so $GL(V)$ acts faithfully on $V$. $\square$

Example 2.7 (Linear groups). We introduce a few important subgroups of $GL(n,\mathbb{F})$. Since $GL(n,\mathbb{F})$ is a subgroup of the symmetric group $S_{\mathbb{F}^n}$, Example 2.4 implies that all such subgroups act faithfully on $\mathbb{F}^n$. The special linear group is the subgroup $SL(n,\mathbb{F})$ consisting of matrices of determinant 1. The orthogonal group is the subgroup $O(n,\mathbb{F})$ consisting of matrices $A \in GL(n,\mathbb{F})$ such that $AA^t = A^t A = I_n$, where $t$ denotes the matrix transpose operation and $I_n$ the $n$-by-$n$ identity matrix. The special orthogonal group is the subgroup $SO(n,\mathbb{F}) = O(n,\mathbb{F}) \cap SL(n,\mathbb{F})$.

If $\mathbb{F} = \mathbb{R}$, then we typically write $O(n)$ and $SO(n)$ to denote $O(n,\mathbb{R})$ and $SO(n,\mathbb{R})$, respectively. Both $O(n)$ and $SO(n)$ act faithfully on $\mathbb{R}^n$.

If $\mathbb{F} = \mathbb{C}$, then we define the unitary group $U(n)$ to be the group of matrices $A \in GL(n,\mathbb{C})$ such that $AA^* = A^*A = I_n$, where $*$ denotes the conjugate transpose operation. The special unitary group is the subgroup $SU(n) = U(n) \cap SL(n,\mathbb{C})$. Both $U(n)$ and $SU(n)$ act faithfully on $\mathbb{C}^n$. $\square$

Example 2.8 (Isometry groups, orthogonal groups, and symmetry groups). Let $\lVert x \rVert$ denote the usual Euclidean norm of $x \in \mathbb{R}^n$. An isometry on $\mathbb{R}^n$ is a function $f:\mathbb{R}^n \to \mathbb{R}^n$ such that $\lVert f(x) - f(y)\rVert = \lVert x-y\rVert$ for all $x,y \in \mathbb{R}^n$. We note that isometries are always injective. Indeed, if $f(x) = f(y)$ for some $x,y \in \mathbb{R}^n$, then $0 = \lVert f(x) - f(y) \rVert = \lVert x-y \rVert$, whence $x=y$.

The key fact is as follows; see, for example, Theorem 6.2.3 in M. Artin’s Algebra for a proof.

Theorem 2.8.1. If $f:\mathbb{R}^n \to \mathbb{R}^n$ is an isometry, then there exists an $M \in O(n)$ such that $f(x) = f(0) + Mx$ for all $x \in \mathbb{R}^n$.

Since the mapping $x \mapsto Mx$ is a bijection, we see that every isometry on $\mathbb{R}^n$ is an element of $S_{\mathbb{R}^n}$, the symmetric group on $\mathbb{R}^n$. It follows from Example 2.4 that any group of isometries on $\mathbb{R}^n$ acts faithfully on $\mathbb{R}^n$. Note also that Theorem 2.8.1 implies that $O(n)$ is, in fact, the group of all isometries $f$ on $\mathbb{R}^n$ such that $f(0) = 0$.

Let us now fix a subset $E$ of $\mathbb{R}^n$. An isometry $f:\mathbb{R}^n \to \mathbb{R}^n$ is said to be a symmetry of $E$ if $f(E) = E$. Since $f|_E:E \to E$ is a restriction of an injective function, it must be injective. The surjectivity is guaranteed by defintion, and so we conclude that a symmetry of $E$ is a permutation on $E$. Therefore, any group of symmetries of $E$ is a subgroup of $S_E$, whence by Example 2.4 it acts faithfully on $E$. In particular, the symmetry group of $E$, which we define below, acts faithfully on $E$:

Definition 2.8.2. Let $E \subseteq \mathbb{R}^n$. The symmetry group of $E$ is the group $G$ of all symmetries of $E$.

For example, $O(n)$ is the symmetry group of the $(n-1)$-sphere

$$\mathbb{S}^{n-1} = \lbrace x \in \mathbb{R}^n : \lVert x \rVert = 1\rbrace$.$

To see this, we first fix $M \in O(n)$. $M$ is invertible, and so the mapping $x \mapsto Mx$ is injective on $\mathbb{S}^{n-1}$. Moreover, for each $y \in \mathbb{S}^{n-1}$, we see that

$$\begin{align*} \lVert m^{-1}y \rVert^2 &= \langle M^{-1}y,M^{-1}y \rangle = \langle M^ty, M^t y \rangle \\ &=\langle MM^t y, y \rangle = \langle I_ny,y \rangle = \lVert y \rVert^2 = 1, \end{align*}$$

whence $x=M^{-1}y$ is an element of $\mathbb{S}^{n-1}$ such that $Mx = y$. It follows that the mapping $x \mapsto Mx$ is a symmetry of $\mathbb{S}^{n-1}$. We conclude that the symmetry group of $\mathbb{S}^{n-1}$ includes $O(n)$.

To show the reverse inclusion, we fix a symmetry $f$ of $\mathbb{S}^{n-1}$ and show that $f(x) = Mx$ for some $M \in O(n)$. It suffices to show that $f(0) = 0$, for then the desired result follows from Theorem 2.8.1.

To this end, we suppose for a contradiction that $f(0) \neq 0$ and let $x = \lVert f(0) \rVert^{-1} M^{-1} f(0)$. Since

$$\lVert x\rVert = \lVert f(0) \rVert^{-1} \lVert M^{-1} f(0) \rVert = \lVert f(0)\rVert^{-1} \lVert f(0)\rVert = 1,$$

we see that $x \in \mathbb{S}^{n-1}$. Now, Theorem 2.8.1 implies that

$$\begin{align*} \lVert f(x)\rVert &= \lVert f(0) + Mx \rVert = \lVert f(0) + \lVert f(0) \rVert^{-1} f(0) \rVert \\ &= (1 + \lVert f(0) \rVert^{-1}) \lVert f(0)\rVert = \lVert f(0) \rVert + 1 > 0, \end{align*}$$

which contradicts the assumption that $f$ is a symmetry of $\mathbb{S}^{n-1}$. It now follows that $O(n)$ is the symmetry group of $\mathbb{S}^{n-1}$. $\square$

Example 2.9 (Dihedral groups). Consider $O(2)$, the orthogonal group of degree 2. We have seen in Example 2.8 that $O(2)$ is the symmetry group of the unit circle $\mathbb{S}^1$. The typical elements of $O(2)$ are the rotation matrices

$$R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$

and the reflection matrices

$$T_\theta = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}.$$

Note that $T_\theta = R_\theta T_0$. Since

$$\displaystyle T_0 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ - y \end{pmatrix},$$

$T_0$ represents reflection with respect to the $x$ axis. Consequently, $T_\theta$ can be thought of as reflection with respect to an appropriate axis of reflection determined by the rotation matrix $R_\theta$.

An important fact about rotation matrices and reflection matrices in $O(2)$ is as follows; see, for example, Section 2.1 of Grove & Benson, Finite Reflection Groups for a proof.

Theorem 2.9.1. $O(2)$ is generated by $\lbrace R_\theta : \theta \in \mathbb{R}\rbrace \cup \lbrace T_0\rbrace$.

We now define the dihedral group of degree $n$, denoted by $D_n$, to be the subgroup of $O(2)$ generated by $\rho = R_{2\pi/n}$ and $\tau = T_0$. Every element of $D_n$ is a symmetry of the regular $n$-gon $G^n$ whose vertices are precisely the elements of the following set:

$$\left\lbrace \left( \cos \frac{2k \pi}{n} , \sin \frac{2 k \pi}{n} \right) \in \mathbb{S}^1 : k \in \lbrace 0,1,\ldots,n-1\rbrace \right\rbrace.$$

By Theorem 2.9.1, every symmetry of $G^n$ is a finite product of rotation matrices and $\tau$. Since the symmetry group of $G^n$ cannot contain rotation matrices other than $\rho,\ldots,\rho^{n-1}$, we see that every symmetry of $G^n$ must be a finite product of $\rho$ and $\tau$. The relation $\rho\tau = \tau\rho^{-1} = \tau \rho^{n-1}$ now implies that all such products can be written in the form $\rho^i \tau^j$ for some $i,j \in \mathbb{N}$. All such isometries are already included in the dihedral group, and so we conclude that $D_n$ is the symmetry group of $G^n$. In particular, it follows from Example 2.8 that $D_n$ acts faithfully on $G^n$. $\square$

Example 2.10 (Actions of dihedral groups on other sets). Recall $D_n$, the dihedral group of degree $n$, from Example 2.9. We think of $D_n$ as a subgroup of $O(2)$, so that each element of $D_n$ is a 2-by-2 matrix. For all $k,n \in \mathbb{N}$, we define $v_k^n = ( \cos \frac{2 k \pi}{n}, \sin \frac{2 k \pi}{n} )$. The set

$$V^n = \lbrace v_0^n,v_1^n,\ldots,v_{n-1}^n\rbrace$$

is a subset of $\mathbb{S}^1 = \lbrace x \in \mathbb{R}^2 : \lVert x \rVert = 1\rbrace$, the unit circle in $\mathbb{R}^2$. Let us define a $D_n$-action on $V^n$ by setting $M \cdot v = Mv$ for each $M \in D_n$ and every $v \in V^n$. To check that this is indeed a group action, we must check that the mapping $v \mapsto Mv$ is a bijection on $V^n$. This can be checked explicitly, using rotation matrices and reflection matrices introduced in Example 2.9. $D_n$ can now be thought of as a subgroup of $S_{V^n}$, whence Example 2.4 implies that $D_n$ acts faithfully on $V^n$.

Let us now assume that $n$ is even, and consider the following collection of two-element sets:

$$P^n = \left\lbrace \lbrace v_k^n,v_{k+n/2}^n\rbrace : k \in \lbrace 0,1,\ldots,n/2-1\rbrace\right\rbrace.$$

The $D_n$-action on $P^n$ given by setting $M \cdot \lbrace v,w\rbrace = \lbrace Mv,Mw\rbrace$ is likewise a group action. Note, however, that rotation by $\pi$ fixes all elements of $X$. It follows that this $D_n$-action is not faithful. $\square$

Example 2.11 (Regular actions). Recall from the proof of Cayley’s theorem that the left-multiplication map $L_g:G \to G$ with respect to $g \in G$ is the map $L_g(h) = gh$. The mapping $L:G \to S_G$ that sends each $g \in G$ to the corresponding left-multiplication map $L_g$ is a permutation representation, called the left regular representation.

Analogously, we define the right-multiplication map $R_g:G \to G$ with respect to $g \in G$ to be the map $R_g(h) = hg^{-1}$. The map $R:G \to S_G$ that sends each $g \in G$ to the corresponding right-multiplication map $R_g$ is a permutation representation, called the right regular representation.

We note that $G$ acts faithfully on itself in both cases. $\square$

Example 2.12 (Regular action on cosets). Let $G$ be a group and $H$ be a subgroup of $G$. We define $G/H = \lbrace xH : x \in G\rbrace$, the collection of all left cosets. We define a $G$-action on $G/H$ by setting $g \cdot xH = (gx)H$. This is indeed a group action, as $1_G \cdot xH = xH$ and

$$(g_1g_2) \cdot xH = (g_1g_2x)H = g_1(g_2H) = g_1 \cdot (g_2 \cdot H)$$

for all $g_1,g_2,x \in G$. $\square$

Example 2.13 (Conjugation on elements). Combining the left-multiplication map $L_g$ and the right-multiplication maps $R_g$ from Example 2.10, we obtain a very important permutation representation. We define the conjugation representation of a group $G$ to be the map $\varphi:G \to S_G$ defined by the formula $\varphi(g)(x) = gxg^{-1}$. Since $gxg^{-1} = (L_g \circ R_g)(x)$, we see at once that each $\varphi(g)$ is a permutation on $G$. Observe that $1_G \cdot x = 1_Gx = x$, and that

$$\begin{align*} (g_1g_2) \cdot x &= (g_1g_2)x(g_1g_2)^{-1} = g_1(g_2xg_2^{-1})g_1^{-1} \\ &= g_1(g_2 \cdot x)g_1^{-1} = g_1 \cdot (g_2 \cdot x)\end{align*}$$

It follows that $\varphi$ is a permutation representation. The resulting action is referred to as the conjugation action. In light of this, we call $gxg^{-1}$ the conjugation of $x$ by $g$.

The conjugation action is not necessarily faithful. For example, if $G$ is abelian, then $gxg^{-1} = x$ regardless of the choice of $g \in G$. $\square$

Example 2.14 (Conjugation on subgroups). Let $G$ be a group, and let $\mathcal{S}(G)$ be the collection of all subgroups of $G$. We claim that the conjugate subgroup $gHg^{-1} = \lbrace ghg^{-1} : h \in H\rbrace$ of an arbitrary subgroup $H$ of $G$ is also a subgroup of $G$, regardless of our choice of $g \in G$. To see this, we fix $g \in G$. Since $1_G = g1_Gg^{-1}$, the identity element $1_G$ is in $gHg^{-1}$. If $gh_1g^{-1},gh_2g^{-1} \in gHg^{-1}$, then

$$(gh_1g^{-1})(gh_2g^{-1}) = gh_1h_2g^{-1} \in gHg^{-1},$$

as $h_1h_2 \in H$. Therefore, $gHg^{-1}$ is closed under group operation. Finally, if $ghg^{-1} \in gHg^{-1}$, then

$$(ghg^{-1})^{-1} = gh^{-1}g^{-1} \in gHg^{-1},$$

as $h^{-1} \in H$. We conclude that $gHg^{-1} \leq G$. Let us now define a mapping $\phi:G \to S_{\mathcal{S}(G)}$ by setting $\phi(g)(H) = gHg^{-1}$ for each $g \in G$ and every $H \in \mathcal{S}(G)$. Observe that $\phi(g)$ is a permutation on $\mathcal{S}(G)$, as $\phi(g^{-1})$ is the inverse of $\phi(g)$. Moreover,

$$\begin{align*} (g_1g_2) \cdot H &= (g_1g_2)H(g_1g_2)^{-1} = g_1(g_2Hg_2^{-1})g_1^{-1} \\ &= g_1(g_2 \cdot H)g_1^{-1} = g_1 \cdot (g_2 \cdot H)\end{align*}$$

and so $\varphi$ is a permutation representation. Similarly as in Example 2.13, the action induced by $\phi$ is not necessarily faithful. For example, if $G$, then $gHg^{-1} = H$ for all $g \in G$ and $H \in \mathcal{S}(G)$. $\square$

3. Counting Principles

Let us now turn to the abstract theory of group actions; some applications of the theory presented in this section can be found in Section 4. The two fundamental objects associated with a group action are as follows:

Definition 3.1. Let $G$ act on $X$, and fix $x \in X$. The $G$-orbit of $x$ is
$$\mathcal{O}(x) = \lbrace gx : g \in G\rbrace.$$
The stabilizer, or the isotropy subgroup, of $x$ is
$$G_x = \lbrace g \in G : gx = x\rbrace.$$

We remark that $\mathcal{O}(x)$ is a subset of $X$, and that $G_x$ is a subgroup of $G$. Example 3.2. Pick a group $G$ and let $G$ act on itself by conjugation (see Example 2.13). The orbit of $x \in G$ is the conjugacy class of $x$

$$Conj(x) = \lbrace gxg^{-1} : g \in G\rbrace,$$

and the stabilizer of $x$ is the centralizer

$$C_G(x) = \lbrace g \in G : gxg^{-1} = x\rbrace.$$

Example 3.3. Pick a group $G$ and let $G$ act on the collection of all subgroups $\mathcal{S}(G)$ by conjugation (see Example 2.14). The orbit of $H \in \mathcal{S}(G)$ is the collection of all conjugates

$$\lbrace gHg^{-1} : g \in G\rbrace,$$

and the stabilizer of $H$ is the normalizer

$$N_G(H) = \lbrace g \in G : gHg^{-1} = H\rbrace.$$

Exercise 3.4. Write down orbits and stabilizers for actions introduced in Example 2.9, Example 2.10, Example 2.11, and Example 2.12.

How are the orbit and the stabilizer of a point $x$ related? Observe that the coset $h G_x$ can be rewriten as follows, by making the substitution $z = hg$:

$$\begin{align*} h G_x &= \lbrace hg : g \in G \text{ such that } gx = x\rbrace \\ &= \lbrace z : z \in G \text{ such that } (h^{-1}z) x = x\rbrace \\ &= \lbrace z : z \in G \text{ such that } z x = h x\rbrace \\ &= \lbrace g : g \in G \text{ such that } gx = hx\rbrace. \end{align*}$$

Therefore, there is a one-to-one correspondence between the cosets $h G_x$ and the points of the form $hx$. Now, the collection of all points of the form $hx$ is precisely the stabilizer $\mathcal{O}(x)$. We therefore conjecture that $Vert \mathcal{O}(x) \rVert$ equals the number of cosets $hG_x$. This turns out to be true:

Theorem 3.5 (Orbit-stabilizer theorem). Let a group $G$ act on a set $X$. For each fixed $x \in X$, we have the formula
$$\vert\mathcal{O}(x)\vert = [G:G_x].$$

Proof. Let $G/G_x$ denote the left coset space of $G_x$. We define a function $f:\mathcal{O}(x) \to G/G_x$ by setting $f(gx) = gG_x$. Let us first check that $f$ is well-defined. Indeed, if $gx = hx$, then $x = g^{-1}hx$, and so $g^{-1}h \in G_x$. Therefore, $g^{-1}hG_x = G_x$, whence $gG_x = hG_x$. It follows that $f$ is well-defined.

We claim that $f$ is bijective. If $gG_x = hG_x$, then $G_x = g^{-1}hG_x$, and so $g^{-1}h \in G_x$. This implies that $x = g^{-1}hx$, whence $gx = hx$. It follows that $f$ is injective. Moreover, $f(gx) = gG_x$ for all $g \in G$, and so $f$ is surjective. We conclude that $Vert \mathcal{O}(x) \rVert = [G:G_x]$. $\square$

The orbit-stabilizer theorem is a counting principle, in the sense that many combinatorial results about finite groups can be derived as simple corollaries. Below, we establish a few.

Corollary 3.6. Let a finite group $G$ act on $X$. For each $x \in X$, we have the identity $\vert\mathcal{O}(x)\vert = \frac{\vert G \vert}{\vert G_x \vert}$. In particular, $\vert \mathcal{O}(x) \vert$ divides $\vert G \vert$.

Proof. We apply Lagrange’s theorem and the orbit-stabilizer theorem to obtain the following:

$$|G| = |G_x|[G:G_x] = |G_x||\mathcal{O}(x)|.$$

Diving through by $|G_x|$, we obtain the desired result. $\square$

Corollary 3.7. Let $G$ be a finite group. For each $x \in G$, the number of conjugates of $x$ equals $[G:C_G(x)]$.

Proof. Apply the orbit-stabilizer theorem to Example 3.2. $\square$

Corollary 3.8. Let $G$ be a finite group. For each $H \leq G$, the number of conjugate subgroups of $H$ equals $[G:N_G(H)]$.

Proof. Apply the orbit-stabilizer theorem to Example 3.3. $\square$

As alluded to above, each of the three corollaries give rise to useful combinatorial results in group theory. Let us first establish a combinatorial restriction on the size of $X$. For this, we need the following result:

Theorem 3.9 (Orbit decomposition theorem). Let a group $G$ act on a set $X$. The collection $\lbrace \mathcal{O}(x) : x \in X\rbrace$ forms a partition of $X$.

Proof. Suppose that $\mathcal{O}(x) \cap \mathcal{O}(y) \neq \varnothing$ and fix $z \in \mathcal{O}(x) \cap \mathcal{O}(y)$. Find $g_0,h_0 \in G$ such that $g_0 \cdot x = z = h_0 \cdot y$. It now follows that $g \cdot x = (gg_0^{-1} h_0) \cdot y$ for all $g \in G$, whence $\mathcal{O}(x) \subseteq \mathcal{O}(y)$. Similarly, we see that $h \cdot y = (hh_0^{-1} g_0) \cdot x$ for all $h \in H$, and so $\mathcal{O}(y) \subseteq \mathcal{O}(x)$. We conclude that $\mathcal{O}(x) = \mathcal{O}(y)$. $\square$

The following combinatorial principle now is an immediate corollary of what we have established thus far.

Corollary 3.10. Let a finite group $G$ act on a finite set $X$. Let $\Gamma$ be a subset of $X$ consisting of one representative from each set in the partition formed by the orbits (see Theorem 3.9). Then
$$\vert X \vert = \sum_{x \in \Gamma} \vert \mathcal{O}(x) \vert =\sum_{x \in \Gamma} [G:G_x] = \vert G \vert \sum_{x \in \Gamma}\frac{1}{\vert G_x \vert}.$$

Proof. The first equality is an immediate consequence of Theorem 3.9. The second equality follows from the orbit-stabilizer theorem (Theorem 3.5). The third equality follows from Corollary 3.6. $\square$ When a group $G$ acts on itself, then Corollary 3.10 becomes a statement about the cardinality of the group itself. Let us carefully translate Corollary 3.10 in the context of conjugation action. For this, we need the following result:

Corollary 3.11. The conjugacy classes of a group $G$ form a partition of $G$.

Proof. From Example 3.2, we know that the orbits of the conjugation action of $G$ on itself are precisely the conjugacy classes. The orbit decomposition theorem (Theorem 3.9) now furnishes the desired result. $\square$

We are now ready to adapt Corollary 3.9 to describe groups acting on themselves by conjugation.

Theorem 3.12 (Conjugacy class equation). Let $G$ be a finite group. Let $\Gamma$ be a subset of $G$ consisting of one representative from each set in the partition formed by the conjugacy classes (see Corollary 3.11). Then
$$\begin{align*} \vert G \vert &= \sum_{x \in \Gamma} [G : C_{G}(x)] \\ &= \vert Z(G) \vert + \sum_{x \in \Gamma \smallsetminus Z(G)} [G:C_{G}(x)], \end{align*}$$
where
$$Z(G) = \lbrace x \in G : gxg^{-1} = x \text{ for all } g \in G\rbrace$$
is the center of $G$.

Proof. The first equality follows from Corollary 3.10 and Corollary 3.7. The second equality follows from the fact that the conjugacy class of each element in the center is a singleton set. $\square$

The class equation, combined with Corollary 3.8, allows us to establish powerful converses to Lagrange’s theorem, which we establish in the next section. We conclude by establishing a combinatorial restriction on the size of $\Gamma$.

Theorem 3.12 (Burnside's lemma). Let a finite group $G$ act on a finite set $X$. Let $\Gamma$ be a subset of $X$ consisting of one representative from each set in the partition formed by the orbits (see Theorem 3.9). Then
$$\vert \Gamma \vert = \frac{1}{\vert G \vert} \sum_{g \in G} X^g,$$
where $X^g$ denotes the number of elements of $X$ fixed by the action of $g$.

Proof. We first observe that

$|\Gamma| = \sum_{x \in X}\frac{1}{|\mathcal{O}(x)|},$

as there are $|\mathcal{O}(x)|$-many orbits in the equivalence class of $\mathcal{O}(x)$ furnished by the orbit decomposition theorem (Theorem 3.9). For each $g \in G$ and every $x \in X$, we define

$$F(g,x) = \begin{cases} 1 & \text{ if } g \cdot x = x; \\ 0 & \text{ if } g \cdot x \neq x. \end{cases}$$

We now invoke the orbit-stabilizer theorem (Corollary 3.6) to deduce that

$$\begin{align*} |\Gamma| &= \displaystyle \frac{1}{|G|} \sum_{x\in X} |G_x| = \displaystyle \frac{1}{|G|} \sum_{x \in X} \sum_{g \in G} F(g,x) \\ &= \displaystyle \frac{1}{|G|} \sum_{g\in G} \sum_{x \in X} F(g,x) = \displaystyle \frac{1}{|G|}\sum_{g \in G} X^g, \end{align*}$$

as was to be shown. $\square$

4. Sylow’s theorems

While Lagrange’s theorem imposes restrictions on the orders of subgroups and elements of a group, it does not guarantee the existence of a subgroup or an element of a certain order.

Example 4.1. Consider the alternating group $A_4$, which is of order 12. We show that $A_4$ has no subgroup of order 6. To this end, we shall make use of the following preliminary result:

Lemma 4.2. Every subgroup of a group $G$ of index 2 is normal in $G$.

We assume for now that the lemma holds. Suppose for a contradiction that $H$ is a subgroup of $A_4$ of order 6, which must be a normal subgroup of $A_4$ by the lemma. The conjugacy classes of $A_4$ are as follows:

$$\begin{align*} &\lbrace \operatorname{id}\rbrace, \lbrace (12)(34), (13)(24),(14)(23)\rbrace, \\ &\lbrace (123),(124),(134),(234)\rbrace, \\ &\lbrace (132),(142),(143),(243)\rbrace. \end{align*}$$

Since $H$ is normal, $H$ must be a union of conjugacy classes. Since $\lbrace e\rbrace$ must be a subset of $H$, this implies that we ought to be able to write $|H|=6$ as a sum of 1 and the cardinalities of conjugacy classes. This is impossible, and so we conclude that $A_4$ has no subgroup of order 6.

It now suffices to establish the lemma. Let $G$ be a group and let $H$ be a subgroup of index 2. Note that the two left cosets of $H$ are $H$ itself and $g_0H$ for an arbitrary $g_0 \in G \smallsetminus H$. Fix $h \in H$ and $g \in G$. If $g \in H$, then $ghg^{-1} \in H$. We therefore suppose that $g \in G \smallsetminus H$. If $g(hg^{-1}) \in gH$, then $hg^{-1}) \in H$, and so $g^{-1} = h^{-1}(hg^{-1})$. This, in particular, implies that $g \in H$, whence $gH \subseteq H$, a contradiction. It follows that $ghg^{-1} \in H$, as was to be shown.

See Brennan and MacHale, “Variations on a Theme: $A_4$ Definitely Has no Subgroup of Order Six!” for alternate proofs. $\square$

When, then, could we establish a converse to Lagrange’s theorem? The solution is intimately tied to the study of $p$-groups, which we now define.

Definition 4.3. Let $p$ be a prime number. A group $G$ is said to be a $p$-group if each element of $G$ is of order $p^n$ for some integer $n$ that depends on the element.

The problem of finding a nontrivial $p$-subgroup of a group $G$ is equivalent to finding an element of $G$ of order $p$. Indeed, if $g \in G$ is an element of order $p$, then the cyclic subgroup $\langle g \rangle$ is a nontrivial $p$-subgroup of $G$. Conversely, if $H$ is a nontrivial $p$-subgroup of $H$, then we can find a non-identity element $g \in H$ of order $p^n$. In this case, $g^{p^{n-1}}$ is an element of order $p$.

Now, in order for a finite group $G$ to have a $p$-subgroup, the order of $G$ must be divisible by $p$. The following result shows that the converse holds.

Theorem 4.4 (Cauchy, 1845). Every finite group $G$ whose order is divisible by a prime number $p$ contains an element of order $p$.

Proof. We first prove the theorem for abelian groups. We shall proceed by induction on $m = |G| / p$. If $m = 1$, then $G$ is a group of prime order, hence cyclic; it follows that $G$ contains an element of order $p$.

Fix $m > 1$ and suppose that every abelian group of order $pn$ for each $1 \leq n < m$ contains an element of order $p$. Let $g$ be a non-identity element of $G$. If $|g|$ is divisible by $p$, then some power of $g$ is of order $p$. If not, then $G / \langle g \rangle$ is an abelian group of order $pn$ for some $1 \leq n < m$, whence it contains an element $h \langle g \rangle$ of order $p$. Now, the canonical projection $\pi:G \to G/\angle x \angle$ is a homomorphism, and so the order of $h$ must be divisible by $p$, order of $\pi(h) = h\langle x \rangle$. It follows that some power of $h$ must be $p$. By induction, every Aelian group whose order is divisible by $p$ contains an element of order $p$.

We now lift the assumption that $G$ is abelian. Once again, we proceed by induction on $m = |G| / p$. The $m=1$ case is established above. Fix $m > 1$ and suppose that every group of order $pn$ for some $1 \leq n > m$ contains an element of order $p$. Let $g \in G$. If $g \notin Z(G)$, then $[G:C_G(g)]$, the number of conjugates of $g$, is larger than $1$, and so $|C_G(g)| < |G|$. By induction, $C_G(g)$ must contain an element of order $p$, whence so must $G$. We may therefore assume that $p \nmid |C_G(g)|$ whenever $g \notin Z(G)$. Lagrange’s theorem now implies that $p \mid [G:C_G(g)]$ whenever $g \notin Z(G)$. We now invoke the conjugacy class equation (Theorem 3.12):

$$|G| = |Z(G)| + \sum_{g \in \Gamma \smallsetminus Z(G)} [G : C_G(g)].$$

Since $|G|$ and all $[G:C_G(g)]$ are divisible by $g$, it follows that $Z(G)$ is divisible by $p$. But $Z(G)$ is abelian, and so $Z(G)$ must contain an element of order $p$. It follows that $G$ contains an element of order $p$. $\square$

Cauchy’s theorem provides a convenient equivalence characterization of finite $p$-groups:

Corollary 4.5. A finite group $G$ is a $p$-group if and only if the order of $G$ is a power of $p$.

Proof. If $|G| = p^n$, then Lagrange’s theorem implies that $G$ is a $p$-group. If $|G|$ is divisible by some other prime $q$, then Cauchy’s theorem implies that $G$ contains an element of order $q$, and so $G$ is not a $p$-group. $\square$

The main result of this section is a substantial strengthening of Cauchy’s theorem, due to Sylow.

Definition 4.6. A Sylow $p$-subgroup of a group $G$ is a maximal $p$-subgroup. In other words, if $P$ is a Sylow $p$-subgroup of $G$, then no other $p$-subgroup of $G$ properly contains $P$.

Sylow’s theorems concern the existence and various other properties of Sylow $p$-subgroups of finite groups whose orders are divisible by $p$.

Theorem 4.7 (Sylow, 1872). Let $G$ be a finite group of order $p^nm$, where $p$ is a prime number and $m$ is an integer relatively prime to $p$. The following holds:
First Sylow theorem. $G$ contains a Sylow $p$-subgroup of order $p^n$.
Second Sylow theorem. All Sylow $p$-subgroups are conjugates of one another. In other words, if $P$ and $Q$ are Sylow $p$-subgroups of $G$, then there exists a $g \in G$ such that $gPg^{-1} = Q$. This, in particular, implies that all Sylow $p$-subgroups are isomorphic to one another
Third Sylow theorem. Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. We have the following combinatorial restrictions on $n_p$:
$n_p$ divides $\vert G \vert$;
$n_p \equiv 1 \text{ mod } p$.

Before we prove the Sylow theorems, we make a few observations. Firstly, the second Sylow theorem, in conjunction with the third Sylow theorem and Corollary 4.5, implies that $G$ has a normal subgroup of order $p^n$ if and only if $n_p = 1$. The third Sylow theorem provides two combinatorial restrictions on $n_p$, which can be exploited in various clever ways to determine $n_p$. Secondly, the second Sylow theorem, in conjunction with the first Sylow theorem, implies that every Sylow $p$-subgroup of $G$ is of order $p^n$. In other words, maximality in subgroup inclusion automatically implies maximality in order. Thirdly, the maximality observation, combined with the second Sylow theorem, implies the following structure theorem:

Corollary 4.8 (Frattini's argument). Let $G$ be a finite group and $K$ a normal subgroup of $G$. If $P$ is a Sylow $p$-subgroup of $K$ for some prime $p$, then
$G = KN_G(P).$

Proof of Corollary. Fix $g \in G$. Since $K$ is normal in $G$, we see that $K = gKg^{-1}$. Now, $gPg^{-1} \leq gKg^{-1} = K$, and so $gPg^{-1}$ is a Sylow $p$-subgroup of $K$ by maximality. The second Sylow theorem therefore furnishes $k \in K$ such that $kPk^{-1} = gPg^{-1}$, so that $P = (k^{-1}g)P(k^{-1}g)^{-1}$. It follows that $k^{-1}g \in N_G(P)$. Since $g \in G$ was arbitrary, the factorization $g = k(k^{-1}g)$ establishes the desired identity. $\square$

Let us now turn to the proof of the Sylow theorems, which are taken from Rotman, An Introduction to the Theory of Groups. As in the statement of the theorem, we assume that $|G| = p^n m$. We shall make use of the following lemma:

Lemma 4.9. Let $P$ be a Sylow $p$-subgroup of a finite group $G$. The following holds:
$\vert N_G(P)/P\vert$ is relatively prime to $p$;
If $g \in G$ is of order $p^k$ for some $k$ and $gPg^{-1} = P$, then $g \in P$.

Proof of lemma. (1) If $|N_G(P)/P|$ is divisible by $p$, then Cauchy’s theorem (Theorem 4.4) furnishes an element $gP$ of order $p$. Therefore, $H = \langle gP \rangle$ is a subgroup of $N_G(P)/P$ of order $p$.

Let $\pi:N_G(P) \to N_G(P)/P$ be the canonical projection map and set $H^* = \pi^{-1}(H)$. If $g,h \in H^*$, then $\pi(gh^{-1}) = \pi(g)\pi(h)^{-1} \in H$, and so $gh^{-1} \in H^*$. It follows that $H^* \leq N_G(P)$. Moreover, the first isomorphism theorem, in conjunction with Lagrange’s theorem, implies that

$$|H^*| = \left\vert\ker \pi\vert_{H^*}\right\vert \left\vert \operatorname{im} \pi\vert_{H^*} \right\vert = \vert P \vert \vert H \vert,$$

whence by Corollary 4.5 $H^*$ is a $p$-subgroup of $G$ containing $P$. This contradicts the maximality of $P$, and so we conclude that $|N_G(P)/P|$ fails to be divisible by $p$.

(2) Note that $g' = g^{p^{k-1}}$ is of order $p$. Since $N_G(P) \leq G$, we see that $g' \in N_G(P)$, and so $g’P \in N_G(P)/P$. If $g' \notin P$, then $|g'| = p$, and so $|gP| = p'$. By Lagrange’s theorem, $|N_G(P)/P|$ must be divisible by $p$, and this contradicts (1). $\square$

Proof of the Second and Third Sylow theorems. Let $X = \lbrace P_1,\ldots,P_k\rbrace$ denote the collection of all conjugates of $P$ in $G$, where $P_1 = P$. Taking a cue from Example 2.14, we define the conjugation action of $G$ on $X$ as follows: $\varphi(g) = (gPg^{-1}$ for each $g \in G$. It is routine to check that $\varphi$ is a group homomorphism from $G$ to $S_X$: a simple modification of the argument in Example 2.14 suffices.

Let $Q$ be an arbitrary Sylow $p$-subgroup of $G$. $Q$ is a subgroup of $G$, and so $Q$ acts on $X$ by conjugation (Example 2.5). By the orbit-stabilizer theorem (Corollary 3.6), the size of each orbit divides $|Q|$, whence by Corollary 4.5 each $|\mathcal{O}(P_i)|$ is a power of $p$.

If $|\mathcal{O}(P_i)| = 1$ for some $1 \leq i \leq k$, then $gP_ig^{-1} = P_i$ for all $g \in Q$. In this case, we must have $Q \leq P_i$ by Lemma 4.9(2). Since $P_i$ is a $p$-group by Corollary 4.5, the maximality of $Q$ implies that $P_i \leq Q$, and so $Q = P_i$. Now, if we take $Q$ to be $P$, then we must have $P \neq P_i$ for all $2 \leq i \leq k$, and so $|\mathcal{O}(P_i)| = p^{t_i}$ for some $t_i \geq 1$ whenever $2 \leq i \leq k$. By the orbit decomposition theorem (Theorem 3.9), $X$ is the disjoint union of distinct orbits, whence $|X|$ equals 1 plus a sum of powers of $p$. In other words, $|X| \equiv 1 \text{ mod } p$.

Now, we assume for a contradiction that $Q$ is a Sylow $p$-subgroup of $G$ not included in $X$. We have shown above that $P_i = Q$ whenever $|\mathcal{O}(P_i)| = 1$. Therefore, we must have $|\mathcal{O}(P_i)| = p^{t_i}$ for some $t_i \geq 1$ for all $1 \leq i \leq k$. This, in particular, implies that $|X| \equiv 0 \text{ mod } p$, contradicting the 1-mod-p relation. Therefore, all Sylow $p$-subgroups are conjugates of one another. This, in particular, implies that $|X| = n_p$, whence $n_p \equiv 1 \text{ mod } p$.

We now invoke the orbit-stabilizer theorem to the conjugation action of $G$ on the set of all subgroups of $G$ to deduce that $[G:N_G(P)] = |X| = n_p$ (Corollary 3.8). Lagrange’s theorem implies that $n_p$ divides $|G|$. This completes the proof of the second and third Sylow theorems. $\square$

Proof of the first Sylow theorem. Cauchy’s theorem (Theorem 4.4) implies that $G$ has a subgroup of order $p$, which is a $p$-subgroup of $G$. This, in particular, shows that $G$ has a Sylow $p$-subgroup $P$. Indeed, if $Q$ is a non-Sylow $p$-subgroup of $G$, then, by definition, there exists a $p$-subgroup $Q'$ of $G$ that properly contains $Q$. Since $|G|$ is finite, this process must terminate in finitely many steps, and we obtain a Sylow $p$-subgroup $P$.

We claim that $[G:P]$ is relatively prime to $p$. Since Corollary 4.5 implies that $|P|$ must be a power of $p$, the claim, in conjunction with Lagrange’s theorem, implies that $[G:P] = m$ and $|P| = p^n$, as was to be shown. To prove the claim, we shall make use of the following lemma:

Lemma 4.9. If $C$ is a finite group and $A \leq B \leq C$, then $[C:A] = [C:B][B:A]$.

Proof of lemma. This is an immediate consequence of the fact that cosets of a subgroup partition a group into equal-sized equivalence classes. $\square$ We let $C = G$, $B = N_G(P)$, and $A = P$ and invoke the lemma to show that both $[G:N_G(P)]$ and $[N_G(P):P]$ are relatively prime to $p$. We apply the orbit-stabilizer theorem to the conjugation action of $G$ on the set of all subgroups of $G$ to deduce that $[G:N_G(P)] = n_p$ (Corollary 3.8). The third Sylow theorem implies that that $n_p \equiv 1 \text{ mod } p$, whence $[G:N_G(P)]$ is relatively prime to $p$.

As for $[N_G(P):P]$, we first note that $P \triangleleft N_G(P)$, so that $[N_G(P):P] = |N_G(P)/P|.$ It follows from Lemma 4.9(1) that $|N_G(P)/P|$ is relatively prime to $p$. This concludes the proof of the first Sylow theorem. $\square$

The Sylow theorems play an important role in the classification of finite groups. In what follows, we study methods of constructing larger groups out of a collection of smaller groups.

5. Direct Products

Recall that the external direct product of two groups $G$ and $H$ is the cartesian product $G \times H$ equipped with coordinatewise group operation: $(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2)$ for all $g_1,g_2 \in G$ and $h_1,h_2 \in H$. The external direct product of $G$ and $H$ is arguably the simplest and the most natural way to construct a new group out of $G$ and $H$. Therefore, it is of interest to determine whether a group can be represented as a direct product of its subgroups.

Example 5.1. The integers mod 6 group $\mathbb{Z}/6\mathbb{Z} = \lbrace 0,1,2,3,4,5\rbrace$ is isomorphic to $\lbrace 0,3\rbrace \times \lbrace 0,2,4\rbrace$. Indeed, the map

$$\varphi:\lbrace 0,3\rbrace \times \lbrace 0,2,4\rbrace \to \lbrace 0,1,2,3,4,5\rbrace$$

given by the formula $\varphi(n,m) = n+m$ is a group isomorphism. $\square$

Example 5.2. The Klein-four group $V = \langle a,b \mid a^2 = b^2 = (ab)^2 = 1 \rangle$ is isomorphic to $\lbrace 1,a\rbrace \times \lbrace 1,b\rbrace$. Indeed, the map

$$\varphi:\lbrace 1,a\rbrace \to \lbrace 1,b\rbrace \to \lbrace 1,a,b,ab\rbrace$$

given by the formula $\varphi(x,y) = xy$ is a group isomorphism. $\square$

Example 5.3. The multiplicative group of complex numbers $\mathbb{C}^\times = \lbrace z \in \mathbb{C} : z \neq 0\rbrace$, equipped with the standard multiplication operation, is isomorphic to the direct product of the circle group $\mathbb{T} = \lbrace z \in \mathbb{C} : |z| = 1\rbrace$ and the group of positive real numbers $\mathbb{R}^+ = \lbrace r \in \mathbb{R} : r > 0\rbrace$, both equipped with the standard multiplication operation. Indeed, every nonzero complex number $z$ can be written uniquely as $|z| e^{i \mathrm{Arg} z}$, and so the map

$$\varphi: \mathbb{R}^+ \times \mathbb{T} \to \mathbb{C}^\times$$

given by the formula $\varphi(r,t) = rt$ is a group isomorphism. $\square$

The question, then, is as follows: given two subgroups $N$ and $K$ of a group $G$, when is $G$ isomorphic to $N \times K$? The above examples suggest that we should be able to create an arbitrary element of $G$ by taking the product of an element of $N$ and an element of $K$. Moreover, in all of the examples above, $N$ and $K$ are effectively disjoint, in the sense that $N \cap K = \lbrace 1_G\rbrace$. These observations lead us to the following definition:

Definition 5.4. Two subgroups $N$ and $K$ of $G$ are permutable complements—or simply complements if there is no danger of confusion—in $G$ if $N \cap K = \lbrace 1_G\rbrace$ and $NK$ equals $G$.

Is this sufficient? Not so, as the next example shows.

Example 5.5. Consider the dihedral group

$D_3 = \langle \sigma, \tau \mid \sigma^3 = \tau^2 = 1, \sigma \tau = \tau \sigma^2 \rangle.$

Let $S = \langle \sigma \rangle$ and $T = \langle \tau \rangle$, so that $S \cap T = \lbrace \operatorname{id}\rbrace$. Since every element of $D_3$ is of the form $\sigma^m \tau^n$, we see that $D_3 = ST$. Nevertheless, $D_3$ cannot be isomorphic to $S \times T$: $D_3$ is nonabelian, but $S \times T$ is abelian. $\square$

What additional conditions do we need?

Proposition 5.6. Let $N$ and $K$ be subgroups of a group $G$. If $N$ and $K$ are permutable complements in $G$, $N \triangleleft G$, and $K \triangleleft G$, then $G \cong N \times K$.

Proof. Define a mapping $\varphi:N \times K \to G$ by setting $\varphi(n,k) = nk$. Since $NK = G$, the mapping is surjective. If $\varphi(n_1,k_1) = \varphi(n_2,k_2)$, then $n_1n_2^{-1} = k_1^{-1}k_2$, so that

$n_1n_2^{-1} ,k_1^{-}1k_2 \in N \cap K$. Since the intersection is trivial, we see that $n_1 = n_2$ and $k_1 = k_2$. It follows that $\varphi$ is injective.

It remains to show that $\varphi$ is a group homomorphism. To this end, we show that $nk = kn$ whenever $n \in N$ and $k \in K$. Indeed, the normality of $N$ in $G$ implies that $nkn^{-1}k^{-1} \in N$, and the normality of $K$ in $G$ implies that $nkn^{-1}k^{-1} \in K$. Since the intersection is trivial, we see that $nkn^{-1}k^{-1} = 1_G$, and so $nk = kn$.

We now observe that

$$\begin{align*} \varphi( n_1n_2,k_1k_2) &= n_1n_2k_1k_2 \\ &= n_1k_1 n_2k_2 \\ &= \varphi(n_1,k_1) \varphi(n_2,k_2) \end{align*}$$

for all $n_1,n_2 \in N$ and $k_1,k_2 \in K$, whence $\varphi$ is a homomorphism. $\square$

The converse also holds.

Proposition 5.7. If $G \cong H_1 \times H_2$, then $G$ has normal subgroups $N$ and $K$ such that $N \cong H_1$, $K \cong K_1$, and $N$ and $K$ are permutable complements in $G$.

Proof. Let $N = H_1 \times \lbrace 1_{H_2}\rbrace$ and $K = \lbrace 1_{H_1}\rbrace \times H_2$. We see at once that $N$ and $K$ are permutable complements in $G$. Moreover, it follows from the definition of the direct product that $nk = kn$ whenever $n \in N$ and $k \in K$.

Fix $n \in N$. For each $g \in G$, we can find $n' \in N$ and $k' \in K$ such that $g = n'k'$. Therefore,

$$\begin{align*} gng^{-1} &= (n'k') n (n'k')^{-1} \\ &= k'(k')^{-1} n' n (n')^{-1} \\ &= n'n (n')^{-1} \in N \end{align*}$$

by the commutativity of elements of $N$ and elements of $K$. Since $g$ and $n$ were arbitrary, we conclude that $N \triangleleft G$. Similarly, we can show that $K \triangleleft G$. $\square$

We can summarize the above two propositions as the equivalence of external direct product and internal direct product.

Definition 5.8. A group $G$ is said to be the internal direct product of subgroups $N$ and $K$ if $N$ and $K$ are permutable complements in $G$, $N \triangleleft G$, and $K \triangleleft G$.

As an application, we show that $GL(n,\mathbb{R})$ can be written as a direct product of two subgroups.

Example 5.9. If $n$ is odd, then the general linear group $GL(n,\mathbb{R})$ is isomorphic to the direct product of the group of scalar matrices $S = \lbrace \lambda I_n : \lambda \neq 0\rbrace$ and the special linear group $SL(n,\mathbb{R})$. Given $M \in GL(n,\mathbb{R})$, we let $\lambda = (\det M)^{1/n}$, so that $N = (\lambda^{-1} I) M$ is of determinant 1. We can then write $M$ as the product $ (\lambda I_n) N$ of a scalar matrix and a matrix of determinant 1. Since $M$ was arbitrary, we see that $GL(n,\mathbb{R}) = S \, SL(n,\mathbb{R})$.

Now, $n$ is odd, and so a scalar matrix $\lambda I_n$ is of determinant 1 if and only if $\lambda = 1$. Therefore, $S \cap SL(n,\mathbb{R}) = \lbrace I_n\rbrace$, and it follows that $S$ and $SL(n,\mathbb{R})$ are permutable complements in $GL(n,\mathbb{R})$.

Since scalar matrices commute with all matrices of compatible size, $S \triangleleft GL(n,\mathbb{R})$. Moreover, if $M \in GL(n,\mathbb{R})$ and $N \in SL(n,\mathbb{R})$, then

$\det(MNM^{-1}) = \det(M)\det(N)\det(M)^{-1} = 1,$

and so $MNM^{-1} \in SL(n,\mathbb{R})$. It follows that $SL(n,\mathbb{R}) \triangleleft GL(n,\mathbb{R})$. We conclude from Proposition 1.6 that $GL(n,\mathbb{R}) \cong S \times SL(n,\mathbb{R})$. $\square$

An important classification theorem regarding direct products is the following:

Theorem 5.10 (Fundamental theorem on finitely generated abelian groups). If $G$ is a finitely generated abelian group, then we can find prime numbers $p_1,\ldots,p_k$ and positive integers $n,n_1,\ldots,n_k$ such that
$$G \cong \mathbb{Z}^n \times \mathbb{Z}/p_1^{n_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_k^{n_k}\mathbb{Z};$$
here $n = 0$ if and only if $G$ is finite.

6. Semidirect Products

Classification results with direct products, even something as strong as Theorem 5.10, does not take into account the decomposition of $D_3$ into the “internal product” $\lbrace \operatorname{id},\sigma,\sigma^2\rbrace\lbrace \operatorname{id},\tau\rbrace$ that we discussed in Example 5.5. It is therefore reasonable to generalize the notion of internal direct products introduced in Definition 5.8 for the sake of completeness.

Definition 6.1. Let $G$ be a group, and fix subgroups $N$ and $K$ of $G$. The group $G$ is said to be an internal semidirect product of $N$ by $K$, denoted by $G = N \rtimes K$, if $N \triangleleft G$ and $N$ and $K$ are permutable complements of $G$.

The notation $\rtimes$ emphasizes the fact that the left component, $N$, is the normal subgroup. Since Definition 2.1 is a straightforward generalization of Definition 1.8, we see that every internal direct product is an internal semidirect product. This, in particular, implies that internal semidirect products are not necessarily isomorphic.

Example 6.2. Consider the dihedral group $D_3$ and the integers mod 6 group $\mathbb{Z}/6\mathbb{Z}$. By Example 5.1 and Example 1.8, each group is an internal semidirect product of a subgroup isomorphic to $\mathbb{Z}/3\mathbb{Z}$ by a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Nevertheless, two groups fail to be isomorphic: $D_3$ is nonabelian, whereas $\mathbb{Z}/6\mathbb{Z}$ is abelian. $\square$

Before we consider more examples of internal semidirect products, we make the following observation: the two “parts” of a semidirect product do not necessarily commute. In the proof of Proposition 5.6, we have shown that if $G$ is an internal direct product of $N$ and $K$, then $nk =kn$ whenever $n \in N$ and $k \in K$. Nevertheless, we know that $\sigma \tau = \tau \sigma^2$ in $D_3$, and so the same principle fails to hold for semidirect products. What we do have, however, is the following:

Proposition 6.3. Let $G$ be a group, $N$ a normal subgroup of $G$, and $K$ a subgroup of $G$. Then $NK$ is a subgroup of $G$ and $NK = KN$.

Proof. Given arbitrary $n_1k_1, n_2k_2 \in NK$, we observe that

$$ \begin{align*} (n_1k_1)(n_2k_2)^{-1} &= n_1k_1 k_2^{-1} n_2 \\ &= n_1 (k_1k_2^{-1}) n_2 (k_1k_2^{-1})^{-1} (k_1k_2^{-1}) \\ &= n_1 n_3 (k_1k_2^{-1}) \in NK \end{align*} $$

for some $n_3 \in N$ by the normality of $N$ in $G$. It follows that $NK \leq G$. Similarly, $KN \leq G$. Now, both $NK$ and $KN$ coincide with the subgroup $\langle N \cup K \rangle$ of $G$ generated by $N \cup K$, and so $NK = KN$. $\square$

Let us now consider two examples of internal semidirect products, and one counterexample.

Example 6.4. Fix $n \geq 2$. The symmetric group $S_n$ is an internal semidirect product of $A_n$ by $T = \lbrace 1,(12)\rbrace \cong \mathbb{Z}/2\mathbb{Z}$. Indeed, $A_n \triangleleft S_n$ because conjugation preserves parity. Moreover, every transposition is in $A_n T$, as $(a b) = (a b) (1 2) (12)$. Note also that $A_n T$ contains all odd permutations. Since every permutation is a product of transpositions, it follows from the subgroup property established in Proposition 2.3 that $S_n = A_n T$. Finally, every permutation in $A_n$ is even, but $(12)$ is an odd permutation, and so $A_n$ and $T$ are permutable complements. It now follows that $S_n$ is an internal semidirect product of $A_n$ by $T$. $\square$

Example 6.5. Generalizing Example 5.5. and Example 6.2., we consider the dihedral group

$$D_n = \langle \sigma, \tau \mid \sigma^n = \tau^2 = 1, \sigma \tau = \tau \sigma^{n-1} \rangle.$$

Let $S = \langle \sigma \rangle$ and $T = \langle \tau \rangle$, so that $S \cap T = \lbrace \operatorname{id}\rbrace$ and $ST = D_n$. Observe now that every element of $D_n$ is of the form $\sigma^k \tau^l$. Given $\sigma^k \tau^l \in D_n$ and $\sigma^m \in S$, we see that

$$ \begin{align*} (\sigma^k \tau^l) \sigma^m (\sigma^k \tau^l)^{-1} &= \sigma^k \tau^l \sigma^m \tau^{-l} \sigma^{-k} \\ &= \sigma^k \tau^l \tau^{-l} \sigma^{-m} \sigma^{-k} \\ &= \sigma^{k-m-k} \in S. \end{align*}$$

It follows that $S \triangleleft D_n$, whence $D_n$ is an internal semidirect product of $S$ by $T$. $\square$

Example 6.6. Consider the unit quaternion group $Q = \lbrace 1,-1,i,-i,j,-j,k,-k\rbrace$. Since every nontrivial subgroup of $Q$ must contain $\lbrace -1,1\rbrace$, it follows that no two nontrivial subgroups of $Q$ are permutable complements in $Q$. Therefore, $Q$ is not an internal semidirect product of its nontrivial subgroups. $\square$

We now turn to the problem of constructing an external semidirect product of two groups, without considering them as subgroups of an ambient group. Note that if $G = N \rtimes K$, then the conjugation action $\varphi:K \to \operatorname{Aut}(N)$ of $K$ on $N$, given by the formula $\varphi(k)(n) = knk^{-1}$, is a well-defined group homomorphism by the normality of $N$ in $G$. In this case, we see that the generic group product in $G = NK$ takes the following form:

$$(6.7) \, \, \, \, \begin{align*} (n_{1}k_{1})(n_{2}k_{2}) &= n_{1}(k_{1}n_{2}k_{1}^{-1})k_{1}k_{2} \\ &= \left(n_{1}\varphi_{k_{1}}(n_{2})\right) (k_{1}k_{2}). \end{align*}$$

Taking a cue from this computation, we make the following definition:

Definition 6.8. Let $N$ and $K$ be groups. Given a homomorphism $\varphi:K \to \operatorname{Aut}(N)$, we define the external semidirect product of $N$ by $K$ with respect to $\varphi$ to be the cartesian product $N \times K$ equipped with the group operation
$(n_1,k_1)(n_2,k_2) = (n_1\varphi_{k_1}(n_2), k_1k_2)$
for all $n_1,n_2 \in N$ and $k_1,k_2 \in K$. We denote the resulting group by $N \rtimes_\varphi K$.

Exercise 6.9. Show that every external semidirect product $N \rtimes_\varphi K$ is a group.

Mimicking Proposition 5.6, we show that an internal semidirect product is isomorphic to the corresponding external semidirect product.

Proposition 6.10. Let $N$ and $K$ be subgroups of a group $G$. If $N$ and $K$ are permutable complements in $G$, $N \triangleleft G$, and $K \leq G$, then $G \cong N \rtimes_\varphi K$, where $\varphi:K \to \operatorname{Aut}(N)$ is the conjugation map $\varphi_k(n) = knk^{-1}$.

Proof. We define a map $\Gamma:N \rtimes_\varphi K \to G$ by setting $\Gamma(n,k) = nk$. Observe that

$$\begin{align*} \Gamma(n_1,k_1)\Gamma(n_2,k_2) &= n_1k_1n_2k_2 \\ &= n_1 \varphi_{k_1}(n_2) k_1k_2 \\ &= \Gamma( (n_1,k_1)(n_2,k_2) ) \end{align*}$$

by what we have shown in (6.7), and so $\Gamma$ is a group homomorphism. If $\Gamma(n,k) = 1_G$, then $n = k^{-1}$, whence $n,k^{-1} \in N \cap K$. Since the intersection is trivial, we see that $n = k^{-1} = 1_G$, whence the kernel of $\Gamma$ is trivial. Finally, $\Gamma$ is surjective, as $NK = G$. $\square$

Conversely, an analogue of Proposition 5.7 holds as well.

Proposition 6.11. If $G \cong H_1 \rtimes_\varphi H_2$, then $G$ has a normal subgroup $N$ and a subgroup $K$ such that $N \cong H_1$, $K \cong K_1$, and $N$ and $K$ are permutable complements in $G$.

Proof. Let $N = H_1 \times \lbrace 1_{H_2}\rbrace$ and $K = \lbrace 1_{H_1}\rbrace \times H_2$, so that $N$ and $K$ are permutable complements in $G$. To show that $N \triangleleft G$, we fix $(n,1) \in N$ and $(n',k') \in G$. We first compute $(n',k')^{-1}$. To this end, we let $(a,b) = (n',k')^{-1}$ and observe that

$$1_G = (n',k')(a,b) = (n' \varphi_{k'}(a) , k' b).$$

Therefore, $b = (k')^{-1}$. On the other hand,

$$1_G = (a,b)(n',k') = (a \varphi_{b}(n'), b k'),$$

so that $a = \varphi_b((n')^{-1}) = \varphi_{(k')^{-1}}((n')^{-1})$. Therefore

$$(n',k')^{-1} = \left(\varphi(\varphi_{(k')^{-1}}((n')^{-1})), (k')^{-1} \right) = (n^{-1},k^{-1}).$$

We now observe that

$$\begin{align*} (n',k')(n,1)(n',k')^{-1} &= \left( n'\varphi_{k'}(n),k' \right) \left( (n')^{-1}, (k')^{-1} \right) \\ &= \left( n'\varphi_{k'}(n) \varphi_{k'}( (n')^{-1}), k' (k')^{-1} \right) \\ &= \left( n' \varphi_{k'}( n(n')^{-1}) , 1 \right) \in N. \end{align*}$$

It follows that $N \triangleleft G$, as was to be shown. $\square$

Therefore, the knowledge of automorphism groups is paramount in determining whether a group can be realized as a semidirect product. Here is a simple example; see Section 7.2 of Rotman, An Introduction to the Theory of Groups for a more comprehensive survey.

Example 6.12. Let $G$ and $H$ be finite groups, and let $\varphi:G \to \operatorname{Aut}(H)$ be a group homomorphism. Lagrange’s theorem, in conjunction with the first isomorphism theorem, implies that $|G| = |\ker \varphi| |\operatorname{im} \varphi|$, and so $|\operatorname{im} \varphi|$ divides $|G|$. Moreover, Lagrange’s theorem implies that $|\operatorname{im} \varphi|$ divides $|\operatorname{Aut}(H)$. Therefore, if $|G|$ and $|\operatorname{Aut}(H)$ are relatively prime, then $\varphi$ must be the trivial homomorphism, and so $G \rtimes_\varphi H$ must be the direct product $G \times H$.

Now, we observe that $\operatorname{Aut}(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/(p-1)\mathbb{Z}$ whenever $p$ is prime. Indeed, Lagrange’s theorem implies that every non-identity element of $\mathbb{Z}/p\mathbb{Z}$ is a generator. Now, every map from a cyclic group to a cyclic group that maps a generator to a generator can be extended to a group homomorphism. Since there are $p-1$ generators, there are $p-1$ such homomorphisms from $\mathbb{Z}/p\mathbb{Z}$ into itself. It is easy to show that all such maps are automorphisms, and that no other map is an automorphism. The isomorphism statement now follows.

Therefore, there is no non-direct semidirect product $\mathbb{Z}/3 \mathbb{Z} \rtimes \mathbb{Z}/5\mathbb{Z}$, as 5 and $3-1 = 2$ are relatively prime. On the other hand, there is precisely one non-direct semidirect product $\mathbb{Z}/3\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$, as there are precisely two ways of mapping $\mathbb{Z}/4\mathbb{Z}$ homomorphically into $\operatorname{Aut}(\mathbb{Z}/4\mathbb{Z})$: trivially or surjectively. $\square$

7. Classification of Groups of Small Order

As an application of the product constructions discussed so far, we establish several classifaction results for finite groups. To this end, we shall make use of the Sylow theorems, which we state here again for ease of reference:

Theorem 7.1 (Sylow, 1872). Let $G$ be a finite group of order $p^nm$, where $p$ is a prime number and $m$ is an integer relatively prime to $p$. The following holds:
First Sylow theorem. $G$ contains a Sylow $p$-subgroup of order $p^n$.
Second Sylow theorem. All Sylow $p$-subgroups are conjugates of one another. In other words, if $P$ and $Q$ are Sylow $p$-subgroups of $G$, then there exists a $g \in G$ such that $gPg^{-1} = Q$. This, in particular, implies that all Sylow $p$-subgroups are isomorphic to one another
Third Sylow theorem. Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. We have the following combinatorial restrictions on $n_p$:
$n_p$ divides $\vert G \vert$;
$n_p \equiv 1 \text{ mod } p$.

We also make use of Lagrange’s theorem and its corollaries repeatedly; let us record a few here:

Theorem 7.2 (Lagrange). Let $G$ be a finite group, $H$ a subgroup of $G$, and $g$ an element of $G$. The following holds:
$\vert g \vert$ divides $\vert G \vert$;
$\vert G \vert = \vert H \vert [G:H]$;
If $H$ is normal in $G$, then $\vert G \vert = \vert H \vert \vert G/H \vert $;
If $\varphi:G \to G\'$ is a homomorphism, then $\vert G \vert = \vert \ker \varphi \vert \\, \vert \operatorname{im} \varphi \vert.$

Finally, we also rely on the fundamental theorem on finitely generated abelian groups (Theorem 6.10).

The first result is a trivial corollary of Lagrange’s theorem

Proposition 7.2. Every finite group of prime order is cyclic.

Proof. Let $G$ be a group of order $p$. By Lagrange’s theorem, every non-identity element of $G$ is of order $p$. It follows that $G$ is cyclic. $\square$

The second result makes use of the fundamental theorem on finitely generated abelian groups.

Theorem 7.4. Let $p$ be a prime number. Every group of order $p^2$ is isomorphic to either $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$.

Proof. We shall make use of two lemmas:

Lemma 7.5. If $G$ is a $p$-group, then $\vert Z(G) \vert > 1$.

Proof of lemma. Consider the conjugacy class equation (Theorem 3.11 in my blog post on group actions):

$$ \displaystyle |G| = |Z(G)| + \sum_{g \in \Gamma \smallsetminus Z(G)} [G:C_G(g)]. $$

Since $C_G(g) \leq G$ for all $g \in G$, Lagrange’s theorem implies that each $[G:G_C(g)]$ is divisible by $p$. Since $|G|$ is divisible by $p$, it follows that $|Z(G)|$ is divisible by $p$. $\square$

Lemma 7.6. If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof of lemma. Let $\pi:G \to G/Z(G)$ be the canonical projection map, and find $g \in G$ such that $\pi(g) = gZ(G)$ is a generator of $G/Z(G)$. Fix $x,y \in G$ and find $m,n \in \mathbb{N}$ such that $x \in \pi(g^m)$ and $y \in \pi(g^n)$. We can find $x',y' \in Z(G)$ such that $x = g^mx'$ and $y = g^n y'$. We now observe that

$$\begin{align*} xy &= g^mx' g^n y' \\ &= y' g^mg^n x' \\ &= y' g^n g^m x' \\ &= y' g^n x'g^m \\ &= yx. \end{align*}$$

Since $x$ and $y$ were arbitrary, we conclude that $G$ is abelian. $\square$

We now let $G$ be a group of order $p^2$. Suppose for a contradiction that $G$ is nonabelian. By Lagrange’s theorem, $|Z(G)|$ is either 1 or $p$. Lemma 3.5 implies that $|Z(G)| = p$. Since $Z(G)$ is normal in $G$, we apply Lagrange’s theorem to conclude that $|G/Z(G)| = p$. It follows from Proposition 3.2 that $G/Z(G)$ is cyclic. Lemma 3.6 now implies that $G$ is abelian, which is evidently absurd. It follows that $G$ is abelian, whence the fundamental theorem on finitely generated abelian groups (Theorem 1.10) implies the desired classification result. $\square$

The next result is another one of general nature, covering the two-factor cases that Theorem 3.4 leaves out.

Theorem 7.7. Let $p$ and $q$ be prime numbers such that $p > q$. Every group of order $pq$ is isomorphic to either the cyclic group $\mathbb{Z}/pq\mathbb{Z}$ or a group given by the presentation
$\langle a,b \mid a^q = b^p = 1, aba^{-1} = b^m \rangle,$
where $m^q \equiv 1 \text{ mod } p$ and $m \not\equiv 1 \text{ and } p$. In particular, if $q \nmid p-1$, then the second case cannot occur.

Proof. We shall make use of three lemmas:

Lemma 7.8. Let $G$ be a finite group containing a subgroup $H$ of index $r$, where $r$ is the smallest prime divisor of $\vert G \vert$. Then $H \triangleleft G$.

Proof of lemma. Let $G/H$ denote the left coset space $\lbrace gH : g \in G\rbrace$, so that $|G/H| = r$. Let $\varphi:G \to S_{G/H}$ be the action of $G$ on $G/H$ by left multiplication (see Example 2.12). By the first isomorphism theorem, the quotient group $G/\ker \varphi$ is isomorphic to a subgroup of $S_{G/H}$. Since $|S_{G/H}| = r!$, Lagrange’s theorem implies that $|G/\ker \varphi|$ divides $r!$.

Now, Lagrange’s theorem also implies that $\vert G/\ker \varphi \vert$ divides $\vert G \vert$. Since $r$ is the smallest prime that divides $\vert G \vert$, we must have $\vert G/ker \varphi \vert = 1 \text{ or } r$. The action $\varphi$ is nontrivial, and so we must have $\vert G/\ker \varphi| = r \vert$.

For each $g \in \ker \varphi$, we see that $gH =\pi(g)(H) = H$, and so $g \in H$. Therefore, $\ker \varphi \leq H$. Lagrange’s theorem implies that

$r = [G:\ker \varphi] = [G: H][H:\ker \varphi] = r [H:\ker \varphi] .$

It follows that $[H:\ker \varphi] = 1$, and so $H = \ker \varphi$. Since $\ker \varphi$ is normal in $G$, the proof is complete. $\square$

Lemma 7.9. Let $G$ be a finite group of order $p^m q^n$. If all of the Sylow subgroups of $G$ are normal, then $G$ isomorphic to the direct product of its Sylow subgroups.

Proof of lemma. The third Sylow theorem, in conjunction with the second Sylow theorem, implies that $G$ has precisely one Sylow $p$-subgroup $A$ and one Sylow $q$-subgroup $B$. By Lagrange’s theorem, $A \cap B = \lbrace 1_G\rbrace$. Moreover, $|AB| = p^m q^n$, an dso $AB = G$. Therefore, $A$ and $B$ are permutable complements in $G$, both of which are normal in $G$ by hypothesis. It follows from Proposition 1.6 that $G \cong A \times B$. $\square$

Lemma 7.10 (Chinese remainder theorem for groups). If $k$ and $n$ are relatively prime positive integers, then $\mathbb{Z}/kn\mathbb{Z} \cong \mathbb{Z}/k\mathbb{Z} \times \mathbb{Z} / n \mathbb{Z}$.

Proof of lemma. Consider $\mathbb{Z}/k\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, which is of order $kn$. Let $x = (1,0)$ and $y = (0,1)$, so that $|x| = k$ and $|y| = n$. Since $\mathbb{Z}/k\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ is abelian, we see that $(xy)^i = x^i y^i$ for all $i \in \mathbb{Z}$. As $k$ and $n$ are relatively prime, $|xy| = kn$, and the product group is cyclic. $\square$

We now let $G$ be a group of order $pq$. The first Sylow theorem implies that $G$ contains a Sylow $p$-subgroup $P$. Since |P| = p, Proposition 7.2 implies that $P$ is cyclic; let $b$ be an element of $P$ of order $p$. Lemma 7.8 implies that $P \triangleleft G$.

The first Sylow theorem also implies that $G$ contains a Sylow $q$-subgroup $Q$. Once again, we can invoke Proposition 3.2 to find an element $a$ of $Q$ of order $q$. If $n_q = 1$, then the second Sylow theorem implies that $Q \triangleleft G$, whence $G \cong P \times Q$ by Lemma 7.9. Proposition 7.2 implies that $P \cong \mathbb{Z}/p\mathbb{Z}$ and $Q \cong \mathbb{Z}/q\mathbb{Z}$, and it follows from Lemma 7.10 that $G \cong \mathbb{Z}/pq \mathbb{Z}$.

We therefore assume that $n_q \neq 1$. The third Sylow theorem furnishes a positive integer $k$ such that $n_q = kq + 1$. The third Sylow theorem also implies that $n_q$ divides $pq$, whence we must have $n_q = p$. This, in particular, implies that $kq = p-1$, and so $q \mid p-1$.

Now, $P \triangleleft G$, and so $aba^{-1} = b^m$ for some $m$. If $m \equiv 1 \text{ mod } p$, then $ab = ba$, and $G$ is abelian. Let us therefore assume that $m \not\equiv 1 \text{ and } p$.

We claim that $a^l b a^{-l} = b^{m^l}$ for all $l \in \mathbb{N}$. This, in particular, implies that $b = a^q b a^{-q} = b^{m^q}$, whence $m^q \equiv 1 \text{ mod } p$, as was to be shown. It therefore suffices to establish the claim.

We proceed by induction. The $l = 1$ case has already been established. We fix $l_0$ and assume that the $l = l_0 – 1$ case holds. Then

$$\begin{align*} a^{l_0} b a^{-l_0} &= a (a^{l_0 – 1} b a^{-(l_0-1)}) a^{-1} = a b^{m^{l_0-1}} a^{-1} \\ &= (aba^{-1})^{m^{l_0-1}} = (b^m)^{m^{l_0-1}} = b^{m^{l_0}},\end{align*} $$

as was to be shown. $\square$

Remark 7.11. By generalizing the notion of internal direct products to more than two subgroups, we can establish the following extension of Lemma 7.9: If $G$ is a finite group such that all of its Sylow subgroups are normal, then $G$ is isomorphic to the direct product of all of its Sylow subgroups.

Corollary 7.12. If $p$ is prime, then every group of order $2p$ is either cyclic or isomorphic to the dihedral group $D_p$.

Proof. If the group is non-cyclic, then it is isomorphic to a group given by the presentation

$\langle a,b \mid a^2 = b^p = 1, aba^{-1} = b^m \rangle,$

where $m^2 \equiv 1 \text{ mod } p$ and $m \not\equiv 1 \text{ and } p$. Among $1 \leq m \leq p$, we see that $m = p-1$ is the only choice that satisfies both restrictions. It now suffices to note that the above presentation with $m = p – 1$ yields the dihedral group $D_p$. $\square$

The only groups of order at most $15$ that are not covered by the results established thus far are groups of order 8 and groups of order 12. We now tackle them “by hand”.

Theorem 7.13. Every group of order 8 is isomorphic to $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z} / 4 \mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, the unit quaternion group $Q$, or the dihedral group $D_4$.

Proof. Let $G$ be a group of order 8. By Lagrange’s theorem, every element of $G$ is of order 1, 2, 4, or 8. If $G$ contains an element of order 8, then $G \cong \mathbb{Z}/8\mathbb{Z}$.

Let us suppose that $G$ has no element of order 8. If $G$ has no element of order 4, then every non-identity element of $G$ is of order 2. Whenever $x,y \in G$, we have the identity

$xyxy = xy(yx)^{-1} = xyyx = x^2 = 1_G,$

whence $G$ is abelian. Fix three non-identity elements $a,b,c$ of $G$ and define a mapping $\varphi:(\mathbb{Z}/2\mathbb{Z})^3 \to G$ by setting $\varphi(k,m,n) = a^kb^mc^n.$ It is routine to check that $\varphi$ is a group isomorphism.

Let us now suppose that $G$ has an element of order $4$. Assume for now that $G$ is abelian. Fix an element $a$ of order 4, and pick an element $b$ of order 2 that is not a power of $a$. Define a mapping $\varphi:\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \to G$ by setting $\varphi(m,n) = a^mb^n$. Once again, it is easy to check that $\varphi$ is a group isomorphism.

Finally, we suppose that $G$ is a nonabelian group of order 8 that contains an element $x$ of order 4. Since $[G:\langle x \rangle] = 2$, we see that $\langle x \rangle \triangleleft G$: see Lemma 4.2. In particular, $G/\langle x \rangle \cong \mathbb{Z}/2\mathbb{Z}$. Therefore, every $g \in G$ such that $g \notin \langle x \rangle$ satisfies the property that $g^2 \in \langle x \rangle$.

We now fix such a $g$. If $g^2 = x$ or $g^2 = x^3$, then $g$ is of order 8, which is absurd. We must therefore have $g^2 = 1_G$ or $g^2 = x^2$.

Moreover, $\langle x \rangle$ is normal in $G$, and so $gxg^{-1} = a^n$ for some $n$. $n$ cannot be 0, as $gx \neq g$. $n$ cannot be 1, as $G$ is nonabelian. $n$ cannot be 2, as as $x$ and $gxg^{-1}$ must have the same order. We conclude that $g x = x^3g$.

Therefore, $G$ is isomorphic to either

$$\langle g,x \mid g^2 = x^4 = 1_G, gx = x^3g \rangle$$

$$\langle g,x \mid x^4 = 1, g^2 = x^2, gx = g^3g \rangle.$$

The first case corresponds to $D_4$; the second case corresponds to $Q$. $\square$

Theorem 7.15. Every group of order 12 is isomorphic to $\mathbb{Z}/12\mathbb{Z}$, $\mathbb{Z}/6 \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $A_4$, $D_6$, or $\mathbb{Z}/3\mathbb{Z} \rtimes_\varphi \mathbb{Z}/4\mathbb{Z}$, where $\varphi:\mathbb{Z}/4\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z}/3\mathbb{Z})$ is the unique nontrivial group homomorphism (see Example 2.12).

Proof. We suppose that $G$ is a nonabelian group of order 12 and show that $G$ is isomorphic to $A_4$, $D_6$, or $\mathbb{Z}/3\mathbb{Z} \rtimes_\varphi \mathbb{Z}/4\mathbb{Z}$. To this end, we assume without loss of generality that $G \not\cong A_4$.

The first Sylow theorem furnishes a Sylow 3-subgroup $P$, which is of order 3. We show that $P \triangleleft G$. We can think of each element of $G$ as a permutation on the set of all left cosets of $P$; since $[G:P] = 4$, this furnishes a homomorphism $\varphi:G \to S_4$ such that $\ker \varphi \leq P$. $|P| = 3$, and so Lagrange’s theorem implies that $|\ker \varphi|$ is either 1 or 3. If $\ker \varphi$ is trivial, then $G$ is isomorphic to a subgroup of $S_4$ of order 12. Since $A_4$ is the only subgroup of $S_4$ of order 12, we see that $G \cong A_4$, which is absurd. Therefore, $|\ker \varphi| = 3$, and so $\ker \varphi = P$. It follows that $P \triangleleft G$.

The first Sylow theorem also furnishes a Sylow 2-subgroup $Q$, which is of order 4. By Lagrange’s theorem $|P \cap Q| = 1$, and so $P \cap Q = \lbrace 1_G\rbrace$. Moreover, $|PQ| = 12$, and so $PQ = G$. It now follows from Proposition 2.10 that $G \cong P \rtimes_\varphi Q$. Since $|Q| = 4$ and $\operatorname{Aut}(P) \cong \mathbb{Z}/2\mathbb{Z}$, there are two possible homomorphisms $\varphi:Q \to \operatorname{Aut}(P)$: the trivial homomorphism, and the surjective homomorphism (see Example 2.12). The first case corresponds to the direct product $P \times Q$. Since $P$ and $Q$ are both abelian, this implies that $G$ is abelian, which is absurd.

The second case corresponds to the semidirect product $P \rtimes_\varphi Q$, where $\varphi:Q \to \operatorname{Aut}(P)$ is the unique surjective homomorphism. If $Q \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, then $P \rtimes Q \cong D_6$. If $Q \cong \mathbb{Z}/4\mathbb{Z}$, then $P \rtimes_\varphi Q \cong \mathbb{Z}/3\mathbb{Z} \rtimes_\varphi \mathbb{Z}/4\mathbb{Z}$. $\square$

8. Exact Sequences and the Extension Problem

We now approach the study of semidirect product from a different angle. We first note that $N \rtimes K$ is obtained by piecing together $N$ and $(N \rtimes K)/N$.

Proposition 8.1. $(N \rtimes_\varphi K)/(N \times \lbrace 1_K\rbrace) \cong K$.

Proof. It suffices to observe that

$$(N \rtimes_\varphi K)/(N \times \lbrace 1_K\rbrace) \cong 1 \times \lbrace 1_K\rbrace \cong K$. $\square$$

In light of this, we make the following definition:

Definition 8.2. Let $N$ and $K$ be groups. We say that $G$ is an extension of $N$ by $K$ if $G$ has a normal subgroup $N'$ such that $N' \cong N$ and $G/N' \cong K$.

Observe that $G$ is an extension of $N$ by $K$ if and only if there exist an injective homomorphism $i:N \to G$ and a surjective homomorphism $p:G \to K$ such that $\operatorname{im} i = \ker p$. Indeed, if such maps exist, then $N' \operatorname{im} i \cong N$ and

$$K = \operatorname{im} p \cong G / \ker p = G / \operatorname{im} i = G/N'.$$

Conversely, if we have isomorphisms $i:N \to N'$ and $\bar{p}:G/N' \to K$, then we can define a mapping $p:G \to K$ by setting $p(g) = \bar{p}(gN')$ and check that $\operatorname{ker} p = N' = \operatorname{im} i$.

We isolate the crucial condition:

Definition 8.3. Let $A \xrightarrow{f} B \xrightarrow{g} C$ be a sequence of group homomorphisms. The sequence is said to be exact in case $\operatorname{im} f = \ker g$. A longer sequence
$$G_1 \xrightarrow{f_1} G_2 \xrightarrow{f_2} \cdots \xrightarrow{f_{n-1}} G_n$$
is said to be exact if each joint $G_{k-1} \xrightarrow{f_{k-1}} G_k \xrightarrow{f_k} G_{k+1}$ is exact.

Definition 8.4. A short exact sequence of groups is an exact sequence of groups of the form
$$1 \to A \xrightarrow{f} B \xrightarrow{g} C \to 1,$$
where $1$ denotes the trivial group, $1 \to A$ denotes the group homomorphism that maps the identity to the identity, and $C \to 1$ denotes the trivial homomorphism.

We make a few observations. Since the image of $1 \to A$ is trivial, the exactness condition on $1 \to A \xrightarrow{f} B$ means that $\ker f = \lbrace 1_B\rbrace$, or that $f$ is injective. The kernel of $C \to 1$ is $C$, and so the exactness condition on $B \xrightarrow{g} C \to 1$ implies that $\operatorname{im} g = C$, or that $g$ is surjective. Finally, the exactness condition on $A \xrightarrow{f} B \xrightarrow{g} C$ states merely that $\operatorname{im} f = \operatorname{g}$. We thus see that short exact sequences are precisely the embodiments of extensions of groups:

Proposition 8.5. $G$ is an extension of $N$ by $K$ if and only if there exists a short exact sequence $1 \to N \to G \to K \to 1$.

The extension problem, then, is to determine all groups $G$ that makes the sequence $1 \to N \to G \to K \to 1$ exact for fixed $N$ and $K$. The extension problem lies at the heart of the classification of finite groups. To see why, we need a few notions from the theory of normal series:

Definition 8.6. A normal series of a group $G$ is a sequence of subgroups
$$G = G_0 \geq G_1 \geq \cdots \geq G_n = 1$$
such that $G_{k+1} \triangleleft G_k$ for all $0 \leq k \leq k-1$. The $k$th factor group of the above normal series is the quotient group $G_k/G_{k+1}$. The length of the above normal series is the number of nontrivial factor groups. The above normal series is said to be a composition series in case each $G_{k+1}$ is either a maximal proper normal subgroup of $G_k$ or $G_{k+1} = G_k$.

The Jordan–Hölder theorem states that each group has a unique composition series in a precise sense. Now, we take a composition series

$$G = G_0 \geq G_1 \geq \cdots \geq G_n = 1$$

of $G$ and let $F_k = G_k/G_{k-1}$ be the corresponding factor groups. Note that, at each stage, $G_k$ is an extension of $G_{k-1}$ by $F_k$. Moreover, we observe that each factor group is either simple or trivial. Therefore, an arbitrary finite group $G$ can be obtained by carrying out the “extending by a finite simple group” process finitely many times—and, by the Jordan–Hölder theorem, this procedure is uniquely determined by $G$.

In other words, we can complete the classification of finite groups by (1) classifying all finite simple groups and (2) solving the extension problem. (1) has been completed, after 60 years of hard work by many group theorists. (2) is still unsolved, as there is no known general theory of classifying all extensions of a given group: see the Wikipedia page on group extensions.

The unresolved state of the group extension problem indicates that there are, in general, many group extensions that fail to be semidirect products. The following exercise sheds light on this phenomenon:

Exercise 8.7. We say that a short exact sequence of groups
$$1 \to N \xrightarrow{f} G \xrightarrow{g} K \to 1$$
is left split if there exists a group homomorphism $G \xrightarrow{f'} N$ such that $f' \circ f = \operatorname{id}_N$. Similarly, the short exact sequence is said to be right split if there exists a group homomorphism $K \xrightarrow{g'} G$ such that $g \circ g' = \operatorname{id}_K$. Show that a group extension $G$ of $N$ by $K$ is a semidirect product of $N$ by $K$ if and only if the corresponding short exact sequence is right split. Show also that a group extension $G$ of $N$ by $K$ is a direct product of $N$ and $K$ if and only if the corresponding short exact sequence is left split. Conclude that left split implies right split.

See Chapter 7 of Rotman, An Introduction to the Theory of Groups for a more detailed survey of the group extension problem. The extension problem for abelian groups (and, in general, abelian categories) is also heavily studied. The proper context for this type of problem is homological algebra: see, for example, Weibel, An Introduction to Homological Algebra.