# Semidirect products and extensions

## 1. Introduction: 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 1.1. The integers mod 6 group $\mathbb{Z}/6\mathbb{Z} = \{0,1,2,3,4,5\}$ is isomorphic to $\{0,3\} \times \{0,2,4\}$. Indeed, the map

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

Example 1.2. The Klein-four group $V = \langle a,b \mid a^2 = b^2 = (ab)^2 = 1 \rangle$ is isomorphic to $\{1,a\} \times \{1,b\}$. Indeed, the map

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

Example 1.3. The multiplicative group of complex numbers $\mathbb{C}^\times = \{z \in \mathbb{C} : z \neq 0\}$, equipped with the standard multiplication operation, is isomorphic to the direct product of the circle group $\mathbb{T} = \{z \in \mathbb{C} : |z| = 1\}$ and the group of positive real numbers $\mathbb{R}^+ = \{r \in \mathbb{R} : r > 0\}$, 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

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 $N \cap K = \{1_G\}$. These observations lead us to the following definition:

Definition 1.4. Two subgroups $N$ and $K$ of $G$ are permutable complements—or simply complements if there is no danger of confusion—in $G$ if $N \cap K = \{1_G\}$ and $NK$ equals $G$.

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

Example 1.5. Consider the dihedral group

Let $S = \langle \sigma \rangle$ and $T = \langle \tau \rangle$, so that $S \cap T = \{\operatorname{id}\}$. 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 1.6. 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

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 1.7. 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 $N = H_1 \times \{1_{H_2}\}$ and $K = \{1_{H_1}\} \times H_2$. 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,

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 1.8. 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 1.9. 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 $S = \{\lambda I_n : \lambda \neq 0\}$ 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, $S \cap SL(n,\mathbb{R}) = \{I_n\}$, 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

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 1.10 (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

here $n = 0$ if and only if $G$ is finite.

## 2. Semidirect Products

Classification results with direct products, even something as strong as Theorem 1.10, does not take into account the decomposition of $D_3$ into the “internal product” $\{\operatorname{id},\sigma,\sigma^2\}\{\operatorname{id},\tau\}$ that was discussed in Example 1.5. It is therefore reasonable to generalize the notion of internal direct products introduced in Definition 1.8 for the sake of completeness.

Definition 2.1. 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 2.2. Consider the dihedral group $D_3$ and the integers mod 6 group $\mathbb{Z}/6\mathbb{Z}$. By Example 1.1 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 1.6, 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 2.3. 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

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 2.4. Fix $n \geq 2$. The symmetric group $S_n$ is an internal semidirect product of $A_n$ by $T = \{1,(12)\} \cong \mathbb{Z}/2\mathbb{Z}$. 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 2.5. Generalizing Example 1.5. and Example 2.2., we consider the dihedral group

Let $S = \langle \sigma \rangle$ and $T = \langle \tau \rangle$, so that $S \cap T = \{\operatorname{id}\}$ 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

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

Example 2.6. Consider the unit quaternion group $Q = \{1,-1,i,-i,j,-j,k,-k\}$, introduced in Example 2.8 of my blog post on generators and relations. Since every nontrivial subgroup of $Q$ must contain $\{-1,1\}$, 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:

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

Definition 2.8. 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

for all $n_1,n_2 \in N$ and $k_1,k_2 \in K$, which we denote by $N \rtimes_\varphi K$.

We show that the construction we have given above yields a group.

Exercise 2.9. Every external semidirect product $N \rtimes_\varphi K$ is a group. (This was given as a homework exercise in class.)

Mimicking Proposition 1.6, we show that an internal semidirect product is isomorphic to the corresponding external semidirect product.

Proposition 2.10. 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

by what we have shown in (2.7), 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 1.7 holds as well.

Proposition 2.11. 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 $N = H_1 \times \{1_{H_2}\}$ and $K = \{1_{H_1}\} \times H_2$, 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

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

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

We now observe that

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 2.12. 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 (see my blog post on homomorphic extensions. 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. $\square$

## 3. 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 was covered in my blog post on group actions Let us state the theorems once again, for ease of reference:

Theorem 3.1 (Sylow). 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 $\vert G \vert$;
• $n_p \equiv 1 \mbox{ mod } p$.

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

Theorem 3.2 (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 1.10). Since we have not proved the theorem in class or in the notes, you will not be responsible for any classification result that makes use of the theorem; all such instances are marked in red.

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

Proposition 3.2. 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 3.4. 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 3.5. 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):

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 3.6. 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

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 3.7. 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 (see my blog post on generators and relations).

where $m^q \equiv 1 \mbox{ mod } p$ and $m \not\equiv 1 \mbox{ and } p$. In particular, if $q \nmid p-1$, then the second case cannot occur.

Proof. We shall make use of three lemmas:

Lemma 3.8. Let $G$ be a finite group containing a subgroup $H$ of index $r$, where $r$ is the smallest prime divisor of $|G|$. Then $H \triangleleft G$.

Proof of lemma. Let $G/H$ denote the left coset space $\{gH : g \in G\}$, 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 in my blog post on group actions). 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 $\vert G/ker \varphi \vert = 1 \mbox{ or } r$. 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

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 3.9. 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, $A \cap B = \{1_G\}$. 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 3.10 (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 3.2 implies that $P$ is cyclic; let $b$ be an element of $P$ of order $p$. Lemma 3.8 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 3.9. Proposition 3.2 implies that $P \cong \mathbb{Z}/p\mathbb{Z}$ and $Q \cong \mathbb{Z}/q\mathbb{Z}$, and it follows from Lemma 3.10 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 $m \equiv 1 \mbox{ mod } p$, then $ab = ba$, and $G$ is abelian. Let us therefore assume that $m \not\equiv 1 \mbox{ and } p$.

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 $m^q \equiv 1 \mbox{ mod } p$, 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

as was to be shown. $\square$

Remark 3.11. By generalizing the notion of internal direct products to more than two subgroups, we can establish the following extension of Lemma 3.9: 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 3.12. 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

where $m^2 \equiv 1 \mbox{ mod } p$ and $m \not\equiv 1 \mbox{ and } p$. 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

1. We now tackle them “by hand”.

Theorem 3.13. 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

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 my blog post on group actions for a proof). 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

or

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

Theorem 3.15. 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. The conclusion in red follows from the fundamental theorem on finitely generated abelian groups (Theorem 1.10) and the Chinese remainder theorem for groups (Lemma 3.10). 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 $P \cap Q = \{1_G\}$. 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 $P \rtimes_\varphi Q \cong \mathbb{Z}/3\mathbb{Z} \rtimes_\varphi \mathbb{Z}/4\mathbb{Z}$. $\square$

## 4. 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 4.1. $(N \rtimes_\varphi K)/(N \times \{1_K\}) \cong K$.

Proof. It suffices to observe that

$(N \rtimes_\varphi K)/(N \times \{1_K\}) \cong 1 \times \{1_K\} \cong K$. $\square$

In light of this, we make the following definition:

Definition 4.2. 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

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 4.3. Let $A \xrightarrow{f} B \xrightarrow{g} C$ be a sequence of group homomorphisms (see my blog post on how to read diagrams). The sequence is said to be exact in case $\operatorname{im} f = \ker g$. A longer sequence

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 4.4. A short exact sequence of groups is an exact sequence of groups of the form

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 $\ker f = \{1_B\}$, 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 4.5. $G$ 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 4.6. A normal series of a group $G$ is a sequence of subgroups

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

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 4.7. We say that a short exact sequence of groups

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, Chapter Weibel, An Introduction to Homological Algebra.

Tags:

Categories:

Updated: