Deducing there exists exactly $5$ isomorphism classes of groups of order $12$.

Question

There's a substantial amount that's been written about the semi-direct products of a group of order $12$ on this website. However, there's something that seems to be taken for granted each time the conversation arises, but I'm having trouble taking this fact for granted.

Let $G$ be a group of order $12$. $12=2^2\cdot 3$, one can use the sylow theorems and a counting argument to show that $G$ must have a normal subgroup of order either $3$ or $4$, and in either case, $G$ is a semi-direct product of subgroups.

In the case that this normal subgroup is order $3$, we get $4$ different semi-direct products (I won't include the work here; as I said, a substantial amount of conversation has been had on this website to identify these isomorphism classes).

My question arises in the case that the Sylow-3 subgroup is not normal, and so the sylow subgroup of order $4$ is. I know that there are $5$ isomorphism classes of groups of order $12$, and so it is a fact that there is only one way to embed a normal subgroup of order $4$ into a group of order $12$. When discussing the semi-direct products of groups of order $12$, the above fact seems to be taken for granted. However, suppose one did not know there were only $5$ isomorphism classes of $G$, then how would one deduce this fact? Namely,

Let $N$ denote the normal subgroup of order $4$ in $G$. Then, $G\cong N \rtimes C_3$. I fail to see any reason why $(C_2\times C_2)\rtimes C_3$ and $C_4\rtimes C_3$ are not both valid constructions.

Answer 1

You wrote

it is a fact that there is only one way to embed a normal subgroup of order 4 into a group of order 12

and that is false. Groups of order $4$ have nontrivial automorphisms, so when $G$ has order $12$ and $H$ is a group with order $4$ that admits an injective homomorphism $H \to G$, there are always multiple such injective homomorphisms since you get a new one by composing with a nontrivial automorphism of $H$.

When classifying groups of order $12$, merely writing them as semidirect products is not the end of the story since you need to be able to determine whether two semidirect products are isomorphic.

Consider the possible semidirect products $(\mathbf Z/(2))^2 \rtimes_\varphi \mathbf Z/(3)$, where $$ \varphi : \mathbf Z/(3) \to {\rm Aut}((\mathbf Z/(2))^2) = {\rm GL}_2(\mathbf Z/(2)) $$ is a homomorphism. When $\varphi$ is trivial we get the direct product $(\mathbf Z/(2))^2 \times \mathbf Z/(3)$. When $\varphi$ is nontrivial, $(\mathbf Z/(2))^2 \rtimes_\varphi \mathbf Z/(3)$ is nonabelian so it isn't isomorphic to $(\mathbf Z/(2))^2 \times \mathbf Z/(3)$, but how do we know such semidirect products with nontrivial $\varphi$ are all isomorphic to each other? The map $\varphi$ is determined by $\varphi(1 \bmod 3)$, which must have order $3$, and ${\rm GL}_2(\mathbf Z/(2))$ has two elements of order $3$, so there are two choices of $\varphi$. Call them $\varphi$ and $\varphi'$. Why are $(\mathbf Z/(2))^2 \rtimes_\varphi \mathbf Z/(3)$ and $(\mathbf Z/(2))^2 \rtimes_{\varphi'} \mathbf Z/(3)$ isomorphic? Write $A$ and $A'$ for the matrices of order $3$ in ${\rm GL}_2(\mathbf Z/(2))$. Since $A' = A^{-1}$, $\varphi'(1 \bmod 3) = \varphi(1 \bmod 3)^{-1}$. Thus $\varphi'(k \bmod 3) = \varphi(k \bmod 3)^{-1} = \varphi(-k \bmod 3)$ for all $k$, so $\varphi' = \varphi \circ f$ where $f : \mathbf Z/(3) \to \mathbf Z/(3)$ is inversion. It turns out that when $\varphi : K \to {\rm Aut}(H)$ is any homomorphism and $f: K \to K$ is any automorphism of $K$, so $\varphi \circ f$ is also a homomorphism $K \to {\rm Aut}(H)$, the semidirect products $H \rtimes_\varphi K$ and $H \rtimes_{\varphi \circ f} K$ are isomorphic groups. Therefore both nontrivial semidirect products $(\mathbf Z/(2))^2 \rtimes_\varphi \mathbf Z/(3)$ are isomorphic to each other. In particular, we do not need to know about the concrete group $A_4$ in order to see that there are groups of order $12$ with a normal $2$-Sylow subgroup isomorphic to $(\mathbf Z/(2))^2$ and that all such groups are isomorphic to each other.

Answer 2

Hint:

(1) If Sylow-$3$ is not normal, then it should be $A_4$. For this, number of Sylow-$3$ must be $4$ and $G$ acts by conjugation on them, giving a homomorphism $G\rightarrow S_4$; show that this is $1$-$1$, hence $G$ is isomorphic to a subgroup of order $12$ in $S_4$; show that it should be $A_4$.

(2) If Sylow-$3$ is normal, the consider possibilities of Sylow-$2$; it is $K_4$ or $C_4$ (cyclic). Also, note that Sylow-$2$ should be not normal if $G$ is non-abelian (in Case (2)).

If Sylow-$2$ is $K_4$ then group is $C_3\rtimes K_4$ and $K_4$ normalizes $C_3$ in a non-trivial way: if $K_4=\{1,a,b,c\}$ and if $a,b$ commute with Sylow-$3$, so will be $c$ and $G$ will become abelian, contradiction; hence one of $a,b,c$ should not commute with Sylow-$3$. Hence, if $x$ is a generator of Sylow-$3$ then $axa^{-1}=x^2$ or $bxb^{-1}=x^2$ (or both). You can decide then possibilities of $cxc^{-1}$ easily. So structure of $G$ is determined.

Similarly, if Sylow-$2$ is $C_4$ then $C_4$ normalizes $C_3$ in a non-trivial way (to get non-abelina group $G$): it must be in the form that $yxy^{-1}=x^2$ if $x$ generates Sylow-$3$ and $y$ generates Sylow-$2$ (since only elements of order $3$ in Sylow-$3$ are $x$ and $x^2$, so if $yxy^{-1}=x$ then $G$ will become abelian.

Summary: In each of the three cases, we get unique group of order $12$, which is non-abelian. Q.E.D.