$A_4$ is not simple

Question

I would appreciate if you could please express your opinion about my proof and maybe give me a hint where you deem suitable.

To prove that $A_4$, an alternating group of even permutations of $S_4$, is not simple, we need to show that there exists a normal subgroup (call it $H$) of $A_4$, such that $H$ is not trivial and $H \neq A_4$.

Now define a homomorphism $\pi: A_4 \to A_4/H$ by $\pi(\sigma)=\sigma H$ for any $\sigma \in A_4$. We need to show that there exists an element $\epsilon \neq \sigma \in A_4$ such that $\pi(\sigma)=H$. That is, we need to show that $\ker\pi$ is not trivial.

Define $H=\ker(\pi):=(\epsilon, (12)(34), (23)(14), (13)(24))$ (where $\epsilon$ is the identity permutation). We then prove that $H$ is closed under subgroup criteria (omitted).

Now we need to show that $\sigma H\sigma^{-1} \subset H$. To do this, we need to consider three cases.

Case 1: trivial.

Case 2: products of disjoint transpositions are in $H$ since $H$ is a subgroup of the said transpositions.

Case 3: 3-cycles. This is where I'm a little lost. In order to prove that $\sigma_3 H\sigma^{-1}_3 \subset H$, we need to show that all 3-cycles in $A_4$ conjugating any element of $H$ (a product of disjoint transpositions) will be again a product of disjoint transpositions.

Other then that, I think, we're done. I'm wondering if a simpler proof is possible.

Answer 1

We will require the following lemma.

$\pmb{Lemma.}\quad$ A subgroup $H$ of a group $G$ is normal if and only if it is the union of conjugacy classes in G of all $h\in H$.

$\pmb{Proof.}\quad$ If $H\trianglelefteq G$ and $h\in H$, then $ghg^{-1}\in H\,\forall\,g\in G$, which implies that the conjugacy class $[h]$ is in $H$. It is then straightforward to see that $H$ is the union of all such conjugacy classes. Conversely, if $H$ is a subgroup of $G$ such that $H=\bigcup\limits_{h\in H}[h]$, then $ghg^{-1}\in H\,\forall\,h\in H$ and $\forall\,g\in G$ whence $H\trianglelefteq G$.$\,\Box$

$\pmb{Proposition.}\quad$ $A_4$ is not simple.

$\pmb{Proof.}\quad$ Define $H=\{e,\,(1\,2)(3\,4),\,(2\,3)(1\,4),\,(1\,3)(2\,4)\}$, a subgroup of $A_4$. Now, recall that conjugation of permutations preserves the cycle structure. This implies that $H$ is a union of conjugacy classes, meaning that it is a normal subgroup of $A_4$. Thus, we have proved that there is a non trivial normal subgroup of $A_4$ that is not $A_4$ itself, whence $A_4$ is not simple.$\,\Box$

Perhaps such a proof involving the production of a non trivial normal subgroup out of seemingly nowhere is unsatisfactory. For this reason, I will proceed to explain where this subgroup comes from.

Let us consider the symmetric group $S_4$. The elements in $S_4$ are the identity, $2$ cycles, $3$ cycles, $4$ cycles, and products of two disjoint transpositions. Selecting the even permutations, we have $1$, $(i\,j\,k)$, and $(i\,j)(r\,s)$, which have orders $1,3$, and $2$ respectively. Because there are no elements of order $4$, it follows that every Sylow $2$ subgroup consists of the identity and $3$ elements of order $2$. And it is straightforward to see that there are only three elements of the form $(i\,j)(r\,s)$. Hence, it is only possible to form one subgroup of order $4$ in $A_4$. If $H$ is this unique subgroup of order $4$, which is also not cyclic, then $A_4/H$ has order $3$, and is therefore abelian. Let $A_4'$ be the commutator subgroup of $A_4$. It follows that $A_4'\subset H$. It is also an easy verification that $H$ has no non trivial proper subgroups that are normal in $A_4$, whence $A_4'=H$. This explains where our subgroup $H$ came from. The reason for $A_4$ not being simple is that its commutator subgroup is proper.

Answer 2

Note that $H:=\{(),(12)(34),(13)(24),(14)(23)\}$ is closed under composition and contained in $A_4$, so $H\le A_4$. Now, conjugation preserves the cycle type and $H$ exhausts all the elements of $S_4$ of cycle type $(ij)(kl)$. Therefore, $H$ is normal in $S_4$ and, in particular, in $A_4$.