Is $S_6$ the derived subgroup of some group?

Question

I know that if $H$ is a complete group (meaning that the homomorphism $H\to\text{Aut}(H)$ is an isomorphism) and if $H$ is not perfect (meaning that $H^\prime\lneq H$) then $H$ is not the derived subgroup of any group. In particular, if $n\geq3$ and $n\neq6$ then $S_n$ is not the derived subgroup of any group. Clearly $S_1$ and $S_2$ are both the derived subgroup of some group.

Is $S_6$ the derived subgroup of some group?

A natural choice would be $\text{Aut}(S_6)$ since $S_6$ is a normal subgroup of index 2. Unfortunately, $A_6$ is a normal subgroup of index 4 so $\text{Aut}(S_6)^\prime\leq A_6$.

Answer 1

Here is a general statement that proves that if $n\geq3$ then $S_n$ is not the derived subgroup of any group.

Theorem: Let $H$ be a group such that $\text{Inn}(H)$ is not a subgroup of $\text{Aut}(H)^\prime$. Then $H$ is not the derived subgroup of any group.

Proof: Suppose that $G^\prime=H$. Then there is a conjugation homomorphism $\varphi\colon G\to\text{Aut}(H)$. Then we have $\text{Inn}(H)=\varphi(H)=\varphi([G,G])=[\varphi(G),\varphi(G)]\leq\text{Aut}(H)^\prime$.

If $n\geq3$ then $\text{Inn}(S_n)=S_n$ and $\text{Aut}(S_n)^\prime\leq A_n$ so the theorem applies.