TITLE: Is $S_6$ the derived subgroup of some group?
QUESTION [9 upvotes]: 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 $\text{Aut}(S_6)^\prime\leq A_6$.

REPLY [6 votes]: 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.