Which groups are derived subgroups? Let $G$ be a group. When is there a group $H$ such that $G$ is isomorphic to its derived subgroup $H'$?
I only know that there is not always such a $H$; for instance, no group has its derived subgroup isomorphic to $\mathfrak{S}_n$ for $n \geq 4$ and $n \neq 6$.
 A: Here's another reason for $\mathfrak{S}_3$:
Proposition: If $H$ is a group with a characteristic subgroup $K$ with abelian automorphism group, but $K$ is not contained in the center of $H$, then $H$ is not the derived subgroup of any group.
Proof: Since $K$ is characteristic in $H$, $K$ is normal in $G$. There is a homomorphism from $G$ to the automorphism group of $K$. Since the automorphism group of $K$ is abelian, we must have that $H$ is in the kernel of this homomorphism. However, that just means $K$ is contained in the center of $H$. $\square$
This shows that dihedral groups (of order at least 6) are not derived subgroups. Their subgroup $K$ of rotations are characteristic and cyclic, so the proposition applies.
A: Here's one case you can rule out. We say that a group is complete when $Z(G) = 1$ and $\operatorname{Aut}(G) = \operatorname{Inn}(G)$. 
Fact: If $G$ is complete and $G' \neq G$, then $G$ is not isomorphic to a commutator subgroup $H'$.
Proof: Suppose that $H' = G$. Since $G \trianglelefteq H$ is complete, it follows that $H = G \times C_H(G)$. Here $C_H(G)$ is abelian because $H' = G$. Thus $H' = G' \times C_H(G)' = G'$, which gives us the contradiction $G = G'$.
Note that this proves that $\mathfrak{S}_n$ is never a commutator subgroup for $n \geq 3$ and $n \neq 6$. So what about $\mathfrak{S}_6$?
