What are the groups $X$ for which there exists a group $G$ such that $G' \cong X$?


My considerations:




---


$\bullet$ If $X$ is perfect we are happy with $G=X$.


$\bullet$ If $X$ is abelian then $G := X \wr C\_2$ verifies $G'=\{(x,x^{-1}): x \in X\} \cong X$.


$\bullet$ If $X$ satisfies the following properties:


(1) $X \neq X'$, (2) The conjugation action $X \to \text{Aut}(X)$ is an isomorphism,


then there is no $G$ such that $G' = X$ (consider the composition $G \to \text{Aut}(X) \cong X \to X/X'$, it is surjective so its kernel contains $X$, contradiction). For instance, the symmetric group $S\_n$ verifies (1) and (2) if $n \neq 2,6$.




---


I have been looking for this problem on the web but I didn't find anything.
Do you have any reference and/or suggestion on how to solve this problem?



Realizing groups as commutator subgroups
A complete answer seems not to be known. Let me give you the following two nearly-contemporaneous references from the mid-70s:


Robert Guralnick, *On groups with decomposable commutator subgroups*


Michael Miller, *Existence of Finite Groups with Classical Commutator Subgroup*


Both Guralnick and Miller call groups which are commutator subgroups $C$-groups (though I don't know who, if either, originated the term) and give partial answers to your general question. For example, Theorem 4 from Miller gives the following:



> 
> Let $G$ be a subgroup of $\operatorname{GL}\_n(K)$ containing $\operatorname{SL}\_n(K)$ for $K$ a finite field of characteristic not equal to 2. Then $G$ is the commutator subgroup of some group unless it is of odd index and $n$ is even.
> 


The groupprops-wiki calls such groups commutator-realizable, and give a basic result on such groups, but mention that this terminology is not standard (though is probably safer than the overloaded term $C$-group.)


Edit: Some googling around led to the following slick argument of Schoof (from his *Semistable abelian varieties with good reduction outside 15*), which is closely related to your observation in bullet (3), and also serves to eliminate the symmetric groups. I'll quote verbatim except for change of variable names:



> 
> Let $G$ be a group and let $G'$ be its commutator subgroup. Conjugation gives rise to
> a homomorphism $G \to \operatorname{Aut}(G')$. On the one hand it maps $G'$ to the commutator subgroup of $\operatorname{Aut}(G')$. On the other hand the image of $G'$ is the group $\operatorname{Inn}(G')$ of inner automorphisms of $G'$. Therefore, if a group $X$ is the commutator subgroup of some group, we must have $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$.
> 


