TITLE: Realizing groups as commutator subgroups
QUESTION [23 upvotes]: 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?

REPLY [7 votes]: Let me quote some well-known results and perhaps related problems which may be illuminating!
Let $G$ be a non-abelian finite $p$-group having cyclic center. Then, there is no finite $p$-group $H$ such that $G$ is isomorphic to a normal subgroup of the derived subgroup $[H,H]$ of $H$. In particular, $G$ cannot be isomorphic to the derived subgoup of some $p$-group $H$. The latter is a famous result due to Burnside. The former is a slight generalization of  problem. See  H. Heineken, On normal embedding of subgroups. 
Geom. Dedicata 83, No.1-3, 211-216 (2000).
Related problem: Let $V$ be a non-empty set of words in the free group on the countable set
$\{x_1,x_2,\dots\}$. We call a group $G$ is {\bf integrable with respect to $V$}, whenever there is a group $H$ such that $G\cong V(H)$, where $V(H)$ is the verbal subgroup of $H$ generated by $V$, i.e., $$V(H)=\langle v(h_1,\dots,h_n) | v\in V, h_i \in H \rangle,$$
(the subgroup of $H$ generated by the values of words of $V$ on the elements of $H$)
For example, if one takes $V=\{[x_1,x_2]=x_1^{-1}x_2^{-1}x_1 x_2\}$, then $V(H)$ is the derived subgroup of $H$ for any group $H$ and the problem is the same as it proposed.
One may write (maybe for some propaganda) 
$$\int G \; dV=H \Longleftrightarrow G=V(H).$$ 
In the case, $V=\{ [x_1,x_2]\}$, $$\int G=H  \Longleftrightarrow G=H'$$ and so
$$\int G=H \Longleftrightarrow \int G=H \times A$$ for any abelian group $A$. (This may remind the constant term in the integral of a function!)
I have used the latter notation which has no benefit but only perhaps inspiring  in 
[A. Abdollahi, Integral of a group, 29th Iranian International Conference on Mathematics, Amirkabir University of Technology (Tehran Polytechnic), Iran, March 28-31, 1998.]
I would like to say that the problem given a group $G$, find groups $H$ such that $G=[H,H]$" have been studied in a more general contex which may be found with key wordsnormal embedding of subgroups" and the above-mentioned paper of Heineken is a good start. 
Also it may be worth-mentioning that by a result of Allenby 
R.B.J.T. Allenby, Normal subgroups contained in Frattini subgroups are Frattini subgroups, Proc. Amer. Math. Soc,  Vol. 78, No. 3,  1980, 318- 
if $N$ is a normal subgroup of a finite group $G$ which is contained in the Frattini subgroup of $G$, then $N=\Phi(U)$ for some finite group $U$. 
Of course, for the class of finite $p$-groups the Frattini subgroup is the verbal subgroup
generated by the words $x_1^p, [x_1,x_2]$.