Realizing groups as commutator subgroups 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?
 A: 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 $x_1^p, [x_1,x_2]$.
A: 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 $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$.

