non-capable groups and their direct products A group $H$ is said to be capable if there is a group $G$ such that $G/Z(G)\cong H$.
It is well known that $G/Z(G)$ can never be cyclic of order $>1$; so cyclic groups are not capable.
More interesting: quaternion group $Q_8$ is not capable: there is no $G$ with 
$G/Z(G)\cong Q_8$.
Now $Z_p$ (cyclic of order $p$) is not capable; but $Z_p\times Z_p$ is capable.
This raises question, is $Q_8\times Q_8$ capable? In general,

Q. Given any group $H$, is $H\times H$ capable?


Edit: Is there a known example of a finite group $H$ such that $H\times H$ is not capable?
 A: In the article "On groups occurring as center factor groups, J. Algebra, 61 (1979)" F. R. Beyl, U. Felgner and P. Schmid present a condition in which the capability of a direct product of finitely many of groups implies the capability of each of the factors. For details see section $6$. So for many classes of groups, the condition that $H\times H$ is capable already implies that $H$ is capable.
For example, if we let $G$ be a $p$-group of class
at most two and odd prime exponent, then we obtain that if $G$ is a nontrivial direct product, then $G$ is capable if and only if each direct factor is either capable or nontrivial cyclic. There are furthermore articles by Arturo Magidin on this subject.
I suppose one can decide the question on direct products of quaternion groups  with these results too, but I did not search "long enough" to find it in the literature. There is also a classification of extra-special capable $p$-groups, see Corollary 8.2 in the paper above. 
A: Shahriar Shahriari proved in On normal subgroups of capable groups, Arch. Math. (Basel) 48 (1987) no.3 193-198, MR 0880078 (88e:20026), among other things, that if a finite group $G$ contains a normal subgroup $H$ which is either generalized quaternion of order $2^n$, $n\gt 2$, or semidihedral of order $2^n$, $n\gt 3$, then $G$ is not capable. This immediately shows that $Q_8\times Q_8$ cannot be capable (and more generally, gives you lots of examples of groups $H$ such that $H\times H$ is not capable; note that if $H$ is capable, then so is $H\times H$, since any witness $K$ to the capability of $H$ yields $K\times K$ as a witness for the capability of $H\times H$). 
Shahriari also proved that if $G$ is finite nilpotent and contains a normal subgroup $H$ which is an extraspecial $p$-groups of order $p^3$ and exponent $p$, $p$ odd, then $G$ is not capable. 
Finally, Shahriari proves the following:

Theorem (Prop. 3.2 in the above-cited paper) Let $G$ be a finite group. If $G = QC_Q(G)$ for some $Q\leq G$, and $1\neq M\subseteq Z^*(Q)\cap[Q,Q]$, then $M\subseteq Z^*(G)$; in particular, $G$ is not capable.

Here, $Z^*(Q)$ is the epicenter of $Q$ (similar with $G$). The epicenter is the obstruction to capability: it is the smallest central subgroup of $G$ such that $G/Z^*(G)$ is capable.
