Let $G\_0$ be a finite group and $G\_j$ a Schur cover of $G\_{j-1}$ for $j=1,2,3\ldots$. Is $G\_2$ equal to $G\_1$? If not, will the sequence stop after finite steps in general?



Successive Schur covers
Let me expand on Geoff Robinson's comment. First, for $G$ perfect, there is a unique Schur cover $R$, and $R$ is perfect. As Geoff indicated, $R$ is its own Schur cover. To see this, let $S$ be a Schur cover for $R$. Then $S$ has a central subgroup $Z$ with $S/Z$ isomorphic to $R$, and thus $S$ has a normal subgroup $Y$ containing $Z$ such that $Y/Z$ is central in $S/Z$ and $S/Y$ is isomorphic to $G$. Also, $S$ is perfect. Now $[Y,S] \subseteq Z$, so $[Y,S,S] = 1$. Thus also $[S,Y,S] = 1$, so by the three-subgroups lemma, $[S,S,Y] = 1$. But $[S,S] = S$, so $[S,Y] = 1$ and $Y \subseteq Z(S)$. Since $S$ is perfect and $S/Y$ is $G$, it follows that $Y$ is a homomorphic image of the Schur multiplier of $G$. But $Y/Z$ is the Schur multiplier of $G$ because $R$ is a Schur cover. This forces $Z = 1$, so $S$ is $R$.


If $G$ is not perfect, the Schur cover is not generally unique. If we start with a Klein group, the two Schur covers are $D\_8$ and $Q\_8$. Now $Q\_8$ is its own cover, but we can take $D\_8$ and consider its Schur covers. There are three of these, and one of them is $D\_{16}$. We can continue like this because one of the Schur covers of an arbitrary dihedral $2$-group is the next larger dihedral $2$-group, and this yields an infinite chain of covers. The key step in proving this is to show that the Schur multiplier of a dihedral $2$-group has order $2$. (It should be obvious that it is at least $2$.)


Here's a sketch of a proof of the needed fact. Suppose $R$ is a cover of the dihedral group $D$. Then $R$ has an abelian subgroup $A$ of index $2$. Also $|R:R'| = |D:D'| = 4$, so $R'$ has index $2$ in $A$. Since $A$ is abelian and $R/A$ is cyclic, we deduce that $|A| = |R'||A \cap Z(R)|$, and it follows that $|A \cap Z(R)| = 2$. Since $Z(R) \subseteq A$, we have $|Z(R)| = 2$, as required.


