number of Schur covering groups Let $G$ be a finite group with derived subgroup $G'$ and Schur multiplier $M(G)$. A Schur cover of $G$ is a group $H$ with a subgroup $Z\le H'\cap Z(H)$ such that $H/Z\cong G$ and $Z\cong M(G)$. Let $G/G'\cong \prod_{i}C_{n_i}$ and $M(G)\cong\prod_j C_{m_j}$. A theorem of Schur states that the number of non-isomorphic Schur covers of $G$ is at most $\prod_{i,j}\gcd(n_i,m_j)$. I have seen a proof in Karpilovsky's book "The Schur multiplier" (Thm 2.5.14), but I did not quite get through, because of some non-trivial cohomology statements. If I understand the strategy correctly, to every Schur cover of $G$ one assigns (injectively) a central extension $L$ of $G/G'$ by $W\cong M(G)$ such that $W\cap L'=1$.

*

*What is the most direct way of doing this?

*Is there an alternative way to prove Schur's theorem (other references)?

I'm particularly interested in the case $(|G/G'|,|M(G)|)=1$ where there is only one Schur cover.


*Is it easier to prove this special case? Preferably with as little cohomology as possible.

 A: Note that in the general theory of Schur multipliers and covering groups, $G$ is not restricted to being a finite group.
For a covering group $H$ of $G$, choose elements $h_i \in H$ (for $i$ in some index set) such that their images in $H/H'$ generate $H/H'$. Since $H/Z(H)$ is the fixed group $G$, we can choose the $h_i$ to map onto specific elements of $G$ (i.e. independently of the choice of covering group $H$). Then, as I think you know, then values of the commutators $[h_i,h_j] \in H'$ do not depend on the choice of $H$. (In other words, the different covering groups of $G$ are all isoclinic.)
So, if $R$ is a set of relators of $H/H'$ that define $H/H'$ as an abelian group, then the isomorphism class of $H$ is determined by the values of $r_j \in R$ in $H'$, and these differ in different covering groups only by elements of $Z$.
In other words, if $r_j = g_j$ in $H$, and $H_1$ is another cover of of $G$, then $r_j = g_jz_j$ in $H_1$, for some $z_j \in Z$.
(Of course, if $H/H'$ is finitely generated, then we can choose the $h_i$ such that their images generate the cyclic direct factors of $H/H'$, and then we can choose powers of $h_i$ as the relators of $H/H'$ as abelian group.)
Consider the abelian group extension $E$ of $Z$ by $H/H'$ defined by the relations $r_j = z$. This extension is split if and only if there exist $z_j' \in Z$ such that, when we replace the generators of $h_i$ of $H$ by $h_iz_j'$, the relations $r_j$ in $E$ become trivial in $E$, but since the $z_j$ are central in $H$, this is equivalent to the relations $r_j = g_jz_j$ in $H_1$ changing to $r_j = g_j$, in which case $H$ and $H_1$ are isomorphic.
So equal elements of ${\rm Ext}(H/H',Z)$ (the equivalence classes of abelian group extensions of $Z$ by $H/H'$) induce isomorphic covering groups, and hence the number of isomorphism classes of covering groups is at most $|{\rm Ext}(H/H',Z)|$ which, in your notation in the finite case, is equal to $\prod_{i,j} \gcd (n_i,m_j)$.
Of course this is an inequality, and it is possible for Schur covers to be isomorphic even when the corresponding elements of ${\rm Ext}(H/H',Z)$ are distinct. I believe that such isomorphisms are induced by automorphisms of $G$. Interesting examples are the symmetric groups $S_n$. We have $|M(S_n)| = 2$ for all $n \ge 4$, and, for $n \ne 6$, there are two isomorphism classes of covering groups, but for $n=6$ there is only one, with the exceptional outer automorphism of $S_6$ inducing an isomorphism between them.
