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Let $G$ be a finite group and let $M(G)=H^2(G,\mathbb{C}^*)$ be its Schur multiplier. For "small" groups I can compute the Schur multiplier by hand in terms of corresponding roots of unity.

However, for "large" groups I have truble even to determine if the Schur multiplier of a group is trivial, or if the Schur multiplier of two groups of the same cardinality are isomorphic.

I thought about using GAP, however I think that my groups are too large for it also.

The groups are of order about $10^{50}$, however they are semidirect product of two Sylow subgroups.

I will be happy to hear any idea.

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    $\begingroup$ @DerekHolt I want to describe the Schur multiplier of the groups in the famous example by Martin Hertweck. But maybe it is aiming too high. I will be happy to show that these groups have non-isomorphic Schur multipliers or at least to show that they have non-trivial Schur multipliers. Here is the link to his paper. jstor.org/stable/… $\endgroup$ – Ofir Schnabel Jun 24 '15 at 12:54
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    $\begingroup$ I had a look at the paper but I don't have time right now to try and construct the groups explicitly. If you can produce definitions in GAP say, then I would happy to experiment - you could e-mail them to me if you like. $\endgroup$ – Derek Holt Jun 24 '15 at 13:26
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    $\begingroup$ As another general comment if $G = P \rtimes Q$ is the semidirect product of a $p$-group and a $q$-group with $p \ne q$ then the $q$-part of the Schur Multiplier $M(G)$ of $G$ is isomorphic to $M(Q)$, whereas $Q$ induces an action on $M(P)$, and the $p$-part of $M(G)$ is equal to the fixed subgroup of this action. This may make the calculation easier. There are very efficient algorithms for finding the $p$-multiplicator (which is related to the multipler but not quite the same thing) of a finite $p$-group. $\endgroup$ – Derek Holt Jun 24 '15 at 17:49
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    $\begingroup$ @DerekHolt , Thanks a lot. I am femiliar with the above mentiond theorem. Even more generaly, if $G=N\rtimes T$ and $(|N|,|T|)=1$ then $$M(G)=M(N)^T\times M(T).$$ Here $M(N)^T$ is the $T$-stable cohomology classes in $M(N)$. As you suggested, I will try to construct the groups explicitly. I will do it first in latex. $\endgroup$ – Ofir Schnabel Jun 25 '15 at 7:36
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    $\begingroup$ @AlexanderKonovalov the presentation of the group $X$ from Hertweck's is given in the paper. For the group $Y$ however I didn't found it's presentation anywhere. I sent the presentation of $X$ to Prof. Derek Holt and he computed the Schur multiplier using MAGMA. For the group $Y$ we need to first know its presentation. More generally, I want to know if there exist a commutative ring $R$ such that the second cohomology group of $X$ and $Y$ are not isomorphic over $R$. It is pretty clear that for $R=\mathbb{Z}$ they are isomorphic. $\endgroup$ – Ofir Schnabel Jan 14 '16 at 7:45

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