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$M_{12}\rtimes \mathbb{Z}_2$ is a maximal subgroup of $M_{24}$, where $M_{24}$ and $M_{12}$ are Mathieu groups . Also, it is known that $H^3(M_{24}, U(1)) \cong \mathbb{Z}_{12}$. I want to find the restriction of a 3-cocycle $\alpha \in H^3(M_{24}, U(1))$ to $M_{12}\rtimes \mathbb{Z}_2$.

The semi-direct product is defined in the usual manner where the action of $\mathbb{Z}_2$ on $M_{12}$ is given by $h(g)=g$ for $h=e$, and $h(g)=\phi(g)$ otherwise, where $h \in \mathbb{Z}_2, g \in M_{12}$ and $\phi$ is an outer automorphism of $M_{12}$.

How do 3-cocycles behave under restriction to maximal subgroups? Would the choice of $\phi$ affect the answer? I would highly appreciate if someone could give me a reference to a result in group cohomology that could help me answer this question.

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  • $\begingroup$ Why do you want to know ? What's the background of the question ? $\endgroup$
    – tj_
    Commented Dec 26, 2015 at 23:37
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    $\begingroup$ I am confused by your terminology. Most people would consider $M_{23}$ to be the second largest Mathieu group wouldn't they? $\endgroup$
    – Geoff Robinson
    Commented Dec 27, 2015 at 0:15
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    $\begingroup$ If anyone knows a way to answer this kind of question (restriction of cohomology to subgroups) with software I'd love to know about it. GAP happily computes cohomology groups, but as far as I know doesn't give you access to the actual cocycles. $\endgroup$
    – Noah Snyder
    Commented Dec 27, 2015 at 1:12
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    $\begingroup$ $M_{24}$ is large enough that it's not obvious how a $3$-cocycle might be presented in reasonable time and space. $\endgroup$
    – Noam D. Elkies
    Commented Dec 27, 2015 at 3:55
  • $\begingroup$ @GeoffRobinson: Thanks for pointing out Geoff. That's a mistake on my part. $\endgroup$
    – user35360
    Commented Dec 27, 2015 at 7:25

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