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Let $G$ be a group and $M$ is a minimal normal subgroup of $G$. Suppose that $M$ is an elementary abelian $2$-group and $M=C_G(M)$, so that $G/M$ acts on $M$. The action is faithful and assume that is also irreducible. Suppose that furthermore $G/M$ is a non abelian simple group (for my purposes, $G/M$ can be $A_5$ or $PSL_2(8)$, by the way).

I want to know some information about the irreducible complex characters of $G$, for example the degree of real characters.

My approach is check all the irreducible representations of $G/M$ over $GF(2)$ (that are known), build the semidirect product of $G/M$ by $M$ on GAP and print the character table. Unfortunately, this may not be the case since the group $M$ may not have a complement in $G$. It would be possible to calculate the second cohomology group $H^2(G/M,M)$ with GAP, check that is trivial (if it is true) and conclude that $M$ has a complement in $G$ (the extension splits, namely). There is a package in GAP, cohomolo, maintained by D. Holt, that does this kind of calculation, but I'm struggling to load it (I'm having some technical problems) and I wandering if there a more theoretical approach.

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Well, if you really only interested in $G/M = A_5$ or ${\rm PSL}(2,8)$, then I can tell you what you want to know.

$A_5$ has just two nontrivial irreducible modules over ${\rm GF}(2)$, both of dimension $4$, and they both have trivial $2$-cohomology.

${\rm PSL}(2,8)$ has four such modules, of dimensions $6,8$ and $12$. Of these, only the one of dimension $6$ (which is the natural module) has a nonsplit extension, and that one is unique.

You can access all of these groups except for the split extension $2^{12}:L_2(8)$ from the perfect groups database in GAP.

For example, the nonsplit extension $2^6\!\cdot\! L_2(8)$ can be accessed by:

> gap> G := PerfectGroup( IsPermGroup,32256,2);
  L2(8) N 2^6

and then you can just compute its character table directly.

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  • $\begingroup$ Thank you! Seems that $2\cdot L_2(8)$ and $2:L_2(8)$ have the same character table (if I used gap correctly ). Is it a coincidence? May I ask you where I can find some information about the cohomology groups $H^2(G,V)$ that you kindly wrote? The representation of degree $12$ can be built if the matrix representations of generetors are given: brauer.maths.qmul.ac.uk/Atlas/lin/L28 $\endgroup$
    – Lorban
    Mar 19, 2021 at 10:00
  • $\begingroup$ There is no nonsplit extension $2 \cdot L_2(8)$. $\endgroup$
    – Derek Holt
    Mar 19, 2021 at 10:55
  • $\begingroup$ sorry I meant $2^6\cdot L_2(8)$ and $2^6: L_2(8)$, it was a mistake $\endgroup$
    – Lorban
    Mar 19, 2021 at 10:57
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    $\begingroup$ Yes, there are $8$ equivalence classes of extensions, but they give rise to only two isomorphism classes of groups, one of which is the split extension, corresponding to the zero element of the cohomology group. The groups corresponding to the other seven extensions are all isomorphic. In this example that is easy to see, because the endomorphism ring of the module is cyclic of order $7$. $\endgroup$
    – Derek Holt
    Mar 19, 2021 at 16:35
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    $\begingroup$ I was also going to say that in GAP 4.11 there is new function $\mathtt{TwoCohomologyGeneric}$, which you could use to calcualte $H^2(G,M)$, but it might be slower than the cohomolo functions. $\endgroup$
    – Derek Holt
    Mar 19, 2021 at 16:37

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