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I am reading section 7.B on Clifford theory in this paper and hope someone can help me understand some of the arguments there.

Let me shortly explain the part of the setup which I need for my question:

$X$ is a normal subgroup of $Y$, $M$ is a $kX$-module ($k$ is a field) which is stable under $Y$ and we set $A = \text{End}_{kX}(M)$. Suppose that the characteristic of $k$ does not divide the order of $Y/X$.

Under these conditions, it is stated that any short exact sequence $$1 \to 1 + J(A) \to Z \to Y/X \to 1$$ is split and the explanation which is given is that $1 + J(A)$ is an extension of abelian groups since $1 + J(A)^i/ (1 + J(A)^{i + 1}) \cong J(A)^i/J(A)^{i+1}$ and these are $k(Y/X)$-modules. I am missing some understanding to infer the splitting of the sequence from this explanation and would appreciate if someone could clarify this.

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The idea is to consider your extension as a factor set $$f:Y/X\times Y/X\to 1+J(A)$$ and show that it must be equivalent to a trivial one. For this you first consider the image of $f$ in $(1+J(A))/(1+J(A)^2)\cong J(A)/J(A)^2$. The last group is abelian, so since $char(k)\nmid |Y/X|$ the resulting sequence $$1\to (1+J(A))/(1+J(A)^2)\to \tilde{Z}\to Y/X\to 1$$ splits. This means that you can choose $f$ above in such a way that the image of $f$ is contained in the subgroup $1+J(A)^2$. You finish by induction.

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  • $\begingroup$ Thank you for your answer. It seems to me I am missing some knowledge here concerning the connection between abelian groups, prime divisors and splitting sequences. I am sorry if this seems trivial to you, but could you explain to me in a bit more detail why the resulting sequence is split? $\endgroup$
    – Matthias Klupsch
    Commented Feb 28, 2017 at 7:24
  • $\begingroup$ sure. The easiest way to see it is that when the kernel $R$ is abelian, then you get an action of $Y/X$ on $R$, and extensions correspond to elements in $H^2(Y/X,R)$. In the last abelian group, all elemenets have finite order dividing the order of $Y/X$. On the other hand, if $R$ is a vector space over $K$, then the same will be true for this abelian group. If $K$ has characteristic zero then this group must be zero. If $K$ has characteristic $p$ then this group will also have characteristic $p$, and if $p\nmid |Y/X|$ then the group is again zero. $\endgroup$
    – Ehud Meir
    Commented Feb 28, 2017 at 9:35
  • $\begingroup$ All these things can be found in any standard book on cohomology of groups, for example the one of Brown. $\endgroup$
    – Ehud Meir
    Commented Feb 28, 2017 at 9:45

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