I am reading section 7.B on Clifford theory in 
[this paper](https://arxiv.org/abs/1511.04714) 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.



Splitting of certain short exact sequences in context of Clifford theory
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.


