In my Group Cohomology class, the professor stated the following theorem
If one takes an extension $1 \rightarrow A \rightarrow E \rightarrow G \rightarrow 1 $ with $A$ abelian and finite and $G$ finite, and $|G|, |A|$ coprime, then $E = A \rtimes G$
We proved that with the identification between the extensions of $G$ by $A$ and $H^2(G, A)$
Then he let as an exercise the following corollary :
If $A$ is finite abelian and $G$ finite, and if $E$ is an extension $1 \rightarrow A \rightarrow E \rightarrow G \rightarrow 1 $, and if for every $p$-Sylow $G_p$ of $G$ there is a split $G_p \rightarrow E$, then $E = A \rtimes G$.
I'm not able to prove that. I suppose that I should do an induction, and by push out and pull back, for some $p$ dividing $|A|$ and $|G|$ get an new extention $ 1 \rightarrow A/A_p \rightarrow E'' \rightarrow G_p \rightarrow 1$ and so on, but first I can't see what $E''$ looks like, and secondly, even if I had $E'' = A/A_p \rtimes G_p$, I don't know how to get back to $E$.
Would one of you have any suggestion or reference to this result?
Thanks in advance!
Ok, after the exam I finally have the solution.
What we want to show is actually that the product of the restrictions
$$ \phi : H^2(G, A) \rightarrow \prod_{p \mid \#G} H^2(G_p, A) $$
is injective.
If one writes $\#G = p_1^{e_1} \cdot \ldots \cdot p_n^{e_n}$, it is known that $\#G_{p_i} = p_i^{e_i}$. By Bezout identity write $\sum_i a_p p_i^{e_i} = 1$ for some $a_i$.
Now consider the sum of the co-restrictions: \begin{align} \psi : \prod_{i} H^2(G_{p_i}, A) &\rightarrow & H^2(G, A)& \\ (x_i) &\mapsto& \sum_i a_i \cdot &Cores_{p_i}(x_i) \end{align}
The composition $\psi \circ \phi $ is equal to $\sum_i \left(a_i p_i^{e_i} Id\right) = Id$. Finally $\phi$ is injective.
Nota: this can be generalized to any $H^i$, but in the case of the exercise I had, I only needed $H^2$.