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Let $A$ and $B$ be finite abelian groups with coprime order, and let $G=A\rtimes{}B$ be a semidirect product, via any action. Let $C\subseteq{}A$ be the subgroup of the elements of $A$ which are fixed by the action of $B$, so that $C=Z(G)\cap{}A$. Then we have $$ A = C \oplus G'. $$

Is there a quick reference for this fact? Please note that i'm NOT asking for a proof of this simple (and well known i guess?) fact, i just need a reference to quickly point to in a note, to avoid making it cumbersome. Unless there is a one-line proof that i missed. Thanks for the attention!

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I don't know what you would consider a oneliner but the usual proof in the context of ordinary representation theory works. We get an action of $e=1/|B|\sum_{b\in B}b$ on $A$ and it gives the projection on $C$. –  Torsten Ekedahl Jul 15 '11 at 14:58
    
Yeah, this is the proof i had in my mind. But i would feel guilty for not adding some verification that the kernel of the projection is exactly the derived subgroup, is it also immediate? without having to talk at all about irreducible representations? I prefer to be quoting some one else's proof, so that if he skips the details i will not feel guilty for him ;) –  Maurizio Monge Jul 15 '11 at 17:44
    
Yes, it is immediate, we have $e^2=e$ and $be=e$. The first shows that $A$ is the direct sum of the image and the kernel of $e$, the second that the image lies in $C$ and and it is clear that $e$ is the identity on $C$. –  Torsten Ekedahl Jul 15 '11 at 19:06
    
My question was: why is the kernel exactly $G'$? $G'$ is clearly contained, but why does equality hold? (again, i know how to prove this, just trying to understand why it is supposed to be obvious) –  Maurizio Monge Jul 15 '11 at 22:27
    
The complement has no quotient on which $B$ acts trivially, which means that it is spanned by elements of the form $ba-a$ which are commutators in $G$ so the complement is contained in $G'$. On the other hand $G$ modulo the complement is clearly commutative so that $G'$ is contained in it. –  Torsten Ekedahl Jul 16 '11 at 14:30

1 Answer 1

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Theorem 2.3 in Chapter 5 of the book "Finite Groups" by Daniel Gorenstein states that if $A$ is a $p'$-group of automorphisms of an abelian $p$-group $P$, then $P = C_P(A) \times [P,A]$ (all groups here are assumed to be finite). You can deduce your result easily from this.

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Thanks, the reference is perfect! –  Maurizio Monge Jul 15 '11 at 17:47

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