The following question proposes
Let $H$ be a p'-group of automorphisms of a finite abelian p-group A. Prove that $A = [A,H] \times C_A(H)$.
There is a theorem that says:
Theorem (10.1.6): Let the finite group G have an abelian Sylow p-subgroup P and let N denote $N_{G}(P)$. Then $P = C_{P}(N) \times [P,N]$.
how can i use this result to solve the above question?
Thank you very much.
Since $H$ is a group of automorphisms of $A$, we can construct a semidirect product $G :=A \rtimes H$ in a natural way. Now, $A$ is an abelian Sylow $p$-subgroup of $G$ (It is a Sylow $p$-subgroup since $H$ is a $p'$-group). The theorem you mention is now directly applicable. We write $A$ as
$$ A = C_A(G) \times [A,G] $$
Now, $G$ has subgroups $\tilde{A}$ and $\tilde{H}$ isomorphic to $A$ and $H$ respectively such that $G = \tilde{A}\tilde{H}$. For simplicity, we write $G = AH$ and rewrite the above equation as
$$ A = C_A(AH) \times [A,AH] $$ Since $A$ is abelian, you can convince yourself that $C_A(AH) = C_A(H)$. Also, for $h \in H$ and $a_1, a_2 \in A$, we have that $[a_1,a_2h] = [a_1,h]$, and from this is follows that $[A,AH] = [A,H]$. We may now write $A$ a direct product of the desired subgroups:
$$ A = C_A(H) \times [A,H] $$