Is ${\rm Core}_G(H) \leqslant C_G(A)$?

Question

Let $G$ be a group and let $A$ be a normal abelian subgroup of $G$. Suppose $G$ splits over $A$ with complement $H$. Prove that ${\rm Core}_G(H) \leqslant C_G(A)$.

I have been trying to prove that the core is contained in the centralizer but I have not been successful. Any hints would be very useful.

I clarify some of the notation:

• $G$ splits over $A$ with complement $H$ means that $G=AH$ and $A\cap H=1$.

• ${\rm Core}_G(H)=\bigcap_{g\in G}H^g$.

• $C_G(A)=\{g\in G:ag=ga, \forall a\in A\}$

Answer 1

If $g\in {\rm Core}_G(H)$ then consider the element $aga^{-1}g^{-1}$: you have $aga^{-1}\in {\rm Core}_G(H)$ and also $ga^{-1}g^{-1}\in A$ since $A$ is normal, so $$aga^{-1}g^{-1}\in {\rm Core}_G(H)\cap A\subseteq H\cap A=\{1\},$$ which gives the result.

Answer 2

Put $C={\rm core}_G(H)$. Then $C$ is normal in $G$ and contained in $H$, so $A \cap C=1$. Since $A$ is also normal, this means that $A$ and $C$ centralize each other, whence $C \subseteq C_G(A)$. Note that you do not need the abelianess of $A$.