Question

Let A be a normal elementary p-subgroup of a finite group G such that the index of the centralizer CG(A) is prime to p. Show that for any normal subgroup B of G lying in A, there exists another normal subgroup C of G lying in A such that A = B × C.

Answer 1

Let $k = \mathrm{F}_p$ and let $G$ act on $A$ by conjugation. The subgroup $C_G(A)$ is also normal in $G$, and acts trivially on $A$. Writing $G' = G / _C G(A)$, we may therefore view $A$ as a $kG'$-module. Since $|G'| = [G : C_G(A)]$ is prime to $p = \operatorname{char} k$, $kG'$ is a semisimple ring. The assumption that $B \triangleleft G$ implies that $B$ is a $kG'$-submodule of $A$. Therefore, $A = B \oplus C$ for a suitable $kG$-submodule $C \subseteq A$. Going back to the multiplicative notation, we have $A = B \times C$, and $C \triangleleft G$.