Normal closure of powerfully embedded subgroups.

Question

I'm reading a paper of Lubotzky and Mann (J. Algebra 105, 1987, 484-505), and Im doubtful in a proof.

Proposition. Let $N$ powerfully embedded subgroup of $G$. If $N$ is the normal closure of some subset of $G$, then $N$ is actually generated by this subset.

Proof. Let $N = \langle A \rangle^{G}$. To find generator for $N$, we may assume $N^{p} = 1$. But then $N$ is central in $G$, and $N = \langle A \rangle^{G} = \langle A \rangle$.

Why we may assume $N^{p} = 1$? The proof follows trivially by it. (In this case, $p$ is odd).

Answer 1

$N$ is powerfully embedded in $G$ and so $[N,G] \leq N^{p}$ and so it follows $[N,N] \leq N^{p}$ and so $N$ is powerful. Hence $ N^{p} $ is the Frattini subgroup, $\Phi(N)$ of $N$.

Informally you might like to think of the Frattini subgroup of $N$ consisting of the "non generators" of $N$.

The Burnside Basis Theorem tells you that you have a generating set for $N$ if and only if the images of that set form a basis in $N/\Phi(N)$ viewed as a vector space, which in this case is $N/N^{p}$.

Hence you may assume $N^{p}$ is trivial as you could quotient out by it, choose your generating set, and then lift back up.

Answer 2

First, note that throughout the paper $G$ is a finite $p$-group and hence $N$ is a finite $p$-group, implying $\Phi(N)=\mho^1(N)N'$. Now look at the quotient $N/\mho^1(N)$, if bar $\bar .$ denotes modding out by $\mho^1(N)$, then $\bar{N}^p=\bar{1}$, $[\bar{G},\bar{N}]=\bar{1}$ and its follows that $\bar{N}=\langle\bar{A}\rangle$, whence $N=\langle A\rangle \mho^1(N)$. Since $\mho^1(N) \subseteq \Phi(N)$, it follows that $N=\langle A \rangle$.