# If $G$ is minimal non $P_1$-group with a maximal subgroup $M \leq G$, then $|\mho_1(M)|\leq p$.

I'm having difficulty with Theorem 3.c from this paper.

I'll begin with some definitions for a finite $p$-group $G$. Define, for each integer $n$, $\mho_n(G) = \langle g^{p^n} \mid g \in G \rangle$ and call $G$ a $P_1$-group if $\mho_n(S) = \{ s^{p^n} \mid s \in S \}$ for all $n$ and every section $S$ of $G$, including $S=G$.

Now suppose $G$ is a minimal non-$P_1$-group, i.e. $G$ is not a $P_1$-group but every proper section of $G$ is. Let $M$ be maximal subgroup of $G$.

Why is $|\mho_1(M)| \leq p$?

It has previously been shown:

1. $G$ has exponent $p^2$ and is generated by two elements.
2. $\Phi(G)$ has exponent $p$.
3. If $H$ is a $P_1$-group with a subgroup $J \leq H$ such that $\exp J = p^n$ and $[H:J]=p^k$, then $|\mho_n(H)| \leq p^k$.

The paper states the result follows from 3, but how do we know $M$ contains a subgroup of exponent $p$ and index $p$?

The inequalites I can see are, $|\mho_1(M)| \leq [M:\Phi(G)]$, and $|\mho_{\log_p(\exp N)}(M)| \leq p$ and $|\mho_1(M)/\mho_1(N)| \leq p$ where $N$ is a maximal subgroup of $M$.

Reference: Mann, A. (1976). The power structure of p-groups. I. Journal of Algebra, 42(1), pp.121-135.

• The $n$ is undefined ion your definition of $P_1$-group. Did you mean $P_n$-group? – Derek Holt Aug 11 '17 at 12:28
• @DerekHolt No, I mean it holds for every positive integer $n$, I will edit accordingly. Please see the paper for definitions of a $P_2$ and $P_3$ group. – Bysshed Aug 11 '17 at 12:40

## 1 Answer

Since $G$ is $2$-generated, $|G:\Phi(G)| = p^2$, so $M$ has $\Phi(P)$ as a subgroup of index $p$ and exponent $p$.