1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The $n$ is undefined ion your definition of $P_1$-group. Did you mean $P_n$-group? $\endgroup$ – Derek Holt Aug 11 '17 at 12:28
  • $\begingroup$ @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. $\endgroup$ – Bysshed Aug 11 '17 at 12:40
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.