0
$\begingroup$

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

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

Now suppose $G$ is a minimal non $P_1$-group of exponent $p^2$ and suppose $\Phi(G)$ has exponent $p^2$ too. Thus there exists $z \in \mho_1(\Phi(G)) \cap Z(G) $ with $z^p =1$.

Why is $G/\langle z \rangle $ a $P_1$ group?

I don't see how this follows, I've tried to show $G/\langle z \rangle $ is isomorphic to a subgroup of $G$ but with no success.

Previously in the paper it's been shown that $G$ is a $P_1$-group if every section of exponent $p^2$ is a $P_1$-group.


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

$\endgroup$
2
$\begingroup$

This seems to follow just from the definition of a minimal non-$P_1$-group.

The definition given in the paper is that $G$ is a minimal non-$P_1$-group, if $G$ is not a $P_1$-group and all proper sections of $G$ are $P_1$-groups.

So being a proper section, the quotient $G/\langle z \rangle$ is a $P_1$-group.

$\endgroup$
  • $\begingroup$ Thanks. I missed that line. I had assumed that minimal meant only proper subgroups enjoyed the property. $\endgroup$ – Bysshed Aug 5 '17 at 14:49

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.