Any hint about the following two questions will be greatly appreciated!

Let $G$ be a finite non-abelian $p$-group with the following presentation: $$G=\langle a, b\mid a^{p^n}=b^{p^m}=1, [a,b]=a^{p^{n-m}}\rangle,$$ where $p$ is an odd prime and $n>m\geq 1$.

How could we show that

  1. The subgroup $N=\langle a^{p^{n-1}}\rangle$ is the only normal subgroup of $G$ of order $p$.

  2. Any non-normal subgroup of $G$ that does not contain $N$ is conjugate to one of the subgroups $\langle b\rangle$, $\langle b^p\rangle$, $\dots$ ,$\langle b^{p^{m-1}}\rangle$.

Edit: About the first question I think we may use this fact that every normal subgroup of a $p$-group intersects the center nontrivially. Thus normal subgroups of prime order in a $p$-group will be central. For example let $|G|=p^3$, then $n=2$ and $m=1$ and we have $|G/Z(G)|=p^2$, because $G$ is non-abelian $p$-group. Thus $|Z(G)|=p$ and by previous remark $K=Z(G)$. I could not generalize this argument for $p$-groups of order greater than $p^3$ and with this presentation. In general if we show that $Z(G)$ is cyclic then assetion 1 holds.


closed as off-topic by Derek Holt, loup blanc, Pragabhava, JonMark Perry, iadvd Sep 14 '16 at 2:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, loup blanc, Pragabhava, JonMark Perry, iadvd
If this question can be reworded to fit the rules in the help center, please edit the question.