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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.

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

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