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Let $|G|=p^n$ and have only one subgroup of order $p^{(n-1)}$.Then G is cyclic.I am trying it in many ways bt get nothing. What I get :The unique subgroup is normal in $G$, Center meets the subgroup non trivially, No element outside the sub group has order $p^{(n-1)}$.Plz help me

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Apparently this comes down to showing that every maximal subgroup of a finite $p$-group has index $p$. We do this inductively. Take $M\leq G$ maximal. Then there are two cases:

If $Z(G)\cap M = 1$, then $M$ and $Z(G)$ generate $G$ since $Z(G)\neq 1$. But $Z(G), M \leq N(M)$, so $G=N(M)$, i.e. $M$ is normal.

If $x\in Z(G)\cap M$, $x\neq 1$, then $M/(x) \leq G/(x)$ is a maximal subgroup. By induction, it is normal, so $M$ is normal.

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    $\begingroup$ So you are using induction on n where |G|=p^n,your first case is for n=1.and final statement like this:by induction hyp since |G/(x):M/(x)|=p we have |G:M|=p.So every maximal subgroups of G has index p,which is enough to prove the problem.by this shall show for any y in G outside of M the subgroup (y) cannot be proper.Beacuse in every finite group every proper subgroup is contained in a maximal subgroup. $\endgroup$ – Via Sep 14 '14 at 5:24
  • $\begingroup$ @Via This is all correct. You can take the base case to be $n=0$ if you want, though :) $\endgroup$ – Slade Sep 14 '14 at 9:32

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