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The prompt is from Herstein's Topics in Algebra. The proof is to be done without using the structure theorem for finite abelian groups. I showed that $H$ was cyclic of order $p$ a prime and G was a $p$-group. I then set $|G| = p^n$ and proceeded by induction on n. The case $n=1$ is clear. Then, by induction every proper subgroup of $G$ is cyclic. By the Sylow theorems $G$ therefore contains an element $a$ of order $p^{n-1}$. We have $G = \langle a\rangle \langle b\rangle $, where $b$ is any element of $G$ not in $\langle a\rangle $ (here $\langle a\rangle $ denotes the subgroup generated by $a$). Quotient by $K$, the intersection of $\langle a\rangle $ and $\langle b\rangle $, to see that $G/K$ is isomorphic to $\Bbb{Z}_{p}^{2}$, since $G/K$ is the direct product of two nontrivial cyclic p-groups ($\langle a\rangle /K$ and $\langle b\rangle /K$) and by homomorphism every one of $G/K$'s subgroups is cyclic. Therefore taking the preimage of each of the cyclic subgroups of $G/K$, each of which is a distinct maximal proper subgroup of $G$, $G$ must have $p+1$ distinct subgroups of order $p^{n-1}$. Also, note that since $b$ was an arbitrary element not in $\langle a\rangle $ and $\langle a\rangle /K$ and $\langle b\rangle /K$ have the same number of elements, $\langle a\rangle $ contains every element of order less than $p^{n-1}$. Since $\langle a\rangle $ has a unique subgroup of order $p^m$ where $m<n-1$, $G$ has a unique cyclic subgroup of order $p^m$.

This is as far as I've got so far. Not really sure where to go from here. Any ideas? Let me know if you need more details.

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marked as duplicate by Nicky Hekster group-theory Aug 30 '16 at 12:26

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  • $\begingroup$ In what page of the Herstein book is? (to know what theorems can be used and which ones are very strong) $\endgroup$ – MonsieurGalois Aug 30 '16 at 6:33
  • $\begingroup$ This is on page 108 of the second edition. Chapter 2, section 13. $\endgroup$ – Vik78 Aug 30 '16 at 6:34
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(1) $H$ should be of prime order (otherwise it will have a proper subgroup $K$ and $H\nsubseteq K$).

(2) Let $|H|=p$. If $|G|$ is divisible by another prime $q$, then there will be a subgroup of order $q$ and $H$ can not be contained in it.

(3) Hence $|G|=p^k$ for some $k$, with condition that there is a subgroup $H$ of order $p$ contained in every subgroup.

(4) Let $H=\langle h\rangle$. If $k=1$ we are done. Let $k>1$. Then every proper subgroup of $G$ satisfies the hypothesis and so is cyclic by induction.

(5) Let $M$ be a maximal subgroup of $G$; it is cyclic by (4), say $M=\langle x\rangle$. Take $y\in G\setminus M$.

(6) Then consider the following situation in $G$: $$ 1 \leq \langle h\rangle \leq \cdots \leq \langle x^p\rangle \leq \langle x\rangle \leq G.$$ Since $[G: \langle x\rangle]=p$, so $y^p\in \langle x\rangle$.

(7) Suppose if possible $y^p$ is in $\langle x^p\rangle$. Then $y^p=x^{ip}$ for some $i$. Then consider $(yx^{-i})$. Its order is $1$ or $p$ or $p^2$ or ...

(8) Since $(yx^{-i})^p=1$ and $yx^{-i}\neq 1$ (o.w. $y\in \langle x\rangle$) hence order of $\langle yx^{-i}\rangle$ is $p$, and it should be equal to the subgroup $H=\langle h\rangle$ by hypothesis.

(9) Since $yx^{-1}\in H\leq \langle x\rangle$, hence $y\in \langle x\rangle=M$ contradiction. Thus assumption (that $y^p\in \langle x^p\rangle$) is wrong, hence $y^p$ is in $\langle x\rangle$ but not in $\langle x^p\rangle$.

(10) Then $|y^p|=|x|$ so $|y|=p.|x|=|G|$.

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  • $\begingroup$ Great proof! I was almost there. $\endgroup$ – Vik78 Aug 30 '16 at 18:54

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