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.


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

(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 $|y|=p.|x|=|G|$.

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

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