Let $G$ be a finite abelian group s.t. it contains a subgroup $H_{0} \neq (e)$ which lies in every subgroup $H \neq (e) $. Prove that $G$ must be cyclic. Also what can be said about $o(G)$ ?

I'm clueless about this problem from Herstein. I tried to think $G$ might be a solvable group and its subgroups are all normal (as $G$ is abelian) . But I'm unable to get further.

  • $\begingroup$ Hint: if the subgroup $H$ is of order $p$ then every other subgroup is of order $p^n$ for some natural value $n$. $\endgroup$ – JonesY Jul 18 '16 at 13:58

Hint : If $G$ is not cyclic, then since it is finite and abelian $\exists n \in \mathbb{N}, \exists U \le G$ such that $G\cong\mathbb{Z}_n \times U$.

Also $\{0\}\times U$ and $\mathbb{Z}_n \times\{e\}$ are two subgroups of $G$. Can you conclude ?

  • $\begingroup$ This means there are two subgroups which intersect trivially, am I right? But how can I show $G\cong\mathbb{Z}_n \times U$ ? $\endgroup$ – null Jul 18 '16 at 14:17
  • $\begingroup$ Itis the fundamental theorem of finitely generated abelian groups : en.wikipedia.org/wiki/Finitely_generated_abelian_group $\endgroup$ – Jennifer Jul 18 '16 at 14:19
  • $\begingroup$ And you conclude that there is no $H_0\neq \{e\}$ wich lies in every subgroup $\endgroup$ – Jennifer Jul 18 '16 at 14:26
  • $\begingroup$ Yeah, thanks @Jennifer. $\endgroup$ – null Jul 18 '16 at 14:27

If there is a nontrivial subgroup $H$ that lies in every other subgroup. Then $o(H)$ divides the order of every subgroup $K$. From Sylow's theorems you can prove that $o(K)=o(H)^n$ and $o(H)=p$ (prime) for every other nontrivial subgroup. (Because otherwise you have contradiction). Therefore $G$ is of order $p^n$. Since every group of prime order is cyclic then...

  • $\begingroup$ Can you explain from which Sylow theorem? $\endgroup$ – null Jul 18 '16 at 14:14
  • $\begingroup$ If G is a finite group, $p$ is a prime that $p^n$ divides $O(G)$ and $p^{n+1}$ does not. then there is a subgroup of order $p^n$. $\endgroup$ – JonesY Jul 18 '16 at 14:18
  • $\begingroup$ Sorry for late reply, but I still can't see why this would imply $o(K)=o(H)^n$ and not $o(K)=o(H)^n * m$ for some m. Can you plz elaborate more? $\endgroup$ – null Jul 19 '16 at 3:14
  • $\begingroup$ cause that m can be factorized to primes $q_1\cdots q_n$ which have subgroups of $q_j^{l_j}$ but $o(H)$ does not divide it. so H is not a subgroup in every subgroup. contradiction to the assumption. $\endgroup$ – JonesY Jul 19 '16 at 6:40
  • $\begingroup$ Thanks a lot, @JonesY , nice answer. $\endgroup$ – null Jul 19 '16 at 14:13

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