# If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic.

Problem: Let $G$ be an infinite abelian group. Show that if $G$ has a nontrivial subgroup $K$ such that $K$ is contained in all nontrivial subgroups of $G$, then $G$ is a $p$- group for some prime $p$. Moreover, $G$ is of type $p^\infty$ (quasicyclic) group.

I have the following result:

If $G$ is an infinite abelian group all of whose proper subgroup are finite, then $G$ is of type $p^\infty$ group for some prime $p$.

So I am trying to show that each of subgroup of $G$ is finite. But I can't see anymore. Please help me to find some direction.

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Show that if $G$ has a nontrivial subgroup that contains in Or is contained in Or contains all? –  Babak S. Dec 6 '12 at 7:11
$G$ is a nontrivial subgroup which contains all nontrivial subgroups for every non-identity group $G$. Please formulate your question better. –  Alexander Gruber Dec 6 '12 at 7:28
Sorry. I want to say : is contained in. More details, $G$ has a nontrivial subgroup $H$ such that $H \le K$, where $K$ is any nontrivial subgroup of G. –  tlquyen Dec 6 '12 at 7:46

Hint: If it is as @Alexander noted, show that the group is torsion. So you can write $G$ for some prime $p$ as $$G\cong G_p$$ where $G_p=\{x\in G\mid p^nx=0\}$ for some $n\geq 0$. This can be show that the group in a $p$ group. For the rest, I think we should focus on proving $G$ is dividable.

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@tlquyen: Could you solve the problem finally? Could you get the gist of Hagen's nite answer? –  Babak S. Dec 8 '12 at 12:00
I can see $G$ is torsion and $G\cong G_p$ for some prime $p$. $G$ is dividable if $pG= G$, so suppose $G/pG\ne 0$. I pass to consider $G/pG$ is a $Z_p$-vector space to get something wrong, but I stuck here. Can you give me some help. –  tlquyen Dec 9 '12 at 14:28
@tlquyen: Let we prove that $G$ is torsion, so we can prove that, according to our assumption, that for some $p$ $G\cong G_p$. If $G$ is divisible so it should be of form $\mathbb Z(p^{\infty})$ (We have a theorem about this fact) and so the proof is complete. –  Babak S. Dec 9 '12 at 17:54
@tlquyen: If $G$ is not divisible so since it is a $p$ group, there is a pure cyclic subgroup like $S$ such that for an $n$, $nS=0$. We have a theorem saying that in this step $S$ is a direct summand in $G$. In other words, there is an $H\leq G$ such that $G=H\oplus S$. But it leads us to a contradiction because according to our assumption $H$ and $S$ have non trivial elements . And the proof is again complete. $G$ is divisible and torsion. Is it useful? –  Babak S. Dec 9 '12 at 17:54
Sorry Babak Sorouh for my late. I have learn many thing from your help. Thank you very much! –  tlquyen Jan 1 '13 at 14:55

Assume $K\le G$ is a non-trivial subgroup of $G$ and that for every non-trivial subgroup $X\le G$, we have $K\le X$. Then $K$ itself has no nontrivial subgroup. Especially, for any $k\in K\setminus\{1\}$, we have $\langle k\rangle=K$ and thus $K\cong \mathbb Z$ or $K\cong \mathbb Z/n\mathbb Z$. But only the case with $n=p$ prime leads to $K$ having no nontrivial subgroups.

Let $g\in G\setminus\{1\}$ be arbitrary. Then $k\in\langle k\rangle = K\le \langle g\rangle$, i.e. $k=g^n$ for some $n$. If we write $n=p^mr$ with $r$ not divisible by $p$, then $h:=g^{p^{m+1}}$ has order dividing $r$ (because $h^r=g^{pn}=k^p=1$) and thus $k\notin \langle h\rangle$, from which we conclude $h=1$, hence $g^{p^{m+1}}=1$ and the order of $g$ is a power of $p$. This shows that $G$ is a $p$-group.

Can you see why the second claim (type $p^\infty$) follows from here if we use that $|G|=\infty$? And where did I use that $G$ is abelian?

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I think the OP should know the structure $\mathbb Z(p^{\infty})$, until find out your first question above. Nice approach. I favorited it for the latter. +1 and :) –  Babak S. Dec 6 '12 at 7:57
@ Hagen von Eitzen: Thank. There are some mistakes in my question. The case that I want to mention is : $H\le G$ is a non-trivial subgroup of $G$ and that for every non-trivial subgroup $X\le G$, we have $H\le X$. What do you think about this case? –  tlquyen Dec 6 '12 at 8:12
@tlquyen Oops, actually that was what my solution is about and I made a typo $X\le H$ instead of $H\le X$. In fact, this is how I concluded that there is no nontrivial subgroup in $H$: If $X\le H$ is nontrivial, then also $H\le X$, hence $X=H$. –  Hagen von Eitzen Dec 6 '12 at 17:51