# $G$ Group, $H< J$, and $\forall J < G \Rightarrow H \subset Z(G)$

Let $G$ be a group and $H$ a subgroup such that $H$ nontrivial is a subgroup of J, for all $J$ nontrivial subgroup of $G$. Show that $H \subset Z (G)$. Thanks

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The center is a subgroup, which may be trivial or not. – Davide Giraudo Jun 2 '12 at 16:01
@DavideGiraudo But davide, how does that help – user9413 Jun 2 '12 at 16:03
I'm probably missing something, but we can take $J=Z(G)$. – Davide Giraudo Jun 2 '12 at 16:05
As long as it's nontrivial – Robert Maschal Jun 2 '12 at 16:06
It helps us have a better discussion if, after stating the problem, you tell us what you have tried. [Even if nothing you've done has quite worked out, or if you can only state some things which you suspect might work.] Welcome, by the way. – Dylan Moreland Jun 2 '12 at 16:06

Hint: $\,\,H\rlap{\,/}{\subset} Z(G)\,\Longrightarrow \exists\,x\in G\,\,s.t.\,\,hx\neq xh$ , for some $\,h\in H\,$ . Now take a closer look at $\,\langle x\rangle\,$

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By your hypothesis, the intersection of all nontrivial subgroups is nontrivial, and the subgroup $H$ is contained in this intersections. In particular, $H$ is contained in every cyclic subgroup $C$ which is not 1. So if $x$ is any non-identity element, we have $H \subseteq \langle x \rangle$. This means that every $h \in H$ is a power of $x$, so it commutes with $x$. Since this works for any $x \neq 1$, we're done.

Corrected according to Matt E's comment:

We can actually deduce much more from this situation. If $G$ contains an element $x$ of infinite order, then $H$ is contained in every subgroup of $\langle x \rangle$, and so $H$ is trivial. Therefore, $G$ is a torsion group. Then since any two cyclic subgroups of $G$ of prime order intersect trivially, $G$ can only contain one such subgroup, and so we can conclude that $H \cong \mathbb Z / p\mathbb Z$ for some prime $p$.

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Dear Dane, Two remarks: firstly, an infinite group need not contain a copy of $\mathbb Z$; it could be an infinite torsion group (e.g. $\mathbb Q/\mathbb Z$). Secondly, if $G$ does contain a copy of $\mathbb Z$, then $H$ is contained in the intersection of all subgroups of $\mathbb Z$, therefore is trivial. Regards, – Matt E Jun 2 '12 at 18:45
@MattE Thank you very much for pointing that out! – Dane Jun 2 '12 at 21:06
G need not be equal to H, indeed G need not be finite or abelian. You are right that H definitely has to be Z/pZ though. – Jack Schmidt Jul 11 '12 at 13:56
@JackSchmidt thanks – Dane Jul 11 '12 at 16:23