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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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    What happens if G si finite?2012-06-06

2 Answers 2

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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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    @JackSchmidt thanks2012-07-11