$G$ 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
 A: 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 $\,\langle x\rangle\,$
A: 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 $p$.
