A nontrivial subgroup of $G$ contained in every other nontrivial subgroup. This question is from a collection of past master's exams. > Let $G$ be a group with a subgroup $H$ as described in the title. I'd like to show that $H$ is in the center of $G$. My intuition is to choose some element $x\in H$ (and thus in every nontrivial subgroup $K$ of $G$) and then apply the counting formula; i.e. the order of $G$ is the product of the order of the centralizer of $x$ with that of its conjugacy class. I feel like this, together with the class equation, should tell me everything I need to know. It's been awhile since I've done a problem like this, so any help would be appreciated.

Consider an arbitrary element $g\in G$. $g$ induces the cyclic subgroup $C:=\\{g^n:n\in\mathbb Z\\}$. Since $C$ is nontrivial (unless $g$ is the neutral element), $H$ must, by definition, be a subgroup of $C$. But all elements of $C$ commute with $g$, by construction. Therefore since $H$ is a subgroup of $C$, all elements of $H$ also commute with $g$. But $g$ was chosen arbitrary, thus all elements of $H$ commute with _every_ element of $G$, thus $H$ is a subgroup of the center.