TITLE: A nontrivial subgroup of $G$ contained in every other nontrivial subgroup.
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: The classification of such groups is simple in two cases:

If G is a finite group, then G has a non-identity subgroup H contained in every non-identity subgroup if and only if G is either non-identity cyclic of prime power order or a (generalized) quaternion 2-group.
If G is an abelian group, then G has a non-identity subgroup H contained in every non-identity subgroup if and only if G is either non-identity cyclic of prime power order or a Prüfer p-group, $\mathbb{Z}[1/p]/\mathbb{Z}$.

The first is well known (for instance, Gorenstein's Finite Groups textbook, Theorem 5.4.10.ii page 199). The second is also easy: H is by definition essential and clearly has order p for some prime p, so G is contained in a maximal essential extension of H, and all such groups are Prüfer p-groups.
I suspect the classification for general groups is also fairly easy: presumably just the infinite generalized quaternion 2-group. However, I neither proved this nor found a classification.
By the above arguments (in other answers), G is a p-group and H is a central subgroup of order p, and both the finite and abelian subgroups of G are known.