Infinite group without a subgroup isomorphic to $\mathbb{Z}$?

Question

I am looking for examples of infinite groups in which no subgroup is isomorphic to $\mathbb{Z}$. Some of the examples I can think of are: $$\prod_{i=1}^\infty \mathbb{Z}_2, \quad \bigcup_{i=1}^\infty S_i, \quad \text{Prüfer groups}.$$ Here $S_n$ denotes the permutation group.

1. The above ones are actually the examples for the given statement
2. The main question I would like to ask is: Is there any characterisation of such groups -- groups that has no subgroup isomorphic to $\mathbb{Z}$.
3. Can you give more such examples?

Thanks in advance,

-- Mike

Answer 1

Having a subgroup isomorphic to $\Bbb Z$ is equivalent to having an element of infinite order, thus the groups without such a subgroup are exactly the torsion groups: groups where every element has finite order.

For abelian groups we also have the following characterisation:

An abelian group is torsion if and only if it is a direct limit of finite groups.

Another characterisation is that an abelian group is torsion if and only if the Pontryagin dual (with respect to the discrete topology) is profinite.