Let $G$ be an abelian group.
If $\{g \in G \mid g=e \text{ or }g \text{ has infinite order}\}$ is a subgroup of $G$, what can we say about the order of the elements of $G$?
My observations:
- It is trivial that any finite abelian group works for $G$.
- Moreover, any group such that all its elements are of finite order is also a valid example of $G$.
- Another possibility is to have a group such that every non-identity element has infinite order. For example, $\mathbb{Z}$.
My question:
Is it possible to have a group with these properties such that it contains both finite and infinite-order non-identity elements?
I have failed to find an example. And I strongly feel that no such example exists.
But how can I prove this? Please help!
Best Answer
It is impossible.
Suppose otherwise. Let $a$ have infinite order and $b$ be nontrivial with finite order. Then $ab$ has infinite order, since $a$ has infinite order, because otherwise $(ab)^n=e$ implies $b^{-n}=a^n$, a contradiction. But observe that
$$b=eb=aa^{-1}b=(ab)a^{-1}$$
is an element of the candidate subgroup $H$ (as $a^{-1}, ab\in H$), a contradiction.