Abelian group $G$ such that the infinite-order elements form a subgroup with the identity.

Question

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!

Answer 1

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.

Answer 2

Let, $H=\{g \in G \mid g=e \text{ or }g \text{ has infinite order}\}$

$G$ contains all elements of finite order or all elements (except identity) of infinite order.

And other possibility is impossible.

Because if , $\exists a, b \in G $ such that $|a|=n(>1) $ and $|b|=\infty$.

then it will contradict that $H$ is a subgroup of $G$. It violates closure property.

$b\in H \implies b^{-1} \in H$

And $ab\in H . $

But, $ab b^{-1} = a \notin H$

$|a|=m \implies a^m=e$

If $ab\notin H$ then, $|ab|=\text{ finite} =n\text{(say)} $

Then, $(ab) ^{mn}=e $

$\implies{( a^m)^n }{b^{mn}}=e$

$\implies b^{mn}=e$

Hence, $|b| $ divides $mn$.

Contradict, $|b|=infinite$

Answer 3

R* is multiplication group. And H={x is positive real number} is subgroup of R* . But R* has a elements in order 2 .