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

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.

• Presumably, you have a theorem or two that gives you concrete tests you can use to check whether a subset is a subgroup. Have you tried those? What happens in the general case? Does it work? If not, what stops it from working? Commented Nov 20, 2021 at 11:44
• "have a group such that every non-identity element has infinite order": a group with this property is called a torsion-free group.
– YCor
Commented Nov 20, 2021 at 15:00

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.

• I don't see why $(ab)^n=e$ implies the order of $b$ divides $n$. I'd say $(ab)^n=e$ implies $a^n=b^{-n}$ but $b^{-n}$ has finite order so $a^n$ has finite order so $a$ has finite order, contradiction. Commented Nov 20, 2021 at 11:55
• I've edited my answer accordingly, @GerryMyerson; thank you.
– Shaun
Commented Nov 20, 2021 at 11:59

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, $$|b|=infinite$$

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