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!

  • 2
    $\begingroup$ 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? $\endgroup$
    – Arthur
    Nov 20, 2021 at 11:44
  • 2
    $\begingroup$ "have a group such that every non-identity element has infinite order": a group with this property is called a torsion-free group. $\endgroup$
    – YCor
    Nov 20, 2021 at 15:00

3 Answers 3


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


is an element of the candidate subgroup $H$ (as $a^{-1}, ab\in H$), a contradiction.

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 20, 2021 at 11:55
  • 1
    $\begingroup$ I've edited my answer accordingly, @GerryMyerson; thank you. $\endgroup$
    – Shaun
    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 .

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Mar 10, 2023 at 2:56
  • 2
    $\begingroup$ This does not answer the question. $\endgroup$ Mar 10, 2023 at 3:01
  • $\begingroup$ H is a subgroup, but it is not the subgroup which we are considering here. $\endgroup$
    – 311411
    Mar 10, 2023 at 3:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .