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Suppose elements $a$, $b\in G$ of group $G$ have infinite order and $ab\neq1$. Can $ab$ have finite order?

Edit: I am interested in Burnside problems and I was thinking about subgroup of a group generated by elements of infinite order. If product of elements of infinite order has infinite order in a group, then subgroup generated by elements of infinite order consist only of elements of infinite order and $1$. Moreover this subgroup would be normal so I would be able to make a quotient group which will be a torsion group. Hence my actual question should be:

Are there conditions for a group which imply that product of elements of infinite order has infinite order and if yes, then what are they?

Ernie060
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Andrzej Karolak
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2 Answers2

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Yes, consider the group $\mathbb{R}^\times$. Simply take $a=-2$ and $b=\frac{1}{2}$. Since both $a$ and $b$ have non-unit magnitudes, they have infinite order. But $ab=-1$, which is non-identity and has order $2$.

Haran
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    Thank you. I was considering torsion groups and I was wondering when a set of elements of infinite order in a group creates a subgroup (together with 1 of course). Is there a condition that implies that $ab$ from the question needs to have infinite order? In that case this subgroup generated by infinite order elements would be normal and the quotient will be a torsion group. – Andrzej Karolak Oct 14 '20 at 09:28
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The answer is always "yes" for nilpotent groups (this includes abelian groups), under the obvious assumption needed to state the problem (elements of finite and infinite order):

Proposition. If $G$ is a nilpotent group containing elements of finite order and of infinite order then yes, there exists elements $a, b\in G$ of infinite order such that $ab\neq1$ has finite order.

Proof. Let $a$ have infinite order and $x$ have finite order. Set $b:=a^{-1}x$. Clearly $ab=x$ has finite order, so we just need to prove that $b$ has infinite order.

Suppose $b$ has finite order. As we are in a nilpotent group, the elements of finite order form a subgroup $\mathrm{Fin}$ of $G$ (see here for a proof), and therefore $\langle b, x\rangle\leq \mathrm{Fin}$. As $a=xb^{-1}\in \langle b, x\rangle$, we have that $a$ has finite order, a contradiction. QED

Example. Let $G=\mathbb{Z}\times\mathbb{Z}_n$ for $n\geq2$. Set $a:=(1, 0)$, $x:=(0, 1)$. Then, using additive notation, $b:=(-1, 1)=-a+x$ has infinite order as $nb=(-n, 0)=-na$. We therefore have $a, b$ of infinite order, while $a+b=(0,1)=x$ has finite order.

The only thing used above was that the elements of finite order form a subgroup. so we actually have the following.

Proposition. If $G$ contains elements of finite order and of infinite order, and if the elements of finite order in $G$ form a subgroup, then yes, there exists elements $a, b\in G$ of infinite order such that $ab\neq1$ has finite order.

user1729
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