# Elements of infinite order in a group and torsion groups.

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?

• What have you tried? Commented Oct 14, 2020 at 8:43
• Is this a homework problem? Commented Oct 14, 2020 at 8:44
• See also this post about the order of a product $ab$ in $G$. Commented Oct 14, 2020 at 9:08
• @HighGPA No, I commented about motivation under the answer. Commented Oct 14, 2020 at 9:31
• Why don't you put it into the question? Commented Oct 14, 2020 at 12:13

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 $$2$$.

• 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. Commented Oct 14, 2020 at 9:28

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.