In a question from an exam in an undergraduate group theory course, we were asked to prove or disprove that the set of all Torsion elements of a group is necessarily a subgroup.

I knew that the set of Torsion elements is closed under the inverse operation, but was later told that it is not closed under multiplication, therefore disproving the claim. However, I couldn't find any examples of a group $G$ and two elements $a,b$ such that both $a$ and $b$ have a finite order, but $ab$ doesn't. I know that for this to happen $G$ must be an infinite and non-Abelian group, but still couldn't find a valid example.

What are some examples of groups/elements fulfilling the aforementioned property?


marked as duplicate by Gerry Myerson, Jyrki Lahtonen abstract-algebra Oct 14 '18 at 6:32

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    $\begingroup$ Think affine transformations of the plane. Let $a$ and $b$ be reflections w.r.t. two parallel lines. Show that $ab$ is a translation, and therefore has infinite order. $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 5:58
  • $\begingroup$ Or, let $a$ and $b$ be reflections w.r.t. two lines thru the origin such that the angle is an irrational multiple of $\pi$. The composition $ab$ is then a rotation by ... $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 6:01
  • $\begingroup$ Thank you. Can you maybe provide actual formulas for the transformations? + what exactly is $G$ in your case? $\endgroup$ – Dean Gurvitz Oct 14 '18 at 6:03

the group of isometries of the real line is generated by translations and reflections, with, for $a \in \mathbb{R}$: $$ T_a: x \to x + a \\ R_a: x \to 2a - x $$

So $R_a$ and $R_b$ are both involutions (elements of order 2), but their product: $R_b \circ R_a: x \to 2b - (2a -x) = x + 2(b - a) = T_{2(b-a)}(x)$. If $a \ne b$ the translation $T_{2(b-a)}(x)$ has infinite order.

  • $\begingroup$ Is it true that these functions would also be a valid example if we consider them to be part of the group of all real bijective functions? $\endgroup$ – Dean Gurvitz Oct 14 '18 at 8:36
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    $\begingroup$ to consider a group as a subgroup of a larger group does not change the order of any element, so the answer to your question is yes $\endgroup$ – David Holden Oct 14 '18 at 20:39
  • $\begingroup$ Thank you, just wanted to make sure. I asked because proving the given functions are bijective is easier for those who don't know much about isometries $\endgroup$ – Dean Gurvitz Oct 15 '18 at 5:32

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