# Torsion elements of a group aren't necessarily a subgroup [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 14 '18 at 6:32

• 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. – Jyrki Lahtonen Oct 14 '18 at 5:58
• 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 ... – Jyrki Lahtonen Oct 14 '18 at 6:01
• Thank you. Can you maybe provide actual formulas for the transformations? + what exactly is $G$ in your case? – 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.