Do there exist groups whose elements of finite order do not form a subgroup?

Question

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Examples and further results about the order of the product of two elements in a group

I was browsing around, and came across the little exercise that elements of finite order in an abelian group form a subgroup. This was easy enough, but now I'm wondering, are there cases of groups where this is not true?

I'm thinking that if there are, these groups are necessarily nonabelian and infinite, since if the group is finite the elements of finite order would be the whole group anyway. I also noticed that if an element $a$ has order $n$, then so does $a^{-1}$, so to find such a group I would need to two elements of finite order whose composition does not have finite order. The two infinite nonabelian groups I could think of off the top of my head were $\textbf{GL}_n(R)$ and $\textbf{SL}_n(R)$, but I had trouble finding a pair of elements satisfying the above property.

Does anyone have an example of such a group and a pair of elements which prove it? Thank you.

Answer 1

Try $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ (order $4$) and $\begin{bmatrix}0 & 1 \\ -1 & 1\end{bmatrix}$ (order $6$). and their product is $\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$ which is of infinite order.

Answer 2

Taking up on one of the comments, consider the group of symmetries of the circle. Each flip is an element of order 2. If you have two flips, and the axis of one meets the axis of the other at an irrational angle (irrational number of degrees, or irrational multiple of $\pi$ radians), then their product is an irrational rotation, hence, of infinite order.