# Do there exist groups whose elements of finite order do not form a subgroup? [duplicate]

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.

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## marked as duplicate by Qiaochu YuanAug 26 '11 at 2:15

If you are familiar with presentations, consider $\langle a,b\mid a^2 = b^2 = 1\rangle$... (and if you are not, this is the infinite dihedral group, I'm sure you'll find it on wikipedia.) – Myself Aug 26 '11 at 0:00
@Myself, I am not, but thanks for mentioning this. I'll look into it. – yunone Aug 26 '11 at 0:04
@yunone: it is actually straightforward to write down explicit matrices with this property even if one doesn't think of the example of incommensurable rotations. See the second half of my answer in the linked question. – Qiaochu Yuan Aug 26 '11 at 2:16
Thanks Qiaochu, I didn't remember something like this had been asked before. I'll take a look. – yunone Aug 26 '11 at 2:33

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.

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Thanks Theo, I wish I had better intuition on what matrices to choose. I was trying things like [1 2 // 0 1] and [1 0 // 1 1] and ending up with matrices of infinite order. – yunone Aug 26 '11 at 0:08
Another example: $\begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}$ and its transpose, both order 2 with product $\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$ of infinite order. – Jonas Meyer Aug 26 '11 at 0:13
The more the merrier, thanks @Jonas. – yunone Aug 26 '11 at 0:16
@yunone:Here's how I came up with an example. Matrices satisfying $a^2=I$ have eigenvalues $\pm 1$, the easiest examples of which are triangular matrices whose diagonal entries are $\pm 1$. If $a$ has this property then so does its transpose, but the product of $a$ with its transpose, being symmetric, has spectral radius equal to its operator "2-norm", which will be larger than 1 unless $a$ is diagonal. Hence the powers of $aa^\mathrm{T}$ will have unbounded spectral radius. So $a=\begin{bmatrix}1&x\\0&-1\end{bmatrix}$ & $a^T$ works if $x\neq 0$. – Jonas Meyer Aug 26 '11 at 0:26
@yunone: that has nothing to do with intuition but rather with memorization :) I just happen to know these by heart because they are standard generators of $SL(2,\mathbb{Z})$ exhibiting it as amalgamated product $\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6$. I wanted to add some explanations on this but unfortunately google messed up the scan of Serre's trees which yields to a psychedelic picture :( – t.b. Aug 26 '11 at 0:27

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.

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Thanks Gerry ${}$. – yunone Aug 26 '11 at 2:33