# Classifying elements of finite and infinite order in $GL_{2}(\mathbb{R})$

A problem on a recent assignment defined the Torsion subset $$F(G)$$ of a Group $$G$$ as the set of elements of G of finite order. It then asked to prove that $$F(GL_2(\mathbb{R}))$$ is not a subgroup of $$GL_2(\mathbb{R})$$, which I did by example (two finite order matrices multiplied to give one with infinite order, proof by induction). However this led me to thinking about, more generally, the requirements separating those elements of finite and infinite order.

My first thought was that, by the definition of finite order, there exists some $$n \in \mathbb{N}$$ such that $$A^n = I$$, and since taking determinants satisfies the conditions for a homomorphism, $$\det{(A^n)} = \det{(A)}^n = \det{(I)} = 1$$, so $$\det{(A)} \in \{ \pm 1 \}$$. However, clearly there is a stronger condition holding that I am not seeing, as taking (for example) $$A = \bigl( \begin{smallmatrix}1 & -1\\ 0 & 1\end{smallmatrix}\bigr)$$, clearly $$\det{(A)} = 1$$, but for any $$n \in \mathbb{N}$$, we have $$\bigl( \begin{smallmatrix}1 & -1\\ 0 & 1\end{smallmatrix}\bigr)^n = \bigl( \begin{smallmatrix}1 & -n\\ 0 & 1\end{smallmatrix}\bigr)$$, so clearly $$A$$ also has infinite order.

My Linear Algebra is not amazing, so I feel like I may be lacking the tools with which to fully analyse the problem.

Any advice on where to advance next is appreciated.

EDIT: My question originally also concerned the notion of abelian groups in relation to $$F(G)$$ being a subgroup but it has been brought to my attention that $$F(G) \le G$$ trivially for abelian groups.

• 1) A matrix has finite order iff it's diagonalizable and its eigenvalues are roots of unity. Your second matrix isn't diagonalizable. 2) The torsion subset of an abelian group is always a subgroup; in the nonabelian case it's almost never a subgroup but it can be. – Qiaochu Yuan Oct 11 '18 at 17:33
• @user6731064: 1) Once you know that the eigenvalues of $A^n$ are the $n^{th}$ powers of the eigenvalues of $A$, the condition $A^n = 1$ immediately implies that the eigenvalues of $A$ must be $n^{th}$ roots of unity. Then once you've learned about Jordan normal form you can check that if $A$ has such eigenvalues but is not diagonalizable then it necessarily has infinite order. The topic of how to compute powers of a matrix using Jordan normal form in this way should come up in any good linear algebra textbook. – Qiaochu Yuan Oct 11 '18 at 17:41
• Well, then you could take some silly example like $S_3 \times \mathbb{Z}$ where the torsion elements are $S_3 \times \{ 0 \}$ which is a subgroup (even a normal subgroup). – Daniel Schepler Oct 11 '18 at 17:41
• Come to think of it - if it happens that $F(G)$ is a subgroup of $G$ then it's automatically a normal subgroup as well. – Daniel Schepler Oct 11 '18 at 17:42
• 2) I'm not sure what you want here. Suppose $g, h$ are torsion elements of some group, with $g^n = h^m = e$. You want to see whether you can prove that $gh$ is torsion. Well, if $g$ and $h$ commuted the proof would be easy, since then we would have $(gh)^{nm} = e$. Otherwise there's no obvious attack and you can look for examples and counterexamples. – Qiaochu Yuan Oct 11 '18 at 17:43