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.