Decomposing an element into product of elements of finite order If $G$ is a group and $g, h\in G$, it is possible that $g$ and $h$ have finite order, yet $gh$ has infinite order. For example, in Algebra: Chapter 0 by Paolo Aluffi, Exercise 1.12, the following is given:
Let $G=GL_2(\mathbb{R})$ be the group of all invertible $2\times 2$ matrices over $\mathbb{R}$. Let
$$
g=\begin{pmatrix}
0 & - 1 \\
1 & 0 \\
\end{pmatrix}
\ \ \ \ \ \ \
h=\begin{pmatrix}
0 & 1 \\ 
-1 & -1 \\
\end{pmatrix}
$$
Then $|g|=4, |h|=3$ but $|gh|=\infty$.
So my question is whether or not, such a decomposition is always possible in the setting of invertible $2\times 2$ matrices. 

Let $G=GL_{2}(\mathbb{R})$. Suppose $x\in G$.
  Does there exist $g, h\in G$ such that $x=gh$ satisfying $|g|<\infty$, $|h|<\infty$?

Is anything known about groups in which every element can be written as product of two elements each with finite order?
 A: If $g,h$ have finite order, then $\det g$, $\det h$ are roots of unity. So if $\left|\det x\right|\ne 1$ (or even: is not a root of unity), we cannot have such a decomposition.
A: Concerning your question Is anything known about groups in which every element can be written as product of two elements each with finite order? : in the abelian case you are basically talking about torsion groups: if every element can be written as a product of two torsion elements, then the torsion subgroup must be the whole group.A torsion abelian group can always be decomposed into its $p$-parts: given an abelian group $A$, and a prime $p$, let
$A_p = \{a\in A\mid \mbox{the order of $a$ is a power of $p$}\}$.
Then $A_p$ is a subgroup of $A$, and $A = \mathop{\oplus}\limits_{p} A_p$.
The infinite dihedral group is another example that can be matrix-interpreted by a non-rational rotation $r$ and reflection $s$, it has presentation $\langle r,s:s^2=1, srs=r^{-1}\rangle$ and $sr$ has order $2$, but $r=s \cdot sr$ has infinite order.
