For a group $G$, let $G'$ denote the commutator of the group $G$, and if $H \leq G$ the left cosets will be denoted as $gH$. Now, I understand the fact that $[SL_2(\mathbb{Z}):GL_2(\mathbb{Z})'] = 2.$
My question is:
Does there exist an element of order $4$ in $GL_2(\mathbb{Z})/GL_2(\mathbb{Z})'$?
What I have worked out so far is that all the elements of $GL_2(\mathbb{Z})$ that have order $4$ will have $i$ and $-i$ as their eigenvalues. Since order of $gH$ divides order of $g$, the existence of an element of order $4$ in $GL_2(\mathbb{Z})/GL_2(\mathbb{Z})'$ would make it isomorphic to $\mathbb{Z}/4\mathbb{Z}$.
So my rephrased question is as follows:
Is $GL_2(\mathbb{Z})/GL_2(\mathbb{Z})' \cong \mathbb{Z}/4\mathbb{Z}? $
${\rm GL}(2,{\mathbb Z})$ (or ${\rm GL}_2({\mathbb Z})$ if you prefer) is defined to be the multiplicative group of $2 \times 2$ invertible matrices over ${\mathbb Z}$ or, equivalently, matrices over ${\mathbb Z}$ with determinant $\pm 1$.
A presentation of ${\rm GL}(2,{\mathbb Z})$, which I believe comes from the paper T. Brady, Automatic structures on Aut($F_2)$, Arch. Math. 63, 97-102 (1994), is $$\langle P,S,U| P^2, S^2, (SP)^4, (UPSP)^2, (UPS)^3, (US)^2 \rangle,$$ (where ${\rm SL}(2,{\mathbb Z}) = \langle PS,U\rangle$) from which you can calculate that $G/[G,G] \cong C_2 \times C_2$, so the answer to your question is no.
Added later: I should have said that an isomorphism from the group defined by the above presentation to ${\rm GL}(2,{\mathbb Z})$ is defined by $$P \mapsto \left(\begin{array}{rr}0&1\\1&0\end{array}\right),\ \ S \mapsto \left(\begin{array}{rr}-1&0\\0&1\end{array}\right),\ \ U \mapsto \left(\begin{array}{rr}1&1\\0&1\end{array}\right). $$