How would I go about showing this, where $p$ is an odd prime? The inclusion $[GL_{2}(\mathbb{Z}/p^{2}\mathbb{Z}),GL_{2}(\mathbb{Z}/p^{2}\mathbb{Z})] \subseteq SL_{2}(\mathbb{Z}/p^{2}\mathbb{Z})$ is relatively clear. I'm wondering if the proof for the other inclusion follows from the result that $[GL_{2}(\mathbb{Z}/p\mathbb{Z}),GL_{2}(\mathbb{Z}/p\mathbb{Z})]=SL_{2}(\mathbb{Z}/p\mathbb{Z})$.
I've tried to use the result above with the natural homomorphism from $GL_{2}(\mathbb{Z}/p^{2}\mathbb{Z})$ to $GL_{2}(\mathbb{Z}/p\mathbb{Z})$ to prove this but to no avail.
Let's work in $\mathrm{GL}_n$, leaving only aside the case $(n,p)=(2,2)$, so that $\mathrm{SL}_n(\mathbf{Z}/p\mathbf{Z})$ is the derived subgroup of $\mathrm{GL}_n(\mathbf{Z}/p\mathbf{Z})$.
Consider a matrix $g$ in $\mathrm{SL}_n(\mathbf{Z}/p^2\mathbf{Z})$. Then its image in $\mathrm{SL}_n(\mathbf{Z}/p\mathbf{Z})$ is a product of commutators of $\mathrm{GL}_n(\mathbf{Z}/p\mathbf{Z})$; lifting this shows that there exist a product $c$ of commutators such that $gc^{-1}$ is in the kernel of the reduction map $\mathrm{SL}_n(\mathbf{Z}/p^2\mathbf{Z})\to\mathrm{SL}_n(\mathbf{Z}/p\mathbf{Z})$, which is exactly the set of matrices of the form $I+pA$ with $A\in\mathrm{M}_n(\mathbf{Z}/p\mathbf{Z})$ of trace zero. Such a matrix is a product of matrices of the form: either $I+pE_{ij}$ with $i\neq j$, or $I+p(E_{ii}-E_{jj})$ with $i\neq j$. That these matrices are product of commutators is a simple exercise.