For prime $p$, normal subgroups of $SL(2, \mathbb Z/p\mathbb Z)$ remains normal in $GL(2, \mathbb Z/p\mathbb Z)$? Let $p$ be a prime. If $G$ is a normal subgroup of $SL(2, \mathbb Z/p\mathbb Z)$ , then is $G$ also normal in $GL(2, \mathbb Z/p\mathbb Z)$
For $p\ge5$, $SL(2,p)/\{\pm I\}$ is simple, so the only nontrivial
normal subgroup of $SL(2,p)$ is $\{\pm I\}$ which is normal in $GL(2,p)$.
You'll then have to treat the cases $p=2$ and $p=3$ separately.
Note the matrix (with $a\neq 0$)
$$m_a=\left[ {\begin{array}{cc}
a &amp;0\\
0 &amp;1
\end{array}}\right]$$
has determinant $a$ and inverse 
$$\left[ {\begin{array}{cc}
a^{-1} &amp;0\\
0 &amp;1
\end{array}}\right]$$
in $GL(2, \mathbb Z/p\mathbb Z)$. Therefore every matrix $g\in GL(2, \mathbb Z/p\mathbb Z)$ can be written as $m_as$ (where $a=\det g$) for some $s\in SL(2, \mathbb Z/p\mathbb Z)$.

Now for all $s\in SL$, $sGs^{-1}=G$. Then for any $g\in GL$, let $g=m_as$, so $gGg^{-1}=m_asGs^{-1}m_a^{-1}=m_aGm_a^{-1}=G$ (somehow)

Can someone help me complete this proof :(