Proving that $SL_2(\mathbb{Z}_5) / \{\pm I\}\simeq A_5$ I see here that one can prove that
$$
SL_2(\mathbb{Z}_5) / \{\pm I\} \simeq A_5
$$
using the First Isomorphism Theorem. 

My question is how one would do that. 

I know that I need a surjective homomorphism 
$$
T: SL_2(\mathbb{Z}_5) \to A_5
$$
with kernel $\{\pm I\}$. The only homomorphism I have come across with matrix groups is the determinant map, so my question is what homomorphism would work here.
 A: The group $PGL(2,\mathbb{F}_5)$ acts faithfully on the set $X$ of lines of the vector space $\mathbb{F}_5^2$. One can show that $X$ has cardinal $6$. Hence the action gives an injective group morphism $\rho$ from $PGL(2,\mathbb{F}_5)$ to $\mathfrak{S}_6$ the symmetric group on $6$ elements.
Now $\rho(PGL(2,\mathbb{F}_5))$ is of cardinal$120$ hence of index $6$ in $\mathfrak{S}_6$ and hence (group theory behind this) isomorphic to $\mathfrak{S}_5$.
Finally $PSL(2,\mathbb{F}_5)$ is of index $2$ in $PGL(2,\mathbb{F}_5)$ and hence isomorphic to $\mathfrak{A}_5$.
A: Let $G=SL_2(\mathbb{Z}_5)$, $S$ be a Sylow $2$-subgroup of $G$. Then $[G:N_G(S)]=5$, hence we have $5$ Sylow $2$-subgroups of $G$, and $G$ acts transitively on them. In such a way we get transitive permutation representation $\rho:G\to S_5$ with $\ker(\rho)=\{\pm I\}$. Since $G$ has not a subgroups of index $2$, containing $\{\pm I\}$, then $\rho(G)\leq A_5$. But $|\rho(G)|=|A_5|$, hence we have epimorphism $\rho:G\to A_5$ and by the fundamental theorem on homomorphisms $SL_2(\mathbb{Z}_5)/\{\pm I\}\cong A_5$.
How to show, that $[G:N_G(S)]=5$? Note, that $|G|=2^3\cdot 3\cdot 5$, hence each Sylow $2$-subgroup of $G$ has order $8$. For natural ring's homomorphism $\mathbb Z\to\mathbb{Z}_5$ we will use a bar convention. Let 
$$
A=
\begin{pmatrix}
\bar 2 & \bar 0 \\
\bar 0      & -\bar 2
\end{pmatrix},
B=\begin{pmatrix}
\bar 0  & \bar 1 \\
-\bar 1 & \bar 0
\end{pmatrix}.
$$
Then $A,B\in G$ and 
$$
A^2=B^2=-I, B^{-1}AB=A^{-1}, o(A)=o(B)=4.
$$
Hence $S:=\langle A,B\rangle\simeq\mathbb{Q}_8$ and $|S|=8$. Therefore $S$ is a Sylow $2$-subgroup of $G$. It is known that quaternion group has $3$ subgroups of order $4$. In $S$ subgroups of order $4$ are 
$\Omega:=\{\langle A\rangle,\langle B\rangle,\langle C\rangle\}$, where $C=AB$. If $X\in N:=N_G(S)$, then $\langle X\rangle$ acts on $\Omega$ by conjugation. This suggests try to find such $X\in G$, that $\langle X\rangle$ acts on $\Omega$ transitively. We may try to find, for example, such $X$, that $A^X=C$ and $B^X=-A$ (it comes down to solving a very simple systems of equations). I did these easy calculations and found one such matrix:
$$
X=
\begin{pmatrix}
\bar 2 & \bar 2 \\
\bar 1 & -\bar 1
\end{pmatrix}.
$$
It follows that $3\shortmid|N|$. If $5\shortmid|N|$, then $S\unlhd G$, it is impossible, since $PSL_2(5)$ simple (in fact, we can avoid the use of simplicity here). In such a way $|N|=2^3\cdot 3$ and $[G:N_G(S)]=5$.
