How to show that a group homomorphism from $SL(2, \Bbb Z/n\Bbb Z)$ is one-to-one? Let $n$ be an odd integer, and let $f:SL(2, \mathbb{Z}/n\mathbb{Z}) \to 
GL(n^2, \mathbb{C})$ be a group homomorphism. The special linear group
$SL(2, \mathbb{Z}/n\mathbb{Z})$ is known to have the following
presentations.
$$
\langle A, B  \mid 
A^n = 1, (AB)^3=B^2, B^4=1, (A^2 B A^{\frac{n+1}{2}} B)^3=1\rangle
$$
$$
\langle A, B  \mid 
A^n = B^2 = (AB)^3 = (A^{\frac{n+1}{2}}BA^4B)^2\rangle
$$
Suppose that the homomorphism $f$ has been defined by specifying 
$f(A)$ and $f(B)$. What are some
techniques to show that $f$ is one-to-one? 
I hope someone can help me out. Thank you. 
 A: If the homomorphism is not injective, there needs to be a nontrivial kernel, which will be a normal subgroup, and thus contains a minimal normal subgroup. Now for each minimal normal subgroup of $SL_2(Z/nZ)$ a word in $A$ and $B$ that represents a nontrivial element in this subgroup. Evaluate these words in the generator images, then the homomorphism is one-to-one if and only if all these words evaluate nontrivial.
As for finding the minimal normal subgroups, if $n$ is prime, this is only the group generated by the scalar matrix for $-1$. If $n$ is a prime poer there also is the kernel of the congruence map modulo $n/p$. If $n$ has multiple prime factors, the group is a subdirect product of the images modulo the prime powers in $n$, then simply do the respective test for every prime power.
Exceptions: If n=2, there are three cyclic groups of order 2 (corresponding to the elements of order 2 in $(Z/2^kZ)^*$) and the congruence kernel splits in the direct sum of one of them and a subgroup of order $2^2$ (instead of $2^3$).
