 Solution

Suppose $N$ were a normal subgroup of index $m$ and put $Q=SL(2,{\mathbb{F}}_p)/N$. The supplied reduction map and part (c) show that $Q$ is generated by the images $s,t$ of $S,ST$. Their orders divide $4$ and $6$, respectively, because $S^4=I$ and $(ST{)}^6=I$. By Lagrange's theorem, their orders also divide $|Q|=m$. Since $\gcd (m,6)=1$, both $\gcd (m,4)$ and $\gcd (m,6)$ equal one. Thus $s=t=1$, forcing $Q$ to be trivial and $m=1$, a contradiction. No such normal subgroup exists for $m>1$.