= 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. \b[No such normal subgroup exists for $m>1$.]
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