Analysis of Small Groups

G −→ Aut(G/N) ≈ Sq. Let K denote the kernel of the map. Clearly K ⊂ N since each g ∈ K must in particular left translate N back to itself. Thus, since q is the smallest prime divisor of |G|, q = |G/N | | |G/K| = q · (product of primes p ≥ q). On the other hand, the first isomorphism theorem says that G/K is isomorphic to the image of G in Sq, a subgroup of Sq. Thus |G/K| | q!, so that |G/K| = q · (product of primes p < q). Comparing the two displays shows that |G/N | = |G/K|, and so the containment K ⊂ N now gives K = N . Thus N is normal because it is a kernel.