The proof of this is usually based on the fact that the transitive multiplication action of G on the left coset space of H has a kernel K given by the largest normal subgroup of G contained in H. This consideration leads to the divisibility of [G : H]! by the index [G : K] = [G : H] [H : K]. The assumption on [G : HI then implies that [H : K] = 1, so H = K is normal in G. (See, for instance, [1, p. 123], [2, p. 44].) This proof, although short, is not suitable for beginning students since they are far from having the theory of group actions at their disposal. In teaching a beginning abstract algebra course, I found a proof for Theorem 1 that is both quick and possible for students at the level of such a course to understand. Moreover, the idea of this proof led to several criteria for the normality of subgroups of prime index (in any group G). Since such criteria do not seem to be available in standard textbooks or even in the