Let be a prime and a positive integer. Prove that if is a group of order then every subgroup of index is normal in . Deduce that every group of order has a normal subgroup of order .
Let be a group of order and a subgroup with . Now acts on the conjugates by conjugation, since and . Moreover, under this action we have . By a previous theorem, we have , a prime.
If , then and we have ; moreover, has exactly conjugates in . Let be the permutation representation induced by the action of on the conjugates of , and let be the kernel of this representation. Now . By the first isomorphism theorem, the induced map is injective, so that divides . Note, however, that is a power of and that the only powers of that divide are and . So is or . If , then so that for all ; then and we have , a contradiction. Now suppose . Again by a previous theorem we have , so that , hence . Again, this implies that is normal so that for all , and we have , a contradiction. Thus .
If , then . That is, for all ; thus is normal.
Now let be a group of order . By Cauchy’s Theorem, contains an element of order . Then , hence is normal.