In a finite group, normal subgroups whose order and index are coprime are unique up to order

Let G be a finite group, N \leq G a normal subgroup, and suppose that |N| and [G:N] are relatively prime. Prove that N is the unique subgroup of order |N| in G.


Let H \leq G be a subgroup of order |N|. By the previous exercise, we have H \leq N. Because H and N are finite sets of the same cardinality, we in fact have H = N.

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