On finite groups, power maps with exponent relatively prime to the group order are surjective

Let G be a cyclic group of order n and let k be an integer relatively prime to n. Prove that the map x \mapsto x^k is surjective. Use Lagrange’s Theorem to prove that the same is true for any finite group of order n.


First let G = \langle \alpha \rangle be cyclic of order n (I.e. |\alpha| = n) and let k be an integer relatively prime to n. Note that since G is abelian, \varphi(x) = x^k is a group homomorphism. There exist integers a and b such that ak + bn = 1, or ak = 1 - bn. Now \varphi(\alpha^{ai}) = (\alpha^{ak})^i = (\alpha^{1-bn})^i = \alpha^i. Since every element of G is of the form \alpha^i, \varphi is surjective. Since G is finite, \varphi is a bijection.

Now let G be any finite group of order n, and k an integer relatively prime to n with ak = 1 - bn. Note that G is a union of cyclic subgroups; in particular, we have G = \cup_{g \in G} \langle g \rangle. Thus we can consider the mapping \varphi : x \mapsto x^k restricted to each of these subgroups. By Lagrange’s Theorem, |\langle g \rangle| divides n for all g \in G, so that by the previous half of this problem the restriction \varphi|_{\langle g \rangle} is a surjection, and in fact a bijection. Thus \varphi is a surjection (hence bijection) on all of G.

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