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


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.25

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 $n$.


Solution: 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 $G$.


Linearity

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