Problem

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 the same is true for any finite group of order \(n\).

Solution

We can write \(G = \langle x \rangle\) and \(|G| = |x| = n\).

Since \(n\) and \(k\) are relatively prime, it follows from Proposition 6 that \(\langle x \rangle = \langle x^k \rangle\). That is, we can express any element of \(G\) in the form \(x^k\) for some \(x \in G\). Therefore, the map is surjective.

$$\tag*{$\blacksquare$}$$

Lagrange’s Theorem states $$ \text{If } G \text{ is a finite group and } H \text{ is a subgroup of } G \text{, then } |H| \text{ divides } |G|. $$

Let \(G\) be a finite group of order \(n\) that is not necessarily cyclic. For an arbitrary \(g \in G\), let \(H = \langle g \rangle\). Since \(H\) is a subgroup of \(G\), by Lagrange’s Theorem, \(|H| \mid n\).

Since \(|H| \mid n\) and \(n\) is relatively prime to \(k\), \(|H|\) is relatively prime to \(k\) as well. Since \(H\) is a cyclic group, as shown previously, the map from \(H\) to itself as defined by \(x \mapsto x^k\) is surjective. This means that \(g = h^k\) for some \(h \in H\) because \(g \in H\).

Note our choice of \(g \in G\) was arbitrary. Therefore, \(x \mapsto x^k\) is surjective as a map from \(G\) to \(G\).

$$\tag*{$\blacksquare$}$$