Show that $K \supseteq \{ g \in G \mid o(g)=n \}$

Question

The question is as follows:

Let $K$ be a normal subgroup of a group $G$ such that $[G : K]=m$. If $n$ is a positive integer such that $\gcd{(m,n)}=1$, then show that $K \supseteq \{ g \in G \mid o(g)=n \}$.

Can anyone please give any hints?

Answer 1

Let $g$ be such that $|g|=n$. Consider the coset $gK$. Then $K=(gK)^m=g^mK$. Thus $g^m \in K$, But $e=g^n \in K$. Now use the fact that $1=\gcd(m,n)=mx+ny$ for some $x,y \in \mathbb{Z}$ to claim $g^1 \in K$.