Prove the group is a direct product Let $G$ be an abelian group of finite order $n = mk$ with gcd$(m,k) = 1$. For $r=m,k$, let $G(r) = \{g \in G: g^r = 1 \}$ . Prove that $G = G(m) \times G(k)$.

It is not hard to show that $G(m)$ and $G(k)$ are subgroups and intersect each other trivially. We need to show then that $G(m)G(k)=G$. Suppose $g\in G$ is an element; then since $m$ and $k$ are relatively prime we have that there are integers $a,b$ such that $am+bk=1$. Then $g=g^{am}g^{bk}$, and $g^{am}\in G(k)$ and $g^{bk}\in G(m)$, so $G(m)G(k)=G$.
