## A finite group is nilpotent if and only if all pairs of elements of relatively prime order commute

Prove that a finite group $G$ is nilpotent if and only if whenever $a, b \in G$ with $\mathsf{gcd}(|a|,|b|) = 1$, then $ab = ba$. [Hint: Use Theorem 3.]

$(\Rightarrow)$ Suppose $G$ is a finite nilpotent group. Let $a,b \in G$ with $\mathsf{gcd}(|a|,|b|) = 1$. By Theorem 6.3, $G$ is the internal direct product $G = P_1P_2 \cdots P_k$ of its Sylow subgroups; thus we may write $a = a_1a_1\cdots a_k$ and $b = b_1b_2 \cdots b_k$, where $a_i,b_i \in P_i$. Moreover, $|a| = \mathsf{lcm}(|a_i|)$ and $|b| = \mathsf{lcm}(|b_i|)$; if some $a_i \neq 1$, then $p_i$ divides $|a|$. so $p_i$ does not divide $|b|$, and $b_i = 1$. Similarly, if $b_i \neq 1$ then $a_i = 1$.

By relabeling the Sylow subgroups of $G$ and collecting factors, we have the internal direct product $G = Q_1 \times Q_2$, with $a \in Q_1$ and $b \in Q_2$. Thus $ab = ba$.

$(\Leftarrow)$ Suppose $G$ is a finite group in which the hypothesis holds. Let $P \leq G$ be a Sylow $p$-subgroup. If $Q \leq G$ is a Sylow subgroup for some other prime $q$, then $Q \leq C_G(P) \leq N_G(P)$. In particular, $N_G(P)$ contains a subgroup of maximal $q$-power order for each $q$ dividing $|G|$; thus $N_G(P) = G$, so that $P \leq G$ is normal. Since $P$ is arbitrary, all Sylow subgroups of $G$ are normal. By Theorem 3, $G$ is nilpotent.