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.

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