Finite subgroups with relatively prime orders intersect trivially

Let G be a group and let H, K \leq G be finite subgroups of relatively prime order. Prove that H \cap K = 1.


Let |H|=p and |K|=q. We saw in a previous exercise that H \cap K is a subgroup of both H and K; by Lagrange’s Theorem, then, |H \cap K| divides p and q. Since \mathsf{gcd}(p,q) = 1, then, |H \cap K| = 1. Thus H \cap K = 1.

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