/sci/ - Science & Math

>>9082340
Let [math]|G|=n[/math] and [math]\gcd(k, n)=1[/math]. If [math]h\in G[/math] is arbitrary, then, by Bezout's lemma, there are integers [math]a, b[/math] such that [math]ak+bn=1[/math], and so [math]h=h^{ak+bn}=h^{ak}=(h^a)^k[/math]. No need for counterassumptions this time either.

>>9082344
Oh right!