Show that in a finite cyclic group G of order n, writtten multiplicatively, the equation xm = e has exactly m solutions x in G for each positive integers m that divides n. Attempt... Proof: Let G be a finite cyclic group of order n, and suppose m is a positive integer that divides n. Let x be an element in G. Then, x generates a cyclic subgroup of G that contains (n/m) elements since the gcd(n,m)=m [we know this by a theorem presented in the section0. Hence, (1) (xm)n/m= xn= e. The positive integer m was picked arbitrarily. Thus, equation (1) will hold for each m that divides n. Hence, xm=e has exactly m solutions. QED?? Does this hold or am I missing something?