ABSTRACT ALGEBRA - cyclic group proof?
Show that in a finite cyclic group G of order n, written multiplicatively, the equation x^m= e has exactly m solutions x in G for each positive integer m that divides n.
** e is the identity element, FYI.
Show that in a finite cyclic group G of order n, written multiplicatively, the equation x^m= e has exactly m solutions x in G for each positive integer m that divides n.
** e is the identity element, FYI.
Comments
If G is cyclic of order n, it has a generating element g. Now what do you suppose
( [g^(n/m)] ) ^ m equals?
( [g^(2n/m)] ) ^ m?
etc.
Steve
It is easy to shows that there are at least m solutions.
if a is a generator, then y=a^(n/m) has order m, so all powers of y are in the order m group generated by y and solve the equation x^m=1.
But you also have to show that there are no more than m solutions. If you already know that there is only one subgroup of order m then use that fact. You may have proved in class that all subgroups of G are cyclic and there is exactly one such for every order m dividing n.