Proving the Existence of Subgroups in Cyclic Groups

screenname123
Messages
1
Reaction score
0

Homework Statement



Let G be a finite cyclic group of order n. If d is a positive divisor of n, prove that the equation x^d=e has d distinct solutions

Homework Equations



n=dk for some k
order(G)=n

The Attempt at a Solution


solved it:
<g^k>={g^k, g^2k,...,g^dk=e} and for all x in <g^k> x^d=e and order(g^k)=d.
 
Last edited:
Physics news on Phys.org
What can we presume that you already know about cyclic groups? Do you know the theorem that if d divides n, then G has a subgroup of order d? If not, then I would start by proving that. Your result will follow immediately from that theorem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top