Modern Algebra Factor Groups question

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Homework Statement


If K is normal in G and |g| = n for some g in G, show that the order of Kg in G/K divides n.

Homework Equations


None

The Attempt at a Solution


Okay so I feel like I have a solution but I don't use all the information given so I'm trying to find holes in it...

g^n = 1

K = Kg^n = (Kg)^n. So it seems If we take Kg to the nth power we get it's identity. This means that the order of Kg must divide n.

What am I missing here?
 
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PsychonautQQ said:

Homework Statement


If K is normal in G and |g| = n for some g in G, show that the order of Kg in G/K divides n.

Homework Equations


None

The Attempt at a Solution


Okay so I feel like I have a solution but I don't use all the information given so I'm trying to find holes in it...

g^n = 1

K = Kg^n = (Kg)^n. So it seems If we take Kg to the nth power we get it's identity. This means that the order of Kg must divide n.

What am I missing here?

If you mean you aren't using that K is normal in G, you need that fact to justify saying (Kg)^n=Kg^n, or even that G/K is a group. Better review why.
 
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...
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