Does a Factor Group Element's Order Imply the Same Order in the Original Group?

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Homework Statement


Let G be a finite group and let K be normal to G. If the factor group G/K has an element of order n, show that G has an element of order n.

Homework Equations


None

The Attempt at a Solution


Lets say Kg is the element in G/K with order n.
That means:
(Kg)^n = K

and from properties of factor groups we know:
(Kg)^n = Kg^n
so Kg^n = K
hence g^n = 1
if g^n = 1 then it must be in G, because G has the identity (1).

Is this correct thinking?
 
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PsychonautQQ said:

Homework Statement


Let G be a finite group and let K be normal to G. If the factor group G/K has an element of order n, show that G has an element of order n.

Homework Equations


None

The Attempt at a Solution


Lets say Kg is the element in G/K with order n.
That means:
(Kg)^n = K

and from properties of factor groups we know:
(Kg)^n = Kg^n
so Kg^n = K
hence g^n = 1
if g^n = 1 then it must be in G, because G has the identity (1).

Is this correct thinking?

No. Haven't you posted this before? Kg^n=K only tells you that g^n is in K.
 
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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