Proving Normality in Group Theory: Guide and Tips for Proving H is Normal in G

moont14263
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Please can someone help me to prove this
If N is a cyclic normal subgroup of G and H is a subgroup of N then H is normal in G.
 
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You'll need to use the following:

If x in N has order k, then gxg-1 also has order n.

So conjugation won't change the order of the elements, and order of the elements determine which subgroup the element will belong to (in cyclic groups).
 
Could you give me more hints because I am not that much in cyclic groups. Thank you.
 
Show that, if G is abelian and if n is a number, then G(n)=\{g\in G~\vert~g^n=e\} is a subgroup of G.

Then show that, if G is cyclic, then all subgroups of G are of the form G(n).
 
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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