Question of lagrange theorem converse.

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Homework Statement


Let G be an abelian group. Suppose p divides ord(G) where p is prime no. Prove G has a subgroup of order p.


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lagrange theorem converse

The Attempt at a Solution


i know the converse is lagrange theorem and easy and this is not the case.
I know abelian has something to do with this and prime no is also something special. I also believe the subgroup of order p is cyclic?
thx
 
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Yes, any group of prime order is cyclic.

The theorem you are being asked to prove is called Cauchy's theorem, and it is true even if G is not abelian. There are numerous ways to prove it, depending on what results you already know. Maybe the most straightforward way is induction on |G|.
 
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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