Very Elementary Group Theory Problem

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Hi. I'm new to Group Theory and wanted to see if I had the right train of thought for this problem.

Homework Statement



Let S be a set with an associative law of composition and with an identity element. Prove that the subset of S consisting of invertible elements is a group.

Homework Equations



A group is a set G together with a law of composition that is associative and has an identity element, and such that every element of G has an inverse.

The Attempt at a Solution



I'll denote my subset of S by G.

I know G has the identity element of S as the identity element is invertible.

The associativity for the law of composition is inherited and the existence of inverses we get for free. All that I'm left with is verifying that G has closure. That is for a,b in G, ab is in G.

It's sufficient to show ab is invertible.

But, (ab)(b^{-1}a^{-1})=e

where e is the identity element. So ab is invertible and thus is an element of G and G is a group.
 
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Not sure why my tex code isn't working...
 
Use [ tex] and [ /tex] (without the spaces). Don't use [ latex] and [/ latex].
 
I think works too. The problem is the wrong slash.
 
Thanks. I think I got it now (the tex code that is).
 
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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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