Abstract Algebra Proof question

Braka
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Homework Statement


The question is:
Let A be a subset of Sn that contains all permutations alpha such that alpha can be written as a product of an even number of transpositions. Prove that A is a group with product of permutations.

I understand what I need to do to prove it, but I am not sure how to start it. Do I use:
Let alpha=(a1a2a3...an) and
beta=(b1b2b3...bn) and try to find closure,

alphabeta=(a1a2a3...an)(b1b2b3...bn)=(a1a2)(a1a3)...(a1an-1)(a1an)(b1b2)(b1b3)...(b1bn-1)(b1bn),

or am I going about it the wrong way?


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The Attempt at a Solution

 
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What is a group? There are some axioms A must satisfy in order for it to be called one. For instance, the product of two things in A must also lie in A. Translating this to the situation at hand: if two permutations can be written as a product of an even number of transpositions, can their product be as well? Obviously yes! Now check the other axioms similarly.
 
Where did you use the fact that they can be written as an even number of transpositions?
 
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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