Group Theory: Proving Subgroup of Elements of Order 2 and e

HuaYongLi
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Homework Statement


G is a commutative group, prove that the elements of order 2 and the identity element e form a subgroup.


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The Attempt at a Solution


I don't know where to even begin.
 
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Given any subset H \subset G, how would you attempt to prove that it is a subgroup of G? What properties of H would you attempt to verify?
 
Well I guess associativity, unique inverse and identity element are all trivial.
What I'm having trouble with is closure, proving that for any elements a and b in the group, ab is also in the group.
 
You need to prove that if a and b are elements of order 2 (i.e. a^{2} = b^{2} = e), then so is c = a b. You need to evaluate c^{2} and use the commutativity of the group.
 
Thank You
 
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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