Abstract Algebra- homomorphisms and Isomorphisms, proving not cyclic

eruth
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1. Suppose that H and K are distinct subgroups of G of index 2. Prove that H intersect K is a normal subgroup of G of index 4 and that G/(H intersect K) is not cyclic.


2. Homework Equations - the back of my book says to use the Second Isomorphism Theorem for the first part which is... If K is a subgroup of G and N is a normal subgroup of G, then K/(K intersect N) is isomorphic to KN/N

3. The Attempt at a Solution - I know that any subgroup with order 2 is normal and that the intersection of 2 normal subgroups is normal. I just wasn't sure how to show it was index 4. Also, I wasn't what to do for the second part.
 
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Hint: what do you know about the product HK?
 
Alright, I think I got the first part of the proof. Now to show that the quotient group isn't cyclic, the back of my book says to observe that it has 2 subgroups of order 2... I'm not seeing what these 2 subgroups are, maybe just because quotients groups confuse me.
 
eruth said:
Alright, I think I got the first part of the proof. Now to show that the quotient group isn't cyclic, the back of my book says to observe that it has 2 subgroups of order 2... I'm not seeing what these 2 subgroups are, maybe just because quotients groups confuse me.

Define L=HnK. You should know that the quotient group is defined by the products of cosets. Take h to be an element of H that's not in L. What coset could (hL)(hL) be?
 
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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