Abstract Algebra- homomorphisms and Isomorphisms, proving not cyclic

In summary, the conversation discusses proving that the intersection of two distinct subgroups H and K of a group G of index 2 is a normal subgroup of index 4. The second part of the conversation focuses on proving that the quotient group G/(H intersect K) is not cyclic. The homework equations mention using the Second Isomorphism Theorem and the attempt at a solution involves showing that the quotient group has two subgroups of order 2. The solution is to define L as H intersect K and to consider the products of cosets.
  • #1
eruth
3
0
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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  • #2
Hint: what do you know about the product HK?
 
  • #3
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.
 
  • #4
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?
 

Related to Abstract Algebra- homomorphisms and Isomorphisms, proving not cyclic

1. What is Abstract Algebra?

Abstract Algebra is a branch of mathematics that deals with the study of algebraic structures, such as groups, rings, and fields, and their properties. It is an abstract approach to algebra, where the focus is on the underlying structure rather than specific numbers or operations.

2. What is a homomorphism in Abstract Algebra?

A homomorphism is a function that preserves the algebraic structure of a mathematical object. In the context of Abstract Algebra, a homomorphism is a function between two algebraic structures that preserves the operations and properties of those structures.

3. What is an isomorphism in Abstract Algebra?

An isomorphism is a bijective homomorphism, which means it is both one-to-one and onto. In Abstract Algebra, an isomorphism is a function between two algebraic structures that preserves both the operations and the structure of those structures.

4. How do you prove that a group is not cyclic?

To prove that a group is not cyclic, you need to show that there is no element in the group that generates the entire group when combined with the group's operation. This can be done by showing that there are elements in the group that cannot be expressed as a power of a single element.

5. How are homomorphisms and isomorphisms related?

Homomorphisms and isomorphisms are both types of functions between algebraic structures that preserve the structure and operations of those structures. The main difference is that isomorphisms are bijective, while homomorphisms do not necessarily have to be. In other words, an isomorphism is a special type of homomorphism.

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