# Abstract Algebra - Cosets

1. Nov 2, 2009

### vwishndaetr

Question:

Prove the following properties of cosets.

Given:

Let H be a subgroup and let a and b be elements of G.

$$H\leq\ G$$

Statement:

$$aH=bH \ if\ and\ only\ if\ a^{-1}b\ \epsilon\ H$$

The statement is what I have to prove.

My issue is I don't know how to start off the problem. When I first looked at the statement. I wanted to say that it is only true when a=b. But there is not talk of the groups being abelian. So what I thought was a start to some thinking, did not take me very far.

2. Nov 3, 2009

### HallsofIvy

Staff Emeritus
First of all, you need to say "H is a subgroup of G" or the problem doesn't make sense. Now, what is the definition of "aH" and "bH"? Your remark that "I wanted to say that it is only true when a=b" indicates that you are not clear on that definition.

3. Nov 3, 2009

### vwishndaetr

$$H\leq\ G$$

That means H is a subgroup of G. So clearly stated.

a and b are elements of G, and "aH" is a left coset with "a" and "bH" is a left coset with "b."

4. Nov 3, 2009

### foxjwill

But then what is the definition of a "left coset with a"?

5. Nov 3, 2009

### vwishndaetr

$$aH= \{a*h\ |\ h\ \epsilon\ H }\$$

H is a subset. Where h is an element of the set H.

6. Nov 3, 2009

### foxjwill

Right. So, what does it mean for the set aH to be equal to the set bH?

Oh, and you can type the "in" symbol using a "\in" and the "not in" symbol using "\notin". I think it formats it better that way.

7. Nov 5, 2009

### Quantumpencil

Remember that aH and bH partition the group (since they are equivalence classes defined by a=b if b = ah for some h in H), if you had any two cosets which had a nonzero intersection, then the transitivity of equivalence classes would automatically make the two equal. So to begin with, you know that if c is in aH, then c equals ah for h in H. if d is in bH, it equals d = bh', for h' in H. Your condition for the equivalence classes being equal is that there exists an element c in aH, such that c=bh' for some h' in H.

8. Nov 5, 2009

### vwishndaetr

Yup thanks. I had one of my professors explain it to me. Forgot to post up.

Thanks though! :)