# Theorem about cosets - what am I doing wrong

1. Jul 18, 2010

### murmillo

1. The problem statement, all variables and given/known
I am reading about cosets and am stuck on this proposition. Let H be a subgroup of a group G. If aH and bH have an element in common, then they are equal.

But let the group be Z with addition as the law of composition. Let H be 5Z, the set of all multiples of 5. Then 4H and 12H are cosets, and 60 is an element of 4H (since 60=4(15) and 15 is in H). But 60 is also an element of 12H (since 60=12(5) and 5 is in H). But according to the proposition, 4H and 12H must then be equal, but they are not, since 20 is an element of 4H but not 12H. What am I doing wrong?

2. Relevant equations

aH = {ah where h is an element of H)

3. The attempt at a solution

I thought that part of the problem has to do with my using multiplication, but I don't think that's right. I've read through the section on cosets several times but I still can't figure out what I'm doing wrong. I'm sure it's something obvious.

2. Jul 18, 2010

### Tedjn

Re: Cosets

Here's a case when notation helps. The law of composition is addition, so cosets are actually 4+H and 12+H.

3. Jul 18, 2010

### murmillo

Re: Cosets

Right, but now I'm confused as to something I've read before:

"Let us denote the subset of Z consisting of all multiples of a given integer b by bZ:
bZ = {n in Z | n = bk for some k in Z}."

When the author writes n = bk, he does not mean b+k, even when the law of composition is addition. In this case bk = b+b+b+b+... (k terms).

This is why I thought that in my example, even if the law of composition of Z is additive, by 4H is meant {n in H } | n = 4h for some h in H} where 4h means h+h+h+h, i.e. 4 times h, not 4+h.

So I'm still confused.

Oh, Artin is so not the best book for self-studying algebra...

4. Jul 18, 2010

### hunt_mat

Re: Cosets

There is an equivalence relation $$a\sim b <=>a\in bH$$ equivalence relations form partitions. there are other details but you will have fun writing the words down.

Mat

5. Jul 18, 2010

### murmillo

Re: Cosets

hunt mat: I think I'm having more trouble with the notation mentioned in the last post. Once I understand the confusing notation, the proof should be easy. But I'm still having trouble with notation.

6. Jul 18, 2010

### vela

Staff Emeritus
Re: Cosets

You can't change the subgroup midstream. Say you have the subgroups H1=5Z and H2=10Z. The proposition says that if H1+m and H1+n have an element in common, they are equal. Similarly, if H2+m and H2+n have an element in common, they are equal. But the proposition doesn't say anything about how the cosets of H1 and the cosets of H2 are related even if they have elements in common.

7. Jul 18, 2010

### vela

Staff Emeritus
Re: Cosets

Say H=5Z. Then you have H={..., -10, -5, 0, 5, 10, ...} and

H+1 = {..., -9, -4, 1, 6, 11,...}
H+2 = {..., -8, -3, 2, 7, 12,...}
H+3 = {..., -7, -2, 3, 8, 13,...}
H+4 = {..., -6, -1, 4, 9, 14,...}
H+5 = {..., -5, 0, 5, 10, 15,...} = H

Does that make sense?

8. Jul 18, 2010

### murmillo

Re: Cosets

What you're saying makes sense. I think the only thing I don't understand is why addition is used instead of multiplication. Let me explain.

In my original post I was not trying to refer to 2 different subgroups. I was referring to only one, H = 5Z. I think my confusion is in this definition:

"A left coset is a subset of the form aH = {ah | h is in H}."

You claim that if H is 5Z, then by aH is meant H+a = {a+h | h is in H}. Since the law of composition on Z is addition. Correct?

But I'm confused because in previous sections he wrote:

"Let us denote the subset of Z consisting of all multiples of a given integer b by bZ:
bZ = {n in Z | n = bk for some k in Z}."

When the author writes n = bk, he does not mean b+k, even when the law of composition is addition. In this case bk = b+b+b+b+... (k terms).

Why does bk mean b+b+b+...(k terms) instead of b+k in this previous section? Why does it mean b+k when referring to cosets?

9. Jul 18, 2010

### hunt_mat

Re: Cosets

Getting an abstract answer is sometimes easier to digest.

10. Jul 18, 2010

### boboYO

Re: Cosets

Because when talking about abstract groups "ab" always means "the law of composition applied to the pair of elements (a,b) ".

When working in Z he uses the normal notation for multiplication because you're used to it.

11. Jul 18, 2010

### vela

Staff Emeritus
Re: Cosets

It's just an unfortunate overloading of notation. Though bZ and aH appear similar notation-wise, they mean different things. The definition of bZ has nothing to do with what aH means.

12. Jul 18, 2010

### murmillo

Re: Cosets

OK, I think I understand. I was mixing up two different notations. By bZ is mean the subset of Z consisting of all multiples of a given integer b by bZ:
bZ = {n in Z s.t. n=bk for some k in Z} and bk meant do b+b+b... k times. That's completely different by what was meant by aH = {ah s.t. h is in H}, since by ah is meant a*h where * is the law of composition on G. Right?

13. Jul 18, 2010

### vela

Staff Emeritus
Re: Cosets

Right.

14. Jul 18, 2010

### murmillo

Re: Cosets

Oh wait. But he also says, "The subgroup H is itself a coset, because H = 1H." By that he doesn't mean 1+H. What's the explanation?

edit: never mind. Z is completely out of the picture.

By 1 he meant the identity! Oh goodness I should've used another textbook.

Now I can finally go on! Thanks, everyone!