Theorem about cosets - what am I doing wrong

In summary: So then a left coset is a subset of the form aH = {ah s.t. h is in H}. Since we're dealing with addition as the law of composition, ah means a+a+a... h times. And since h is in H, h must be a multiple of b, and h=bk for some k in Z. So then ah means a+b+b+... b times. Is that correct?Yes, that's correct. The notation can be confusing, but it's important to keep in mind the context in which it's being used. In the case of cosets, the notation follows the law of composition, which in this case is addition.
  • #1
murmillo
118
0
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?

Homework Equations



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

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.
 
Physics news on Phys.org
  • #2


Here's a case when notation helps. The law of composition is addition, so cosets are actually 4+H and 12+H.
 
  • #3


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


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

Mat
 
  • #5


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


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


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


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


Getting an abstract answer is sometimes easier to digest.
 
  • #10


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


murmillo said:
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?
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


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


Right.
 
  • #14


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!
 

Related to Theorem about cosets - what am I doing wrong

1. What is a coset in relation to group theory?

A coset is a subset of a group that is formed by multiplying all elements of a subgroup by a single element of the group.

2. How is the Theorem about cosets applied in group theory?

The Theorem about cosets is used to prove the Lagrange's Theorem, which states that the order of a subgroup divides the order of the group. It is also used to show the relationship between the index of a subgroup and the number of distinct cosets.

3. What are the conditions for the Theorem about cosets to hold?

Theorem about cosets holds when the subgroup is a normal subgroup of the group. This means that the subgroup is invariant under conjugation by any element of the group.

4. Can the Theorem about cosets be extended to other algebraic structures?

Yes, the Theorem about cosets can be extended to other algebraic structures such as rings, where the cosets are formed by multiplying all elements of a subring by a single element of the ring.

5. What are common mistakes while applying the Theorem about cosets?

Common mistakes while applying the Theorem about cosets include forgetting to check if the subgroup is normal, using an incorrect index for the subgroup, and not considering the possibility of overlapping cosets.

Similar threads

  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
975
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
6
Views
845
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
4K
Back
Top