Understanding Left Cosets in Abstract Algebra

In summary: N is just a subset of S3, right.In summary, the conversation discusses the set G/N, which is the set of all left cosets of N in G. The notation and concept of left cosets is explored through examples using permutations. The subgroup N is identified as the subgroup generated by the permutation (123) and contains the elements (123), (321), and the identity permutation. It is noted that each subgroup must have an order that is a multiple of the order of G. The process of finding left cosets by operating on the subgroup N with elements of G is explained. The conversation also briefly touches on a division algorithm problem involving polynomials and arithmetic mod 3.
  • #1
Shackleford
1,656
2
http://i111.photobucket.com/albums/n149/camarolt4z28/untitled.jpg [Broken]

G/N is the set of all left cosets of N in G.

I don't understand the notation.

a) The permutations are (1,2), (2,3), (3,1). What are the left cosets - <1>, <2>, <3>? That doesn't make sense with permutations.

b) I have no idea. N = mZ is simply all integer multiples of m. How do I find the left cosets?
 
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  • #2
Shackleford said:
http://i111.photobucket.com/albums/n149/camarolt4z28/untitled.jpg [Broken]

G/N is the set of all left cosets of N in G.

I don't understand the notation.

a) The permutations are (1,2), (2,3), (3,1). What are the left cosets - <1>, <2>, <3>? That doesn't make sense with permutations.

b) I have no idea. N = mZ is simply all integer multiples of m. How do I find the left cosets?

You find left cosets by operating on the subgroup on the left by members of your group. For the first one beta isn't three permutations, it's only one. In cycle notation it's (123). And the group N is the subgroup generated by beta. What's that?
 
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  • #3
Dick said:
You find left cosets by operating on the subgroup on the left by members of your group. For the first one beta isn't three permutations, it's only one. In cycle notation it's (123). And the group N is the subgroup generated by beta. What's that?

It's the same thing. (123) is simply more concise. The left cosets are generated by gN, i.e. each element of g is composed with each element of N. Are you saying it's <123>?
 
  • #4
Shackleford said:
It's the same thing. (123) is simply more concise. The left cosets are generated by gN, i.e. each element of g is composed with each element of N. Are you saying it's <123>?

No, I'm asking what N is. It's should be a subgroup with three permutations in it.
 
  • #5
dick said:
no, I'm asking what n is. It's should be a subgroup with three permutations in it.

(1), (2), (3)?
 
  • #6
Shackleford said:
(1), (2), (3)?

Are you using cycle notation? Aren't those all the identity permutation?
 
  • #7
Dick said:
Are you using cycle notation? Aren't those all the identity permutation?

Yes. For these two problems, I don't see how you can take the composition gN.
 
  • #8
Shackleford said:
Yes. For these two problems, I don't see how you can take the composition gN.

For the first problem, you haven't figured out what N is yet. It's the subgroup generated by the permutation (123). That's a permutation of order three and it's one of the permutations in N. What are the other two?
 
  • #9
Dick said:
For the first problem, you haven't figured out what N is yet. It's the subgroup generated by the permutation (123). That's a permutation of order three and it's one of the permutations in N. What are the other two?

What about (1,3) and (1,2)?
 
  • #10
Shackleford said:
What about (1,3) and (1,2)?

Are you just guessing? The product of (123) and (123) must be in the subgroup, mustn't it?
 
  • #11
Dick said:
Are you just guessing? The product of (123) and (123) must be in the subgroup, mustn't it?

Oh, shoot. We're taking the generation <(123)>, right?
 
  • #12
Shackleford said:
Oh, shoot. We're taking the generation <(123)>, right?

Sure are.
 
  • #13
Dick said:
Sure are.

Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.
 
  • #14
Shackleford said:
Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.

I'm not sure what you are saying. It's the subgroup generated by (123). For any element a, <a> is just all of the powers of a and its inverse.
 
  • #15
Dick said:
I'm not sure what you are saying. It's the subgroup generated by (123). For any element a, <a> is just all of the powers of a and its inverse.

The permutation beta is fully defined by (123). You're sticking that in the generator. I need to to look at the (123)n until I find all the distinct left cosets.

This abstract algebra modulus is part of Survey for Undergrad Math. It's a preparatory class for the math field test. I took Abstract last summer and did well, but some of this stuff we didn't cover.

Let f = 2x + 2 and g = x3 + 2x + 2.

Find q,r in Z3 such that g = fq + r.

I just did the problem where the field is R, but I haven't dealt with a polynomial defined over a congruence class.
 
  • #16
Shackleford said:
The permutation beta is fully defined by (123). You're sticking that in the generator. I need to to look at the (123)n until I find all the distinct left cosets.

You are still thinking about this wrong. Finding (123)^n is just going to give you one coset. N itself. There's another one.

Shackleford said:
This abstract algebra modulus is part of Survey for Undergrad Math. It's a preparatory class for the math field test. I took Abstract last summer and did well, but some of this stuff we didn't cover.



I just did the problem where the field is R, but I haven't dealt with a polynomial defined over a congruence class.

I would just do the division algorithm and substitute arithmetic mod 3 for real arithmetic.
 
  • #17
Dick said:
You are still thinking about this wrong. Finding (123)^n is just going to give you one coset. N itself. There's another one.
I would just do the division algorithm and substitute arithmetic mod 3 for real arithmetic.

How about the inverse - (321)?

Okay. Let me try that for the division algorithm problem.
 
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  • #18
Shackleford said:
How about the inverse - (231)?

Okay. Let me try that for the division algorithm problem.

Ok! Yes, N={e,(123),(321)} where e is the identity permutation. That's the coset eN. Now there's another coset, pick another element g in S3 that's not in N and find gN.
 
  • #19
Dick said:
Ok! Yes, N={e,(123),(321)} where e is the identity permutation. That's the coset eN. Now there's another coset, pick another element g in S3 that's not in N and find gN.

How did you know the subgroup N has only three elements - because of the order of S3? Each subgroup must have an order that's a multiple of the order of S3. So of course each subgroup has to have the identity and an inverse of a given element.

If S3 is the set of all permutations, then I can pick any permutation that's not in N. (231)?
 
  • #20
Shackleford said:
How did you know the subgroup N has only three elements - because of the order of S3? Each subgroup must have an order that's a multiple of the order of S3. So of course each subgroup has to have the identity and an inverse of a given element.

If S3 is the set of all permutations, then I can pick any permutation that's not in N. (231)?

I knew it because (123) is a 3-cycle, (123)^3=e. (123)^2=(321). And (231) IS in N. It's the same as (123).
 
  • #21
Dick said:
I knew it because (123) is a 3-cycle, (123)^3=e. (123)^2=(321). And (231) IS in N. It's the same as (123).

Oops. You're right. I suppose I could pick any of (1,2), (1,3), (2,3), etc.
 
  • #22
Shackleford said:
Oops. You're right. I suppose I could pick any of (1,2), (1,3), (2,3), etc.

That would be a great idea.
 
  • #23
Dick said:
That would be a great idea.

Those are simply the rest of the left cosets.

For the other problem, I'm getting 5x2-5x + 2. It's not multiplying out exactly right. In the long division, I'm getting 3x near the end. If I treat that as 0x or 3x, the quotient and remainder still don't give the g.
 
  • #24
Shackleford said:
Those are simply the rest of the left cosets.

For the other problem, I'm getting 5x2-5x + 2. It's not multiplying out exactly right. In the long division, I'm getting 3x near the end. If I treat that as 0x or 3x, the quotient and remainder still don't give the g.

There are only two cosets of H. And 5 isn't in Z3. How did you get that? Can you sort of show what you did? At least what is x^3 divided by 2x in mod 3?
 
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  • #25
Dick said:
There are only two cosets of H. And 5 isn't in Z3. How did you get that?

Those are the other possible permutations of S3. They should be in the other one coset, right?

(x3 + 2x + 2) / (2x+2)

In getting the first term, I multiplied 5*2x = 10x = x.
 
  • #26
Shackleford said:
Those are the other possible permutations of S3. They should be in the other one coset, right?

(x3 + 2x + 2) / (2x+2)

In getting the first term, I multiplied 5*2x = 10x = x.

Right, the cosets are {e,(123),(132)} and {(12),(23),(13)}. And for the other one the first term of the quotient is x^3 divided by 2x mod 3. What's that?
 
  • #27
Dick said:
Right, the cosets are {e,(123),(132)} and {(12),(23),(13)}. And for the other one the first term of the quotient is x^3 divided by 2x mod 3. What's that?

I'm not sure what to make of that. Is it 2x congruence modulo 3? 2x - 1 = 3k, k is an integer.
 
  • #28
Shackleford said:
I'm not sure what to make of that. Is it 2x congruence modulo 3? 2x - 1 = 3k, k is an integer.

If you doing it in the real numbers x^3 divided by 2x is (1/2)x^2. If you are doing it mod 3 you want (1 divided by 2 mod 3)*x^2.
 
  • #29
ok, we have a subgroup N = nZ. let's write N (somewhat incompletely) as:

N = nZ = {0,n,-n,2n,-2n,3n,-3n,...}

one possible (left, or right, it really doesn't make any difference because "+" in Z is commutative) coset is:

1+N = 1+nZ = {1,n+1,1-n,2n+1,1-2n,3n+1,1-3n,...}

now, do 1+N and N have any elements in common? if not, then they are distinct cosets. what about k+N and (k+1)+N? are those different?

what might be the smallest positive integer k such that:

k+N = N?

after answering that question, see if you can figure out when k+N and m+N have the same elements. actually DO this for some small value of n (6 works pretty well, most people have the multiples of 6 memorized).
 

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