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Abstract algebra module

  • #1
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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
Dick
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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
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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
Dick
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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
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no, i'm asking what n is. It's should be a subgroup with three permutations in it.
(1), (2), (3)?
 
  • #6
Dick
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(1), (2), (3)?
Are you using cycle notation? Aren't those all the identity permutation?
 
  • #7
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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
Dick
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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
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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
Dick
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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
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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
Dick
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Oh, shoot. We're taking the generation <(123)>, right?
Sure are.
 
  • #13
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Sure are.
Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.
 
  • #14
Dick
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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
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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
Dick
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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.

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
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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
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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.
 

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