# Abstract algebra module

by Shackleford
Tags: abstract, algebra, module
 P: 1,532 http://i111.photobucket.com/albums/n...8/untitled.jpg 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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P: 24,454
 Quote by Shackleford http://i111.photobucket.com/albums/n...8/untitled.jpg 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?
P: 1,532
 Quote by Dick 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>?

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## Abstract algebra module

 Quote by Shackleford 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.
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 Quote by dick no, i'm asking what n is. It's should be a subgroup with three permutations in it.
(1), (2), (3)?
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 Quote by Shackleford (1), (2), (3)?
Are you using cycle notation? Aren't those all the identity permutation?
P: 1,532
 Quote by Dick 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.
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P: 24,454
 Quote by Shackleford 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?
P: 1,532
 Quote by Dick 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?
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 Quote by Shackleford 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?
P: 1,532
 Quote by Dick 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?
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P: 24,454
 Quote by Shackleford Oh, shoot. We're taking the generation <(123)>, right?
Sure are.
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 Quote by Dick Sure are.
Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.
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 Quote by Shackleford 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.
P: 1,532
 Quote by Dick I'm not sure what you are saying. It's the subgroup generated by (123). For any element 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.
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 Quote by Shackleford 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.

 Quote by Shackleford 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.
P: 1,532
 Quote by Dick 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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