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Abstract algebra module

by Shackleford
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Shackleford
#1
Feb27-12, 03:49 PM
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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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Dick
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Feb27-12, 03:57 PM
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Quote Quote by Shackleford View Post
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?
Shackleford
#3
Feb27-12, 04:05 PM
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Quote Quote by Dick View Post
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>?

Dick
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Feb27-12, 04:10 PM
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Abstract algebra module

Quote Quote by Shackleford View Post
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.
Shackleford
#5
Feb27-12, 04:18 PM
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Quote Quote by dick View Post
no, i'm asking what n is. It's should be a subgroup with three permutations in it.
(1), (2), (3)?
Dick
#6
Feb27-12, 04:21 PM
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Quote Quote by Shackleford View Post
(1), (2), (3)?
Are you using cycle notation? Aren't those all the identity permutation?
Shackleford
#7
Feb27-12, 04:27 PM
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Quote Quote by Dick View Post
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.
Dick
#8
Feb27-12, 04:38 PM
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Quote Quote by Shackleford View Post
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?
Shackleford
#9
Feb27-12, 04:44 PM
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Quote Quote by Dick View Post
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)?
Dick
#10
Feb27-12, 04:46 PM
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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?
Shackleford
#11
Feb27-12, 05:10 PM
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Quote Quote by Dick View Post
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?
Dick
#12
Feb27-12, 05:12 PM
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Quote Quote by Shackleford View Post
Oh, shoot. We're taking the generation <(123)>, right?
Sure are.
Shackleford
#13
Feb27-12, 05:17 PM
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Quote Quote by Dick View Post
Sure are.
Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.
Dick
#14
Feb27-12, 06:51 PM
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Quote Quote by Shackleford View Post
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.
Shackleford
#15
Feb27-12, 07:02 PM
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Quote Quote by Dick View Post
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.
Dick
#16
Feb27-12, 07:12 PM
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Quote Quote by Shackleford View Post
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 Quote by Shackleford View Post
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.
Shackleford
#17
Feb27-12, 07:20 PM
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Quote Quote by Dick View Post
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.
Dick
#18
Feb27-12, 07:24 PM
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Quote Quote by Shackleford View Post
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.


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