View Full Version : Abstract Algebra Help
Shaun.P
Mar 12, 2009, 12:43 PM
Hi all,
I was wondering if any maths people could help me with a question that I am stuck at that is really bothering me. [See picture below].
I know the order of the set S15 is 15! (15 factorial).
I also know how to prove the set H is a supgroup of S15.
Can anyone tell me though, what the order of the subgroup is? Is it the same as the order of the set S15?
Any help would be greatly appreciated.
RHutch
Apr 10, 2009, 07:55 AM
I have not done abstract algebra for several years, so there may be a better solution, but I think this is correct.
In this case it is easy enough to directly compute the powers of alpha. Since there are only 6 permutations of three numbers (1, 14, 15), there are no more than 6 powers of alpha.
All numbers except 1, 14, and 15 are fixed.
Consider alpha squared: (1 14 15)(1 14 15).
1 -> 14 -> 15
14 -> 15 -> 1
15 -> 1 -> 14
So (1 14 15)(1 14 15) = (1 15 14)
Then alpha cubed is (1 14 15)(1 15 14) and
1 -> 15 -> 1
15 -> 14 -> 15
14 -> 1 -> 14
So alpha cubed is the identity.
So H = {alpha, alpha^2, and alpha^3}. The order of the subgroup H is 3.
Shaun.P
May 26, 2009, 05:06 PM
I never thanked you for your help.
Thank you.
I somehow managed to unsubscribe from this thread by mistake, and never noticed there was a reply.
acearchie
May 26, 2009, 05:16 PM
So is this a form of statistics or something completely different altogether?
dukebound85
May 26, 2009, 05:19 PM
So is this a form of statistics or something completely different altogether?
i covered it when i took "linear algebra and matrices" in college
and yea, i forgot how to do that crap lol
Shaun.P
May 26, 2009, 05:25 PM
So is this a form of statistics or something completely different altogether?
This is something completely different.
The actual name of the course at my Uni is Algebraic Structures.
This is the course plan:
Mappings
Composition
Binary operations
Composition as a binary operation
Examples of groups
Permutations
Subgroups
Equivalence relations
Congruence. The division algorithm
Integers modulo n
Euclidean algorithm
Definition and examples of rings
Ideals
Integral domains. Subrings
Fields
Isomorphism. Characteristic
Ordered integral domains
The integers
Field of quotients
Field of rational numbers
Extensions of fields
Definition and elementary properties of polynomials
The division algorithm
Factorisation of polynomials
Unique factorisation domains
Simple extensions
Finite fields
Partially ordered sets
Lattices
Boolean algebras
jecapaga
May 26, 2009, 05:27 PM
Abstract gets his very own thread request.
Frisco
May 26, 2009, 06:48 PM
Wow Abstract Algebra is definitely a few steps above Calculus, which I was lucky to pass.
emmawu
May 26, 2009, 07:45 PM
The order of the subgroup H must be a divisor of 15! AND equals the minimal n such that alpha^n = alpha^n+1.
Mambochicken (emmawu's so).
RHutch
May 27, 2009, 07:25 AM
The order of the subgroup H must be a divisor of 15! AND equals the minimal n such that alpha^n = alpha^n+1.
Mambochicken (emmawu's so).
I think you made a typo or some other error. Don't you mean that it is the minimal n for which alpha = alpha^(n + 1)? As you have written it, you would have consecutive powers of alpha equal to each other: n and n + 1.
magamo
May 27, 2009, 08:06 AM
I think you made a typo or some other error. Don't you mean that it is the minimal n for which alpha = alpha^(n + 1)? As you have written it, you would have consecutive powers of alpha equal to each other: n and n + 1.
Well, I don't want to be an ass, but since you corrected the error, I think the order is the minimal positive integer n >= 2 s.t. alpha = alpha^(n + 1) in this case. Otherwise, a smartass might say alpha = alpha^1 and hence the order has to be 0. An even more annoying ass could argue alpha = alpha^(-2) = alpha^(-5) =... because H happened to be the prime field of order 3.
RHutch
May 27, 2009, 08:12 AM
Well, I don't want to be an ass, but since you corrected the error, I think the order is the minimal positive integer n >= 2 s.t. alpha = alpha^(n + 1) in this case. Otherwise, a smartass might say alpha = alpha^1 and hence the order has to be 0. An even more annoying ass could argue alpha = alpha^(-2) = alpha^(-5) =... because H happened to be the prime field of order 3.
I'm confused; are you saying that I should not have commented on the powers?
magamo
May 27, 2009, 08:51 AM
I'm confused; are you saying that I should not have commented on the powers?
No. Sorry, I expressed myself poorly... You're absolutely right about "alpha^n = alpha^n+1" is a typo or an error. I just wanted to make it complete because you kindly corrected part of emmawu's post.
P-Worm
May 27, 2009, 08:54 AM
Guys, he specifically asked for Abstract algebra help. Why are we answering the question when it is intended for him? Where is Abstract, anyway?
P-Worm
Shaun.P
May 27, 2009, 10:04 AM
Abstract gets his very own thread request.
Guys, he specifically asked for Abstract algebra help. Why are we answering the question when it is intended for him? Where is Abstract, anyway?
P-Worm
Abstract algebra is the name of the branch of maths this is.
jecapaga
May 27, 2009, 12:27 PM
Guys, he specifically asked for Abstract algebra help. Why are we answering the question when it is intended for him? Where is Abstract, anyway?
P-Worm
We will have to wait for his eventual arrival.
emmawu
May 27, 2009, 09:07 PM
Hi,
Mambochicken (a mathematics professor) admitted he made a typographical error. Maybe he should be taken to the parking lot and shot, huh? Or would that be too harsh?
P-Worm
May 28, 2009, 02:34 PM
Abstract algebra is the name of the branch of maths this is.
I know. I guess the joke didn't come out so well.
P-Worm
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