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Non-abelian groups

  • #1
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Homework Statement:

Prove the following matrix is non abelian

Relevant Equations:

matrix below
I know for a group to be abelian a*b=b*a
I tried multiplying the matrix by itself also but I’m not sure what I’m looking for.
picture is below of the matrix
View attachment 255812
 

Answers and Replies

  • #2
Math_QED
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Multiplying the matrix with itself won't be useful. Every matrix commutes with its powers.

You just have to find two matrices in this set that do not commute, i.e. you have to find two matrices ##A,B \in P## such that ##AB \neq BA##. Try matrices with ones and zeros at well-chosen places and you will get such an example rather quickly.
 
  • #3
PeroK
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@Lauren1234 you also have to prove it's a group.
 
  • #4
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@Lauren1234 you also have to prove it's a group.
oh yeah... I’ve proved now it’s non-abelian is there a way to prove it’s a group.
additional information I was given is to Let F = {0,1,...,p−1} be the field of order p (where p is a prime, and we perform arithmetic modulo p)
 
  • #5
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Multiplying the matrix with itself won't be useful. Every matrix commutes with its powers.

You just have to find two matrices in this set that do not commute, i.e. you have to find two matrices ##A,B \in P## such that ##AB \neq BA##. Try matrices with ones and zeros at well-chosen places and you will get such an example rather quickly.
Brilliant and that just shows it’s non abelian? Shown below
 

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  • #6
Math_QED
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oh yeah... I’ve proves now it’s non-abelian is there a way to prove it’s a group.
Depends what you know about group theory. You can check that the group axioms are satisfied.

Another way is to note that the determinant of a matrix in ##P## is non-zero, by the condition defining ##P##. Thus ##P## is a subset of the invertible matrices ##GL_3(\mathbb{F})##, so you can try to prove it is a subgroup of the group ##GL_3(\mathbb{F})##.

The latter approach is the least amount of work.

Brilliant and that just shows it’s non abelian? Shown below
First, this attempt is wrong as your chosen matrix is not in ##P##.

Second, to be really formal and correct, you would need to pick appropriate values of ##a,b,c,d,e,f## such that it is 100% clear that the matrices don't commute.
 
  • #7
PeroK
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Brilliant and that just shows it’s non abelian? Shown below
The matrix you chose is not a member of the set ##P##. Note the condition ##dg -ef \ne 0##.
 
  • #8
PeroK
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@Lauren1234 note that the matrices in ##P## are of the form where you can simplify the multiplication process. Hint: consider the four entries in the bottom right as a 2x2 matrix.

Or, you'll just have to grind through the algebra.
 
  • #9
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The matrix you chose is not a member of the set ##P##. Note the condition ##dg -ef \ne 0##.
got you! Is this correct now I’ve changed it
image.jpg
 
  • #10
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Depends what you know about group theory. You can check that the group axioms are satisfied.

Another way is to note that the determinant of a matrix in ##P## is non-zero, by the condition defining ##P##. Thus ##P## is a subset of the invertible matrices ##GL_3(\mathbb{F})##, so you can try to prove it is a subgroup of the group ##GL_3(\mathbb{F})##.

The latter approach is the least amount of work.



First, this attempt is wrong as your chosen matrix is not in ##P##.

Second, to be really formal and correct, you would need to pick appropriate values of ##a,b,c,d,e,f## such that it is 100% clear that the matrices don't commute.
Thank you I think I’ve got it now and in terms of the group I’m pretty sure I have some knowledge for both methods but the later option seems to be the best option in this case I agree
 
  • #11
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I think that the commutativity property of groups in abstract algebra should not be named after a person, and that similarly, 'algebra' should not, as it is, be named after a person, but should be, instead, named after something of its characteristics, wherefore, I propose 'stitution', without hope that such a terminologic substitution will advance very far.
 
  • #12
PeroK
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Thank you I think I’ve got it now and in terms of the group I’m pretty sure I have some knowledge for both methods but the later option seems to be the best option in this case I agree
Okay, but you do realise that ##P## is the group and not just that one matrix?
 
  • #13
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  • #14
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  • #15
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Okay, but you do realise that ##P## is the group and not just that one matrix?
Oh is it I didn’t no so I need to consider the whole group when doing the group part? Not just the matrix
 
  • #16
PeroK
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Oh is it I didn’t no so I need to consider the whole group when doing the group part? Not just the matrix
The matrix in the defintion of ##P## is giving you the general form of all matrices in the group ##P##.
 
  • #17
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It’s very hard to use it on an iPad sorry 😂
Oh, spoo (oops backwards) -- I didn't know about your affliction. :oops:
 
  • #18
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The matrix in the defintion of ##P## is giving you you the general form of all matrices in the group ##P##.
Right I got you so the first thing I need to do is see if the determinate of the matrix is non zero
 
  • #19
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Also please note, as long as non-Abelian things are being discussed on a Physics Forum, that the non-Abelian ##\text {SU(3)}## homology group is fundamental in quantum chromodynamics (QCD) theory.
 
  • #20
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The matrix in the defintion of ##P## is giving you the general form of all matrices in the group ##P##.
I’ve managed to complete it. Would you mind just telling me where I should start for the next part(I don’t have any notes regarding this)
06B6B2BA-01C7-4FB5-9A1C-F4CE272A45A2.jpeg
 
  • #21
PeroK
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I’ve managed to complete it. Would you mind just telling me where I should start for the next part(I don’t have any notes regarding this)View attachment 255818
To be honest, I think you have conceptual difficulties before you get to group homomorphisms and kernels. You need some serious work and help on the basics of abstract algebra.
 
  • #22
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Ok thanks anyways it’s homework due this week so I don’t really have time to work on it hence why I haven’t but thank you
 
  • #23
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I’ve managed to complete it. Would you mind just telling me where I should start for the next part(I don’t have any notes regarding this)View attachment 255818
You must show an attempt. Definitely since this is homework! I can provide you the definitions though.

To show it is a group morphism, you must show that for all ##A,B \in P## we have
$$\varphi(AB) = \varphi(A) \varphi(B)$$
By definition ##\ker \varphi= \{A \in P: \varphi(A) = I_2 \}## and you are asked to give a nicer description of this set.

Checking that ##\varphi## is onto should be trivial.
 
  • #24
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Hey @Lauren1234, for what it might be worth, I'm firmly convinced that @PeroK is a nice person who is truly trying to be helpful to you, and I share his idea to the effect that some background work on your part in abstract algebra is in order here. The concept of the distinction between Abelian versus non-Abelian homology groups can be as simple as commutativity or non-commutativity was in HS Algebra, but the definitions of groups and rings probably weren't part of your HS Algebra curriculum, and apparently some understanding of the associated concepts, more than could be picked up over a weekend other than by a prodigy, is required for a full grasp of some of the material that your assignment appears to reference.
 
  • #25
SammyS
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Homework Statement:: Prove the following matrix is non abelian
Homework Equations:: matrix below

I know for a group to be abelian a*b=b*a
I tried multiplying the matrix by itself also but I’m not sure what I’m looking for.
picture is below of the matrix
Is there a problem statement? I can't find one. Maybe it has vanished.

( @PeroK knows that I have a way of missing obvious things in a thread. )
 

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