# Is this matrix a non-abelian group?

• Lauren1234
In summary: Yes, the group is the whole set of matrices.Oh is it I didn’t no so I need to consider the whole group when doing the group part? Not just the... one matrix?Yes, the group is the whole set of matrices.

#### Lauren1234

Member warned that some effort must be shown
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
https://www.physicsforums.com/attachments/255812

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.

@Lauren1234 you also have to prove it's a group.

etotheipi and sysprog
PeroK said:
@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)

Math_QED said:
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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Lauren1234 said:
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.

Lauren1234 said:
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.

etotheipi, Lauren1234 and sysprog
Lauren1234 said:
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##.

sysprog
@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.

sysprog
PeroK said:
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

sysprog
Math_QED said:
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

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.

Lauren1234 said:
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 realize that ##P## is the group and not just that one matrix?

sysprog
sysprog said:
Hey @Lauren1234, ##\LaTeX## isn't so bad.
It’s very hard to use it on an iPad sorry

sysprog
PeroK said:
Okay, but you do realize 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

Lauren1234 said:
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##.

sysprog
Lauren1234 said:
It’s very hard to use it on an iPad sorry
Oh, spoo (oops backwards) -- I didn't know about your affliction.

Lauren1234
PeroK said:
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

sysprog
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.

PeroK said:
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)

Lauren1234 said:
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.

member 587159
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

Lauren1234 said:
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.

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.

Lauren1234 said:
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. )

SammyS said:
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. )

Can't see it either. Probably the OP deleted it because this was (graded) homework...

SammyS
Math_QED said:
Can't see it either. Probably the OP deleted it because this was (graded) homework...
Wow, @MathQED, that seems to me to be rather a severe accusation to be presented without evidence. Did you even find that @Lauren1234 previously presented a version of her post that included material that you now do not find? Might you be wrongfully accusing?

sysprog said:
Wow, @MathQED, that seems to me to be rather a severe accusation to be presented without evidence. Did you even find that @Lauren1234 previously presented a version of her post that included material that you now do not find? Might you be wrongfully accusing?

I might be and I apologise to the OP if I'm wrong. I'm here since the beginning of the thread and yes, the post was edited. In the beginning it contained an image. Last time I looked it said the top post was edited around 7pm (several hours after the thread was created). Now, I am not able to see anymore that it was edited though.

This was the link to the original image:

https://www.physicsforums.com/attachments/255812/

which leads to a "Black Hole" now. Maybe you can PM me if you want to discuss this further.

Ooo @Math_QED, maybe I just falsely accused you of falsely accusing. I hope and trust that you will trust that I am sincere in saying that I think that you are great. I am confident that I'm not alone here on PF in thinking so.

Math_QED said:
I might be and I apologise to the OP if I'm wrong. I'm here since the beginning of the thread and yes, the post was edited. In the beginning it contained an image. Last time I looked it said the top post was edited around 7pm (several hours after the thread was created). Now, I am not able to see anymore that it was edited though.

This was the link to the original image:

https://www.physicsforums.com/attachments/255812/

which leads to a "Black Hole" now. Maybe you can PM me if you want to discuss this further.
Hi this wasn’t the case I changed it to add the image of my working so far under the original question but have come back to see nothings there I’ll edit it back in now! Sorry for any issues

Math_QED said:
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.
ive attempted to do this but can’t show it’s true can you see where I’ve gone wrong? My attempt is below

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• FDDF4020-15FA-4715-83B1-C863E736BE86.jpeg
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Lauren1234 said:
ive attempted to do this but can’t show it’s true can you see where I’ve gone wrong? My attempt is below

Why are you calculating determinants? ##\varphi(p), \varphi(q)## are matrices and so their product is also a matrix. Basically showing that ##\varphi## is a group morphism is an easy exercise on matrix multiplication.

Math_QED said:
Why are you calculating determinants? ##\varphi(p), \varphi(q)## are matrices and so their product is also a matrix. Basically showing that ##\varphi## is a group morphism is an easy exercise on matrix multiplication.
Well, @Math_QED, could you please explain it to the person, instead of just saying that it's easy? I could probably explain it, but i think that you could do so better than I could. Thank you, Sir.

I might try to drag in some irrelevant stuff:

Does ##\prod = \Pi \ ? ## They are similar, but not congruent -- \prod is bigger than \Pi -- what's up with that?

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sysprog said:
Well, @Math_QED, could you please explain it to the person, instead of just saying that it's easy? I could probably explain it, but i think that you could do so better than I could. Thank you, Sir.

I might try to drag in some irrelevant stuff:

Does ##\prod = \Pi \ ? ## They are similar, but not congruent -- \prod is bigger than \Pi -- what's up with that?

Saying more than I did is just giving away the solution. We must check that ##\varphi(AB) = \varphi(A) \varphi(B)##. So, calculate the matrix product ##AB##, evaluate it using the map ##\varphi##. Next, calculate the matrix product ##\varphi(A) \varphi(B)## and see that it equals ##\varphi(AB)##. It's just your typical "calculate both sides and see that they are equal"-exercise.

I'm not sure where your ##\prod## and ##\Pi## comes from though.

Math_QED said:
Saying more than I did is just giving away the solution. We must check that ##\varphi(AB) = \varphi(A) \varphi(B)##. So, calculate the matrix product ##AB##, evaluate it using the map ##\varphi##. Next, calculate the matrix product ##\varphi(A) \varphi(B)## and see that it equals ##\varphi(AB)##. It's just your typical "calculate both sides and see that they are equal"-exercise.

I'm not sure where your ##\prod## and ##\Pi## comes from though.
Wow, @Math_QED, thanks for your quick and cogent reply -- the nice young gal sent to me a PM and in my response thereto part of what I said included not only that I think that you're a brilliant math guy but also that I think that you are a really nice person. Regarding the ##\LaTeX## jest, well, please see it as an at-humor attempt on my part -- I was hoping that a nice math guy would see my jibe in the direction of Prof. Don, whom I and other persons deeply admire, as not entirely devoid of smileworthiness.

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