Is this matrix a non-abelian group?

[Lauren1234]

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

 

[member 587159]

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.

 

[sysprog]

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?

 

[member 587159]

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.

 

[sysprog]

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.

 

[SammyS]

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).
. . .

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

(I changed that to bold face for emphasis.)

This thread continues to suffer for lack of a problem statement. (There's not even a poor problem statement.) I have continued the search for one ... in vain.
It seems that perhaps @Lauren1234 has not been able to edit the Original Post in this thread. This is very possible.
I have also searched for any image uploaded (as well as being still available) during the time span in question. None of the images I found was relevant.

In going over this thread quite a few times, I think I get the gist of it. It seems that both @Math_QED and @PeroK read this thread in its original form.

It seems that $P$ is a group. The elements of the group are matrices of the form: $\displaystyle \begin{pmatrix} a & b & c \\ 0 & d & e \\ 0 & f & g \end{pmatrix}$ . The group operation is matrix multiplication.

Since this is formally a group, each element must have an inverse, thus for such such matrix to be in the group, it must be invertible. Therefore, the determinant of any matrix in this group is nonzero. (This says something about $a$ as well as about the relationship between $d,\ e,\ f, \text{and } g$ ).

The rest is sort of clear.
Show that this group in non-abelian.

Then part b), which is much more sophisticated.

 

[member 587159]

(I changed that to bold face for emphasis.)

This thread continues to suffer for lack of a problem statement. (There's not even a poor problem statement.) I have continued the search for one ... in vain.
It seems that perhaps @Lauren1234 has not been able to edit the Original Post in this thread. This is very possible.
I have also searched for any image uploaded (as well as being still available) during the time span in question. None of the images I found was relevant.

In going over this thread quite a few times, I think I get the gist of it. It seems that both @Math_QED and @PeroK read this thread in its original form.

It seems that $P$ is a group. The elements of the group are matrices of the form: $\displaystyle \begin{pmatrix} a & b & c \\ 0 & d & e \\ 0 & f & g \end{pmatrix}$ . The group operation is matrix multiplication.

Since this is formally a group, each element must have an inverse, thus for such such matrix to be in the group, it must be invertible. Therefore, the determinant of any matrix in this group is nonzero. (This says something about $a$ as well as about the relationship between $d,\ e,\ f, \text{and } g$ ).

The rest is sort of clear.
Show that this group in non-abelian.

Then part b), which is much more sophisticated.

View attachment 256051

The original statement was something like "Show that $P$ is a non-abelian group", but I can't recall exactly.

 

[PeroK]

It seems that $P$ is a group. The elements of the group are matrices of the form: $\displaystyle \begin{pmatrix} a & b & c \\ 0 & d & e \\ 0 & f & g \end{pmatrix}$ . The group operation is matrix multiplication.

Since this is formally a group, each element must have an inverse, thus for such such matrix to be in the group, it must be invertible. Therefore, the determinant of any matrix in this group is nonzero. (This says something about $a$ as well as about the relationship between $d,\ e,\ f, \text{and } g$ ).

Then part b), which is much more sophisticated.

View attachment 256051

The original problem stated that $a \ne 0$ and $dg - ef \ne 0$. Part a) was to show that this is a non-abelian group.

The problem is simplified if you recognise that:

$\displaystyle \begin{pmatrix} d & e \\ f & g \end{pmatrix}$ .

is an invertible matrix and that the matrix elements in the first row do not affect these four elements (bottom right) when two members of the set are multiplied. This anticipates part b) as the homomorphic properties start to emerge.

It looks like a good problem because if you plough in without recognising the structure of the multiplication within $P$ you would generate a lot of algebra.

For example, showing the group is non-abelian reduces to finding non-commuting invertible 2x2 matrices (or stating this as a known result).

 

[SammyS]

The original problem stated that $a \ne 0$ and $dg - ef \ne 0$. Part a) was to show that this is a non-abelian group.

Thanks for clearing that up. Now, the earlier discussions in the thread make sense; in particular, the comments made by you, @PeroK, and by @Math_QED .