Is this matrix a non-abelian group?

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.
  • #36
Lauren1234 said:
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).
. . .
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.

06b6b2ba-01c7-4fb5-9a1c-f4ce272a45a2-jpeg.jpg
 
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  • #37
SammyS said:
(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.
 
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  • #38
SammyS said:
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).
 
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  • #39
PeroK said:
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 .
 
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