MHB Multiplicative group of matrices is nilpotent

[mathmari]

Hey!

Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.

I want to show that $G$ is nilpotent.

Could you maybe give me some hints what we could do in this case? (Wondering)

Do we have to find the order of this group? Or isn't this possible? (Wondering)

 

[Euge]

Hi mathmari,

First start by computing the center $Z(G)$ of $G$. It will be something recognizable. ;)

 

[mathmari]

Hi mathmari,

First start by computing the center $Z(G)$ of $G$. It will be something recognizable. ;)

We have that $Z(G)=\{z\in G \mid \forall g\in G, zg=gz\}$.

Let $z=\begin{pmatrix}
1 & \tilde{x} & \tilde{y} \\
0 & 1 & \tilde{z} \\
0 & 0 & 1
\end{pmatrix}$ and $g=\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$.

Then $$zg=\begin{pmatrix}
1 & \tilde{x} & \tilde{y} \\
0 & 1 & \tilde{z} \\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}=\begin{pmatrix}
1 & x+\tilde{x} & y+\tilde{x}z+\tilde{y} \\
0 & 1 & z+\tilde{z} \\
0 & 0 & 1
\end{pmatrix}$$ and $$gz=\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & \tilde{x} & \tilde{y} \\
0 & 1 & \tilde{z} \\
0 & 0 & 1
\end{pmatrix}=\begin{pmatrix}
1 & \tilde{x}+x & \tilde{y}+x\tilde{z}+y \\
0 & 1 & \tilde{z}+z \\
0 & 0 & 1
\end{pmatrix}$$

So, it must be $$\begin{pmatrix}
1 & x+\tilde{x} & y+\tilde{x}z+\tilde{y} \\
0 & 1 & z+\tilde{z} \\
0 & 0 & 1
\end{pmatrix}=\begin{pmatrix}
1 & \tilde{x}+x & \tilde{y}+x\tilde{z}+y \\
0 & 1 & \tilde{z}+z \\
0 & 0 & 1
\end{pmatrix} \Rightarrow \tilde{x}z=x\tilde{z}$$ right? (Wondering)

 

[Opalg]

We have that $Z(G)=\{z\in G \mid \forall g\in G, zg=gz\}$.

Let $z=\begin{pmatrix}
1 & \tilde{x} & \tilde{y} \\
0 & 1 & \tilde{z} \\
0 & 0 & 1
\end{pmatrix}$ and $g=\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$.

Then $$\vdots$$ $$ \tilde{x}z=x\tilde{z}$$ right? (Wondering)

So if that is true for all $x,z\in M$, what does that tell you?

 

[mathmari]

So if that is true for all $x,z\in M$, what does that tell you?

That $Z(G)=G$, or not? (Wondering)

 

[Opalg]

That $Z(G)=G$, or not? (Wondering)

No, that would mean that the whole group $G$ is commutative, which it certainly isn't. For example, the matrices $$\begin{pmatrix}1&1&0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}, \quad \begin{pmatrix}1&0&0 \\ 0&1&1 \\ 0&0&1\end{pmatrix}$$ do not commute.For the matrix $z$ to be in the centre $Z(G)$, it has to commute with every element $g\in G$. (In your notation, one of the elements of the matrix $g$ is $z$, which also denotes the whole matrix $z\in Z(G)$. Not the best choice of notation, so be careful not to confuse the two $z$s.) So we are looking at a (fixed) matrix $z$ with (fixed) entries $\tilde{x}, \tilde{y}, \tilde{z}$ above the diagonal. We want to know whether it commutes with every matrix in $G$. As you showed, this means that $\tilde{x}$ and $\tilde{z}$ must satisfy $\tilde{x}z = x\tilde{z}$ for every $x$ and $z$ in $M$. In particular, you could choose $x$ and $z$ to be the zero or identity elements $0,1\in M$. What does that tell you about $\tilde{x}$ and $\tilde{z}$?

 

[mathmari]

No, that would mean that the whole group $G$ is commutative, which it certainly isn't. For example, the matrices $$\begin{pmatrix}1&1&0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}, \quad \begin{pmatrix}1&0&0 \\ 0&1&1 \\ 0&0&1\end{pmatrix}$$ do not commute.

Ah ok... (Thinking)

For the matrix $z$ to be in the centre $Z(G)$, it has to commute with every element $g\in G$. (In your notation, one of the elements of the matrix $g$ is $z$, which also denotes the whole matrix $z\in Z(G)$. Not the best choice of notation, so be careful not to confuse the two $z$s.)

So, not to get confused, let's define $Z(G)=\{a\in G \mid \forall g\in G, ag=ga\}$, so we want that the matrix $a$ is in the center $Z(G)$.

So we are looking at a (fixed) matrix $z$ with (fixed) entries $\tilde{x}, \tilde{y}, \tilde{z}$ above the diagonal. We want to know whether it commutes with every matrix in $G$. As you showed, this means that $\tilde{x}$ and $\tilde{z}$ must satisfy $\tilde{x}z = x\tilde{z}$ for every $x$ and $z$ in $M$. In particular, you could choose $x$ and $z$ to be the zero or identity elements $0,1\in M$. What does that tell you about $\tilde{x}$ and $\tilde{z}$?

So that the (fixed) matrix $a$ with (fixed) entries $\tilde{x}, \tilde{y}, \tilde{z}$ above the diagonal, commutes with every matrix in $G$, $\tilde{x}$ and $\tilde{z}$ must satisfy $\tilde{x}z = x\tilde{z}$, $\forall x , z \in M$.

Do you mean do take $x=0$ and $z=1$ ? (Wondering)

If you mean this, then we have $\tilde{x}=0$. But what information do we get from that for $\tilde{x}$ and $\tilde{z}$ in general? (Wondering)

I got stuck right now...

 

[Opalg]

Do you mean do take $x=0$ and $z=1$ ? (Wondering)

If you mean this, then we have $\tilde{x}=0$. But what information do we get from that for $\tilde{x}$ and $\tilde{z}$ in general? (Wondering)

I got stuck right now...

Good. So you know that if $a\in Z(G)$ then $\tilde{x}=0$. Similarly, if you take $x=1$ and $z=0$ then you get $\tilde{z}=0$. In that case, the matrix $a$ becomes $\begin{pmatrix}1&0&\tilde{y} \\ 0&1&0 \\ 0&0&1\end{pmatrix}.$ So $Z(G)$ consists of all matrices of that form ($1$s on the main diagonal, anything at all in the top right-hand corner, and zeros everywhere else).

So now that you have followed Euge's hint and identified the centre of $G$, you need to remind yourself of the definition of nilpotency and how $Z(G)$ might be relevant there.

 

[Euge]

The next thing to do is to compute the factor group $G/Z(G)$. Mathmari, what can you say about $G/Z(G)$?

 

[mathmari]

The next thing to do is to compute the factor group $G/Z(G)$. Mathmari, what can you say about $G/Z(G)$?

How can we compute the factor group $G/Z(G)$ ? (Wondering)
Could you give me a hint? (Wondering)

 

[Euge]

What are the left cosets of $Z(G)$?

 

[mathmari]

What are the left cosets of $Z(G)$?

The left cosets of $Z(G)$ are $$Z(G)g=\{ag \mid a\in Z(G)\}, g\in G$$

Since $Z(G)=\{a\in G \mid \forall g\in G, ag=ga\}$, we have that for $g\in G$ $$Z(G)g=\{ag \mid a\in Z(G)\}=\{ga \mid a\in Z(G)\}=gZ(G)$$ right? (Wondering)

 

[Euge]

You actually wrote down right cosets in the first statement, but it doesn't particularly matter. What do the cosets look like, more explicitly? Don't just use $g$ to denote an element of $G$.

 

[mathmari]

You actually wrote down right cosets in the first statement, but it doesn't particularly matter.

Oh yes... (Blush)

What do the cosets look like, more explicitly? Don't just use $g$ to denote an element of $G$.

What do you mean? (Wondering)

 

[Euge]

Your $g$ is a certain kind of matrix, isn't it? What form does it take?

 

[mathmari]

Your $g$ is a certain kind of matrix, isn't it? What form does it take?

It is a matrix of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$, right? (Wondering)

 

[Euge]

Correct. So a typical coset of $Z(G)$ looks like

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

where $x,y,z\in M$. Note that for every $y\in M$ the above coset is the same as the coset

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

since

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & -y \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\in Z(G)$$

Furthermore, two cosets

$$\begin{pmatrix}1 & x_1 & y_1\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}Z(G) \quad \text{and}\quad \begin{pmatrix}1 & x_2 & y_2\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}Z(G)$$

are equal if and only if $x_1 = x_2$ and $z_1 = z_2$. Hence, every coset of $G/Z(G)$ is uniquely represented by an element of the form

$$\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$$

If $g$ and $h$ are matrices of that form, then $gh$ and $hg$ have the same $(1,2)$- and $(2,3)$-entries. So they represent the same coset of $Z(G)$. Consequently, $G/Z(G)$ is abelian.

I asked you to consider $G/Z(G)$ element-wise because I wanted you to see exactly how the elements look like. But we could have proceeded as follows. The mapping $\Phi : G \to M \times M$ given by

$$\Phi\left(\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}\right) = (x,z)$$

is a surjective homomoprhism with kernel $Z(G)$. So $G/Z(G)$ is isomorphic to $M \times M$. Since $M \times M$ is abelian, so is $G/Z(G)$.

Knowing that $G/Z(G)$ is abelian, what can you say about $Z(G/Z(G))$?

 

[Deveno]

Correct. So a typical coset of $Z(G)$ looks like

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

where $x,y,z\in M$. Note that for every $y\in M$ the above coset is the same as the coset

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

since

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & -xz-y \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\in Z(G)$$

When I do the calculations, I get:

$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -x & zx - y\\0 & 1 & -z\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$

$=\begin{pmatrix}1 & 0 & -y\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$

This is still in $Z(G)$, so your argument isn't affected, just wanted to point this out.

Furthermore, two cosets

$$\begin{pmatrix}1 & x_1 & y_1\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}Z(G) \quad \text{and}\quad \begin{pmatrix}1 & x_2 & y_2\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}$$

are equal if and only if $x_1 = x_2$ and $z_1 = z_2$. Hence, every coset of $G/Z(G)$ is uniquely represented by an element of the form

$$\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$$

Matrices of the this form commute, so $G/Z(G)$ is abelian.

I asked you to consider $G/Z(G)$ element-wise because I wanted you to see exactly how the elements look like. But we could have proceeded as follows. The mapping $\Phi : G \to M \times M$ given by

$$\Phi\left(\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}\right) = (x,z)$$

is a surjective homomoprhism with kernel $Z(G)$. So $G/Z(G)$ is isomorphic to $M \times M$. Since $M \times M$ is abelian, so is $G/Z(G)$.

Knowing that $G/Z(G)$ is abelian, what can you say about $Z(G/Z(G))$?

One of the things about this particular group that should "pop out" at you when you actually multiply two elements in it, is that the $(1,2)$- and $(2,3)$-entries ADD, and the $(1,3)$-entry is just a beastly mess (everything else is a forgone conclusion).

What taking the quotient by $Z(G)$ actually "does" for us, is to send that pesky $(1,3)$-entry to $0$, making what's left of $G$ "additive" (abelian).

 

[Euge]

When I do the calculations, I get:

$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -x & zx - y\\0 & 1 & -z\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$

$=\begin{pmatrix}1 & 0 & -y\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$

This is still in $Z(G)$, so your argument isn't affected, just wanted to point this out.

Thanks, I've fixed the typo. :D

 

[mathmari]

So a typical coset of $Z(G)$ looks like

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

where $x,y,z\in M$. Note that for every $y\in M$ the above coset is the same as the coset

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

since

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & -y \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\in Z(G)$$

I haven't really understood why these two cosets are the same for every $y\in M$. (Wondering)
Could you explain it to me?

Furthermore, two cosets

$$\begin{pmatrix}1 & x_1 & y_1\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}Z(G) \quad \text{and}\quad \begin{pmatrix}1 & x_2 & y_2\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}$$

are equal if and only if $x_1 = x_2$ and $z_1 = z_2$. Hence, every coset of $G/Z(G)$ is uniquely represented by an element of the form

$$\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$$

How have we concluded that every coset of $G/Z(G)$ is uniquely represented by an element of the form $\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$ ? (Wondering)

$$\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$$

Matrices of the this form commute, so $G/Z(G)$ is abelian.

$$\begin{pmatrix} 1 & x_1 & 0\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 & x_2 & 0\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & x_1+x_2 & x_1z_2\\0 & 1 & z_2+z_1\\0 & 0 & 1\end{pmatrix}$$

$$\begin{pmatrix} 1 & x_2 & 0\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 & x_1 & 0\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & x_2+x_1 & x_2z_1\\0 & 1 & z_1+z_2\\0 & 0 & 1\end{pmatrix}$$

Are these two results the same? (Wondering)

 

[Euge]

I haven't really understood why these two cosets are the same for every $y\in M$. (Wondering)
Could you explain it to me?

Recall that if $G$ is a group and $N$ is a normal subgroup of $G$, then $gN = hN$ if and only if $g^{-1}h\in N$. This was applied your given matrix group $G$, $N = Z(G)$, $g = \begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$, and $h = \begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$.

How have we concluded that every coset of $G/Z(G)$ is uniquely represented by an element of the form $\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$ ? (Wondering)

First I showed that every coset can be represented in the form

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

Then I showed that two cosets are the same if and only if their matrix representatives have the same $(1,2)$-entry and $(2,3)$-entry. That is,

$$\begin{pmatrix}1 & x_1 & y_1\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}Z(G) = \begin{pmatrix}1 & x_2 & y_2\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}Z(G)$$

if and only if $x_1 = x_2$ and $z_1 = z_2$. From these two facts, we deduce that the cosets

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G) \quad (x,z\in M)$$

are distinct for every $x,z\in M$, and these are the only cosets of $Z(G)$.

$$\begin{pmatrix} 1 & x_1 & 0\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 & x_2 & 0\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & x_1+x_2 & x_1z_2\\0 & 1 & z_2+z_1\\0 & 0 & 1\end{pmatrix}$$

$$\begin{pmatrix} 1 & x_2 & 0\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}\begin{pmatrix} 1 & x_1 & 0\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & x_2+x_1 & x_2z_1\\0 & 1 & z_1+z_2\\0 & 0 & 1\end{pmatrix}$$

Are these two results the same? (Wondering)

No. Sorry, I meant to say that the cosets

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

commute. :p I've made the appropriate edit to that post.

Note that in your calculations, the $(1,2)$-entry and the $(2,3)$-entry of matrices on the right-hand sides of your equations are the same. So they represent the same coset of $Z(G)$. Consequently, $G/Z(G)$ is abelian.

 

[mathmari]

Recall that if $G$ is a group and $N$ is a normal subgroup of $G$, then $gN = hN$ if and only if $g^{-1}h\in N$. This was applied your given matrix group $G$, $N = Z(G)$, $g = \begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$, and $h = \begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$.

Why do we take this $h$ ? (Wondering)

First I showed that every coset can be represented in the form

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

How exactly have we shown that? (Wondering)

Then I showed that two cosets are the same if and only if their matrix representatives have the same $(1,2)$-entry and $(2,3)$-entry. That is,

$$\begin{pmatrix}1 & x_1 & y_1\\0 & 1 & z_1\\0 & 0 & 1\end{pmatrix}Z(G) = \begin{pmatrix}1 & x_2 & y_2\\0 & 1 & z_2\\0 & 0 & 1\end{pmatrix}Z(G)$$

if and only if $x_1 = x_2$ and $z_1 = z_2$.

Do we conclude to that because of the following?

So a typical coset of $Z(G)$ looks like

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

where $x,y,z\in M$. Note that for every $y\in M$ the above coset is the same as the coset

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)$$

since

$$\begin{pmatrix}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}^{-1}\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 & -y \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\in Z(G)$$

(Wondering)

From these two facts, we deduce that the cosets

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G) \quad (x,z\in M)$$

are distinct for every $x,z\in M$, and these are the only cosets of $Z(G)$.

Why do we have that these are the only cosets? (Wondering)

 

[Deveno]

Why do we take this $h$ ? (Wondering)

How exactly have we shown that? (Wondering) Do we conclude to that because of the following?



(Wondering)


Why do we have that these are the only cosets? (Wondering)

Note that:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix} = \begin{pmatrix}1&x&0\\0&1&z\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&y\\0&1&0\\0&0&1\end{pmatrix}$

If we call the matrix on the LHS, $A$, and the matrices on the RHS $B$ and $Z$, we have:

$A = BZ$, where $Z \in Z(G)$.

This shows that $A \in BZ(G)$, and thus $AZ(G) = BZ(G)$.

Since every coset is of the form $AZ(G)$ for (some) $A \in G$, every coset is of the form $BZ(G)$ (for some $B$, which we can get from $A$). The "special form" matrices $B$ (derived from $A$) form what is called a "traversal" of the cosets, they are "representative representatives" (much like the cosets $k + n\Bbb Z$ for $k \in \{1,2,\dots,n-1\}$ are a traversal of $\Bbb Z/n\Bbb Z$).

When are two cosets the same? That is, when does $AZ(G) = A'Z(G)$? We require:

$AA'^{-1} \in Z(G)$, so:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}\begin{pmatrix}1&-x'&x'z' - y'\\0&1&-z'\\0&0&1\end{pmatrix} \in Z(G)$

If we do the matrix multiplication, we get:

$AA'^{-1} = \begin{pmatrix}1&x-x'&x'z'-y'-xz'+y\\0&1&z-z'\\0&0&1\end{pmatrix}$

For this to be in the center, $Z(G)$, all we require is that the $(1,2)$-entry and the $(2,3)$-entry be 0. This is equivalent to setting:

$x - x' = 0$
$z - z' = 0$

so we see that two cosets are the same if $x = x'$ and $z = z'$ (the $(1,3)$-entry is a mess, but we don't care, it's in $M$, which is all that matters).

On the other hand, if $x = x'$ and $z = z'$, running the argument in reverse shows $AZ(G) = A'Z(G)$.

So if the $(1,2)$ and $(2,3)$ entries match, we get the same coset. So we may as well pick a $(1,3)$ entry which is easy to work with. 0 is good.

Look hard at the product decomposition at the beginning of my post. It says:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}$

is the same as:

$\begin{pmatrix}1&x&0\\0&1&z\\0&0&1\end{pmatrix}$

"up to multiplication by an element of $Z(G)$".

This is really the same type of thing as saying, for an integer $a = qn + r$, that $r$ is the same as $a$ "up to adding an element of $n\Bbb Z$", that is:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}$ is congruent to $\begin{pmatrix}1&x&0\\0&1&z\\0&0&1\end{pmatrix}$

modulo the center of $G$.

In other words, we might as well regard $G/Z(G)$ as the set of matrices:

$\left\{\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}: y = 0\right\}$

where we multiply "like usual" except for zeroing out the $(1,3)$-entry after we're done (because anything in $Z(G)$, including the matrix:

$\begin{pmatrix}1&0&xz'\\0&1&0\\0&0&1\end{pmatrix}$

"acts like the identity" in $G/Z(G)$-just like $kn$ "acts like the identity (zero)" in $\Bbb Z_n$).

 

[mathmari]

Note that:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix} = \begin{pmatrix}1&x&0\\0&1&z\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&y\\0&1&0\\0&0&1\end{pmatrix}$

If we call the matrix on the LHS, $A$, and the matrices on the RHS $B$ and $Z$, we have:

$A = BZ$, where $Z \in Z(G)$.

This shows that $A \in BZ(G)$, and thus $AZ(G) = BZ(G)$.

Since every coset is of the form $AZ(G)$ for (some) $A \in G$, every coset is of the form $BZ(G)$ (for some $B$, which we can get from $A$). The "special form" matrices $B$ (derived from $A$) form what is called a "traversal" of the cosets, they are "representative representatives" (much like the cosets $k + n\Bbb Z$ for $k \in \{1,2,\dots,n-1\}$ are a traversal of $\Bbb Z/n\Bbb Z$).

When are two cosets the same? That is, when does $AZ(G) = A'Z(G)$? We require:

$AA'^{-1} \in Z(G)$, so:

$\begin{pmatrix}1&x&y\\0&1&z\\0&0&1\end{pmatrix}\begin{pmatrix}1&-x'&x'z' - y'\\0&1&-z'\\0&0&1\end{pmatrix} \in Z(G)$

If we do the matrix multiplication, we get:

$AA'^{-1} = \begin{pmatrix}1&x-x'&x'z'-y'-xz'+y\\0&1&z-z'\\0&0&1\end{pmatrix}$

For this to be in the center, $Z(G)$, all we require is that the $(1,2)$-entry and the $(2,3)$-entry be 0. This is equivalent to setting:

$x - x' = 0$
$z - z' = 0$

so we see that two cosets are the same if $x = x'$ and $z = z'$ (the $(1,3)$-entry is a mess, but we don't care, it's in $M$, which is all that matters).

On the other hand, if $x = x'$ and $z = z'$, running the argument in reverse shows $AZ(G) = A'Z(G)$.

So if the $(1,2)$ and $(2,3)$ entries match, we get the same coset. So we may as well pick a $(1,3)$ entry which is easy to work with. 0 is good.

Ah ok... I see... (Thinking)

 

[mathmari]

Hence, every coset of $G/Z(G)$ is uniquely represented by an element of the form

$$\begin{pmatrix} 1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}$$

If $g$ and $h$ are matrices of that form, then $gh$ and $hg$ have the same $(1,2)$- and $(2,3)$-entries. So they represent the same coset of $Z(G)$. Consequently, $G/Z(G)$ is abelian.

So, we have that $ghZ(G)=hgZ(G)$.

Let $A=gZ(G)$ and $B=hZ(G)$, then $$AB=(gZ(G))(hZ(G))=ghZ(G)=hgZ(G)=(hZ(G))(gZ(G))=BA$$ So, we conclude that $G/Z(G)$ is abelian, right? (Wondering)

From these two facts, we deduce that the cosets

$$\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G) \quad (x,z\in M)$$

are distinct for every $x,z\in M$, and these are the only cosets of $Z(G)$.

Do we have that these are the only cosets of $Z(G)$, since $\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)=gZ(G), g\in G$ ? (Wondering)

 

[Euge]

So, we have that $ghZ(G)=hgZ(G)$.

Let $A=gZ(G)$ and $B=hZ(G)$, then $$AB=(gZ(G))(hZ(G))=ghZ(G)=hgZ(G)=(hZ(G))(gZ(G))=BA$$ So, we conclude that $G/Z(G)$ is abelian, right? (Wondering)

Yes, that's right.

Do we have that these are the only cosets of $Z(G)$, since $\begin{pmatrix}1 & x & 0\\0 & 1 & z\\0 & 0 & 1\end{pmatrix}Z(G)=gZ(G), g\in G$ ? (Wondering)

Yes.