Subgroups/Internal Direct Product

[Ted123]

Homework Statement

The Attempt at a Solution

< denotes a subgroup and \triangleleft denotes a normal subgroup throughout.

Can anyone tell me what I've done right/wrong? I've posted all my working below:

To prove that A<G, I can say that:

A (2n+1) \times (2n+1) matrix is invertible if and only if it has non-zero determinant so A \subset G.

Furthermore, A is non-empty since I_{2n+1} \in A since \text{det}(I_{2n+1})=1.

To prove that CD^{-1} \in A for all C,D \in A is this correct?:

Let C,D \in A. Then \text{det}(C)=\text{det}(D)=1. Now by the properties of determinants,

\displaystyle \text{det}(CD^{-1}) = \text{det}(C)\text{det}(D^{-1}) = \frac{\text{det}(C)}{\text{det}(D)} = \frac{1}{1} = 1 .

So CD^{-1} \in A and A<G.

Now suppose P \in G and Q \in A

Then \text{det}(PQP^{-1}) = \text{det}(P)\text{det}(Q)\text{det}(P^{-1}) = \text{det}(P)\text{det}(Q)\text{det}(P)^{-1} = \text{det}(Q) = 1 .

Therefore A \triangleleft G .

Now B \neq \emptyset since I\in B (set u=1) .

If U = \begin{bmatrix}u & 0 & \ldots & 0 \\ 0 & u & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0 & 0 &\ldots & u\end{bmatrix} \in B and if V = \begin{bmatrix}v & 0 & \ldots & 0 \\ 0 & v & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0 & 0 &\ldots & v\end{bmatrix} \in B

Then UV^{-1} = \begin{bmatrix}uv^{-1} & 0 & \ldots & 0 \\ 0 & uv^{-1} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0 & 0 &\ldots & uv^{-1}\end{bmatrix} \in B

so that B < G .

If P\in G is a (2n+1) \times (2n+1) matrix of the same size of U with arbitrary entries then UP=PU and it follows that PUP^{-1} = U . Hence B \triangleleft G .

Now since \text{det}(U) = u^{2n+1} and u\neq 0 it follows that if U \in A then u=1 .

Therefore A \cap B = \{1\} .

Let \text{det}(P) = r \neq 0 .

Then P = [The matrix P with every element divided by \sqrt{r} ] \begin{bmatrix}\sqrt{r} & 0 & \ldots & 0 \\ 0 & \sqrt{r} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0 & 0 &\ldots & \sqrt{r} \end{bmatrix} = QU (say)

We have Q \in A for \displaystyle \text{det}(Q)= \frac{\text{det}(P)}{r} = 1 and U \in B.

Therefore G=AB and G is the internal direct product of A and B.

 

[micromass]

We have Q \in A for \displaystyle \text{det}(Q)= \frac{\text{det}(P)}{r} = 1 and U \in B.

This is not true. It would rather be

\text{det}(Q)=\frac{\text{det}(P)}{(\sqrt{r})^{2n+1}}.

So you'll need something else then \sqrt{r}, but you're on the right track.

Also, nobody says that r\geq 0, thus it might be that \sqrt{r} is not real. In that case, U is not an element of GL_{2n+1}(\mathbb{R})...

 

[Ted123]

This is not true. It would rather be

\text{det}(Q)=\frac{\text{det}(P)}{(\sqrt{r})^{2n+1}}.

So you'll need something else then \sqrt{r}, but you're on the right track.

Also, nobody says that r\geq 0, thus it might be that \sqrt{r} is not real. In that case, U is not an element of GL_{2n+1}(\mathbb{R})...

What could I use that's not \sqrt{r}?

 

[micromass]

Instead of dividing through \sqrt{r}, you could divide through something else. So that det(Q)=1. Try another kind of root...

 

[Ted123]

Instead of dividing through \sqrt{r}, you could divide through something else. So that det(Q)=1. Try another kind of root...

Could I use \sqrt[2n+1]{r} .

Then \text{det}(Q) = \frac{\text{det}(P)}{(\sqrt[2n+1]{r})^{2n+1}} = \frac{r}{r} = 1 .

But then r could still be negative?

 

[micromass]

Yes, that is correct. And r being negative doesn't hurt here. 2n+1 is an odd number, so the 2n+1-th root is real, even if r is negative...

 

[Ted123]

Yes, that is correct. And r being negative doesn't hurt here. 2n+1 is an odd number, so the 2n+1-th root is real, even if r is negative...

Ah yes. So does everything look OK now?

Is there a more explicit way of putting this bit:

If P\in G is a (2n+1) \times (2n+1) matrix of the same size of U with arbitrary entries then UP=PU and it follows that PUP^{-1} = U .

i.e. how can I explicitly represent the matrix P in latex and show UP=PU?

 

[micromass]

Yes, this looks good!

I think your statement about the P and U is clear enough. If you really want to represent P as a matrix, then maybe you can present it as an arbitrary matrix with arbitrary entries, i.e.

P=\left(\begin{array}{cccc}<br /> p_{1,1} &amp; p_{1,2} &amp; \hdots &amp; p_{1,2n+1}\\<br /> p_{2,1} &amp; p_{2,2} &amp; \hdots &amp; p_{2,2n+1}\\<br /> \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\<br /> p_{2n+1,1} &amp; p_{2n+1,2} &amp; \hdots &amp; p_{2n+1,2n+1}<br /> \end{array}\right)