# Subgroups/Internal Direct Product

## 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.

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 $$GL_{2n+1}(\mathbb{R})$$...

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 $\sqrt{r}$?

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

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?

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...

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?

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.

$$P=\left(\begin{array}{cccc} p_{1,1} & p_{1,2} & \hdots & p_{1,2n+1}\\ p_{2,1} & p_{2,2} & \hdots & p_{2,2n+1}\\ \vdots & \vdots & \ddots & \vdots\\ p_{2n+1,1} & p_{2n+1,2} & \hdots & p_{2n+1,2n+1} \end{array}\right)$$