- #1

Ted123

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## Homework Statement

## The Attempt at a Solution

*[itex]<[/itex] denotes a subgroup and [itex] \triangleleft[/itex] 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 [itex]A<G[/itex], I can say that:

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

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

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

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

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

So [itex]CD^{-1} \in A[/itex] and [itex]A<G[/itex].

Now suppose [itex]P \in G[/itex] and [itex]Q \in A[/itex]

Then [itex]\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[/itex] .

Therefore [itex]A \triangleleft G[/itex] .

Now [itex]B \neq \emptyset[/itex] since [itex]I\in B[/itex] (set u=1) .

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

Then [itex]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[/itex]

so that [itex]B < G[/itex] .

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

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

Therefore [itex]A \cap B = \{1\}[/itex] .

Let [itex]\text{det}(P) = r \neq 0[/itex] .

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

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

Therefore [itex]G=AB[/itex] and G is the internal direct product of A and B.