ystael said:
For part (a), you don't need to do a calculation to prove that A and B are normal in G; you could simply observe that G is abelian so all its subgroups are normal.
Also, your proof that AB = G has the right idea but is phrased wrong. You need to put it in the form: given g = x + iy \in G, I exhibit the following a = c + ic \in A and b = d - id \in B so that g = ab, namely: (insert computation of c, d in terms of x, y here). The way you phrased it above, it proves that AB \subset G, which is obvious; you need to prove that AB is all of G.
For part (b), think about how you can modify the determinant of a matrix in G using a matrix in B.
So c=\frac{x+y}{2} and \frac{x-y}{2} then g=ab .
For part (b) 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 det(C)=det(D)=1. Now by the properties of determinants,
\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[/itex]
We have Q \in A for \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.