Groups - Internal Direct Product

[Ted123]

Homework Statement

[PLAIN]http://img689.imageshack.us/img689/3047/directproduct.png

$<$ denotes a subgroup.
$\triangleleft$ denotes a normal subgroup.

The Attempt at a Solution

Have I done (a) correctly?

$0 \in A$ so $A \neq \emptyset$

If $a=x+ix$ and $b=y+iy$

then $ab^{-1} = x-y + ix - iy = x-y + i(x-y) \in A$

$\therefore A < G$

Again $0 \in B$ so $B \neq \emptyset$

If $c=x+ix$ and $d=y+iy$

then $cd^{-1} = x-y - ix - iy = x-y - i(x+y) \in B$

$\therefore B < G$

If $g=x+iy \in G$ and $a=c+ic \in A$

then $gag^{-1} = x+iy + c+ic -x-iy = c+ic \in A$

$\therefore A \triangleleft G$

If $b=c-ic \in B$

then $gbg^{-1} = x+iy + c-ic -x-iy = c-ic \in B$

$\therefore B \triangleleft G$

It is clear that $A \cap B = \{0\}$ .

If $a=c+ic \in A$ and $b=d-id \in B$

than $ab = c+ic +d-id = c+d + i(c-d)$

so if $g=x+iy \in G$ then $g=ab$ with $x=c+d \in \mathbb{R}$ and $y=c-d \in \mathbb{R}$

$\therefore G=AB$ and G is the internal direct product of A and B.

Can anyone help with (b)?

 

[ystael]

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 $$B$$.

 

[Ted123]

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.