Groups - Internal Direct Product

[Ted123]

Homework Statement

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

[itex]<[/itex] denotes a subgroup.
[itex]\triangleleft[/itex] denotes a normal subgroup.

The Attempt at a Solution

Have I done (a) correctly?

[itex]0 \in A[/itex] so [itex]A \neq \emptyset[/itex]

If [itex]a=x+ix[/itex] and [itex]b=y+iy[/itex]

then [itex]ab^{-1} = x-y + ix - iy = x-y + i(x-y) \in A[/itex]

[itex]\therefore A < G[/itex]

Again [itex]0 \in B[/itex] so [itex]B \neq \emptyset[/itex]

If [itex]c=x+ix[/itex] and [itex]d=y+iy[/itex]

then [itex]cd^{-1} = x-y - ix - iy = x-y - i(x+y) \in B[/itex]

[itex]\therefore B < G[/itex]

If [itex]g=x+iy \in G[/itex] and [itex]a=c+ic \in A[/itex]

then [itex]gag^{-1} = x+iy + c+ic -x-iy = c+ic \in A[/itex]

[itex]\therefore A \triangleleft G[/itex]

If [itex]b=c-ic \in B[/itex]

then [itex]gbg^{-1} = x+iy + c-ic -x-iy = c-ic \in B[/itex]

[itex]\therefore B \triangleleft G[/itex]

It is clear that [itex]A \cap B = \{0\}[/itex] .

If [itex]a=c+ic \in A[/itex] and [itex]b=d-id \in B[/itex]

than [itex]ab = c+ic +d-id = c+d + i(c-d)[/itex]

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

[itex]\therefore G=AB[/itex] 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 [tex]A[/tex] and [tex]B[/tex] are normal in [tex]G[/tex]; you could simply observe that [tex]G[/tex] is abelian so all its subgroups are normal.

Also, your proof that [tex]AB = G[/tex] has the right idea but is phrased wrong. You need to put it in the form: given [tex]g = x + iy \in G[/tex], I exhibit the following [tex]a = c + ic \in A[/tex] and [tex]b = d - id \in B[/tex] so that [tex]g = ab[/tex], namely: (insert computation of [tex]c, d[/tex] in terms of [tex]x, y[/tex] here). The way you phrased it above, it proves that [tex]AB \subset G[/tex], which is obvious; you need to prove that [tex]AB[/tex] is all of [tex]G[/tex].

For part (b), think about how you can modify the determinant of a matrix in [tex]G[/tex] using a matrix in [tex]B[/tex].

 

[Ted123]

For part (a), you don't need to do a calculation to prove that [tex]A[/tex] and [tex]B[/tex] are normal in [tex]G[/tex]; you could simply observe that [tex]G[/tex] is abelian so all its subgroups are normal.

Also, your proof that [tex]AB = G[/tex] has the right idea but is phrased wrong. You need to put it in the form: given [tex]g = x + iy \in G[/tex], I exhibit the following [tex]a = c + ic \in A[/tex] and [tex]b = d - id \in B[/tex] so that [tex]g = ab[/tex], namely: (insert computation of [tex]c, d[/tex] in terms of [tex]x, y[/tex] here). The way you phrased it above, it proves that [tex]AB \subset G[/tex], which is obvious; you need to prove that [tex]AB[/tex] is all of [tex]G[/tex].

For part (b), think about how you can modify the determinant of a matrix in [tex]G[/tex] using a matrix in [tex]B[/tex].

So [itex]c=\frac{x+y}{2}[/itex] and [itex]\frac{x-y}{2}[/itex] then [itex]g=ab[/itex] .

For part (b) 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, A 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 det(C)=det(D)=1. Now by the properties of determinants,

[itex]\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 A<G.

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 P = [The matrix P with every element divided by [itex]\sqrt{r}[/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]

We have [itex]Q \in A[/itex] for [itex]\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.