Subgroups/Internal Direct Product

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Homework Statement



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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.
 

Answers and Replies

  • #2
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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].

This is not true. It would rather be

[tex]\text{det}(Q)=\frac{\text{det}(P)}{(\sqrt{r})^{2n+1}}[/tex].

So you'll need something else then [tex]\sqrt{r}[/tex], but you're on the right track.

Also, nobody says that [tex]r\geq 0[/tex], thus it might be that [tex]\sqrt{r}[/tex] is not real. In that case, U is not an element of [tex]GL_{2n+1}(\mathbb{R})[/tex]...
 
  • #3
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This is not true. It would rather be

[tex]\text{det}(Q)=\frac{\text{det}(P)}{(\sqrt{r})^{2n+1}}[/tex].

So you'll need something else then [tex]\sqrt{r}[/tex], but you're on the right track.

Also, nobody says that [tex]r\geq 0[/tex], thus it might be that [tex]\sqrt{r}[/tex] is not real. In that case, U is not an element of [tex]GL_{2n+1}(\mathbb{R})[/tex]...

What could I use that's not [itex]\sqrt{r}[/itex]?
 
  • #4
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Instead of dividing through [tex]\sqrt{r}[/tex], you could divide through something else. So that det(Q)=1. Try another kind of root...
 
  • #5
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Instead of dividing through [tex]\sqrt{r}[/tex], you could divide through something else. So that det(Q)=1. Try another kind of root...

Could I use [itex]\sqrt[2n+1]{r}[/itex] .

Then [itex]\text{det}(Q) = \frac{\text{det}(P)}{(\sqrt[2n+1]{r})^{2n+1}} = \frac{r}{r} = 1[/itex] .

But then r could still be negative?
 
  • #6
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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...
 
  • #7
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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 [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] .

i.e. how can I explicitly represent the matrix P in latex and show UP=PU?
 
  • #8
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3,296
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.

[tex]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)[/tex]
 

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