- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
For $n\in \mathbb{N}$ let $GL_n(\mathbb{Q})$ the group of all invertible matrices in $\mathbb{Q}^{n\times n}$. We have the subset \begin{equation*}G=\left \{\begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\right \}\end{equation*} of $GL_3(\mathbb{Q})$ and the map $\Phi :G\rightarrow GL_2(\mathbb{Q})$ with $\Phi : \begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\mapsto A$.
I want to show that the kernel of $\Phi$ is isomorphic to $\mathbb{Q}^2$ with the componentwise addition. The kernel of $\Phi$ is the following:
\begin{align*}\ker \Phi&=\{g\in G\mid \Phi (g)=I_2\} \\ & = \left \{ \begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} \ \text{ with } \ A\in GL_2(\mathbb{Q}), \ a, b, \in \mathbb{Q} \mid A=I_2\right \} \\ & = \left \{ \begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\ \text{ with } \ a, b, \in \mathbb{Q} \right \}\end{align*}
To show that the kernel of $\Phi$ is isomorphic to $\mathbb{Q}^2$ with the componentwise addition, we have to show that the map $i:\ker (\Phi )\rightarrow \mathbb{Q}^2$ with $\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\mapsto \begin{pmatrix}a \\ b\end{pmatrix}$ is an isomorphism, i.e. a bijective homomorphism, right? First we have to show that $\ker\Phi$ is a group homomorphism.
Let $g=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1\\
b_1\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} ,\ h=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_2\\
b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} \in \ker (\Phi )$.
We have that:
\begin{equation*}i\left (g\cdot h\right )=i\left (\begin{pmatrix}
\begin{matrix}
2\cdot I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&2
\end{pmatrix}\right )=\begin{pmatrix}
a_1+a_2\\
b_1+b_2\\
\end{pmatrix}=\begin{pmatrix}
a_1\\
b_1\\
\end{pmatrix}+\begin{pmatrix}
a_2\\
b_2\\
\end{pmatrix}=i(g)+i(h)\end{equation*} Is this correct? Or have I calculated $g\cdot h$ wrong? (Wondering)
For $n\in \mathbb{N}$ let $GL_n(\mathbb{Q})$ the group of all invertible matrices in $\mathbb{Q}^{n\times n}$. We have the subset \begin{equation*}G=\left \{\begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\right \}\end{equation*} of $GL_3(\mathbb{Q})$ and the map $\Phi :G\rightarrow GL_2(\mathbb{Q})$ with $\Phi : \begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\mapsto A$.
I want to show that the kernel of $\Phi$ is isomorphic to $\mathbb{Q}^2$ with the componentwise addition. The kernel of $\Phi$ is the following:
\begin{align*}\ker \Phi&=\{g\in G\mid \Phi (g)=I_2\} \\ & = \left \{ \begin{pmatrix}
\begin{matrix}
A
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} \ \text{ with } \ A\in GL_2(\mathbb{Q}), \ a, b, \in \mathbb{Q} \mid A=I_2\right \} \\ & = \left \{ \begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\ \text{ with } \ a, b, \in \mathbb{Q} \right \}\end{align*}
To show that the kernel of $\Phi$ is isomorphic to $\mathbb{Q}^2$ with the componentwise addition, we have to show that the map $i:\ker (\Phi )\rightarrow \mathbb{Q}^2$ with $\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\mapsto \begin{pmatrix}a \\ b\end{pmatrix}$ is an isomorphism, i.e. a bijective homomorphism, right? First we have to show that $\ker\Phi$ is a group homomorphism.
Let $g=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1\\
b_1\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} ,\ h=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_2\\
b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix} \in \ker (\Phi )$.
We have that:
\begin{equation*}i\left (g\cdot h\right )=i\left (\begin{pmatrix}
\begin{matrix}
2\cdot I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&2
\end{pmatrix}\right )=\begin{pmatrix}
a_1+a_2\\
b_1+b_2\\
\end{pmatrix}=\begin{pmatrix}
a_1\\
b_1\\
\end{pmatrix}+\begin{pmatrix}
a_2\\
b_2\\
\end{pmatrix}=i(g)+i(h)\end{equation*} Is this correct? Or have I calculated $g\cdot h$ wrong? (Wondering)