MHB Is the Kernel of a Group Homomorphism isomorphic to $\mathbb{Q}^2$?

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The discussion centers on proving that the kernel of the homomorphism $\Phi: G \rightarrow GL_2(\mathbb{Q})$ is isomorphic to $\mathbb{Q}^2$ under componentwise addition. The kernel consists of matrices in $G$ where the upper left block is the identity matrix, allowing for the representation of elements as pairs $(a, b) \in \mathbb{Q}^2$. The participants confirm that the mapping $i: \ker(\Phi) \rightarrow \mathbb{Q}^2$ defined by $i(g) = \begin{pmatrix} a \\ b \end{pmatrix}$ is a homomorphism, demonstrating that it preserves the group operation. The conversation also emphasizes the importance of correctly interpreting the group operations involved, particularly distinguishing between matrix multiplication and componentwise addition. The conclusion affirms the correctness of the calculations and the isomorphism.
mathmari
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Hey! :o

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)
 
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mathmari said:
Hey! :o

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)
The "componentwise addition" refers only to the group operation in $\mathbb{Q}^2$. It does not apply to the kernel of $\Phi$, which is a subgroup of $GL_3(\mathbb{Q})$ and inherits the group operation of that group, namely matrix multiplication.
 
Opalg said:
The "componentwise addition" refers only to the group operation in $\mathbb{Q}^2$. It does not apply to the kernel of $\Phi$, which is a subgroup of $GL_3(\mathbb{Q})$ and inherits the group operation of that group, namely matrix multiplication.

Ah so we have the following, or not?
\begin{equation*}g\cdot h=\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}\cdot \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}=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\end{equation*}

And so we get
\begin{equation*}i\left (g\cdot h\right )=i\left (\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\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*} which means that the map is a group homomorphism, right? (Wondering)
 
mathmari said:
Ah so we have the following, or not?
\begin{equation*}g\cdot h=\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}\cdot \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}=\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\end{equation*}

And so we get
\begin{equation*}i\left (g\cdot h\right )=i\left (\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a_1+a_2\\
b_1+b_2\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\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*} which means that the map is a group homomorphism, right? (Wondering)
That is correct. But the matrix multiplication might look more transparent if you wrote matrices such as $$\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}$$ in the form $$\begin{pmatrix}
1&0&a \\ 0&1&b \\ 0&0&1
\end{pmatrix}$$.
 
Opalg said:
That is correct. But the matrix multiplication might look more transparent if you wrote matrices such as $$\begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}$$ in the form $$\begin{pmatrix}
1&0&a \\ 0&1&b \\ 0&0&1
\end{pmatrix}$$.

Ok! Thanks a lot! (Smile)
 

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