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

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Discussion Overview

The discussion revolves around the kernel of a group homomorphism defined from a subset of invertible matrices in $GL_3(\mathbb{Q})$ to $GL_2(\mathbb{Q})$. Participants are exploring whether this kernel is isomorphic to $\mathbb{Q}^2$ under componentwise addition, focusing on the properties of the kernel and the nature of the homomorphism.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the kernel of the homomorphism $\Phi$ consists of matrices of a specific form, leading to the assertion that it is isomorphic to $\mathbb{Q}^2$.
  • Others argue that to establish isomorphism, one must demonstrate that the map $i:\ker (\Phi )\rightarrow \mathbb{Q}^2$ is a bijective homomorphism, questioning the correctness of the calculations involved.
  • A later reply clarifies that the "componentwise addition" applies only to the group operation in $\mathbb{Q}^2$ and not to the kernel of $\Phi$, which operates under matrix multiplication.
  • Participants discuss the multiplication of matrices in the kernel and its implications for the homomorphism, with some confirming the calculations and others seeking further clarification.

Areas of Agreement / Disagreement

There is no consensus on the isomorphism of the kernel to $\mathbb{Q}^2$. While some participants agree on the structure of the kernel, others challenge the interpretation of operations involved and the nature of the homomorphism.

Contextual Notes

Participants express uncertainty regarding the operations in the kernel and the implications for the proposed isomorphism, highlighting the need for careful consideration of group operations and definitions.

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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