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

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SUMMARY

The discussion centers on the kernel of the homomorphism $\Phi: G \rightarrow GL_2(\mathbb{Q})$, where $G$ is a subset of $GL_3(\mathbb{Q})$. The kernel of $\Phi$ is shown to be isomorphic to $\mathbb{Q}^2$ under componentwise addition. The participants confirm that the map $i: \ker(\Phi) \rightarrow \mathbb{Q}^2$ defined by $i\left(\begin{pmatrix} I_2 & \begin{pmatrix} a \\ b \end{pmatrix} \\ 0 & 1 \end{pmatrix}\right) = \begin{pmatrix} a \\ b \end{pmatrix}$ is indeed a group homomorphism, establishing the isomorphism.

PREREQUISITES
  • Understanding of group theory, specifically group homomorphisms.
  • Familiarity with the general linear group $GL_n(\mathbb{Q})$.
  • Knowledge of matrix operations and their properties.
  • Concept of isomorphism in the context of algebraic structures.
NEXT STEPS
  • Study the properties of the general linear group $GL_n(\mathbb{Q})$.
  • Explore the concept of kernels in group homomorphisms.
  • Learn about isomorphic structures in abstract algebra.
  • Investigate the implications of componentwise addition in vector spaces.
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, group theory, and linear algebra, will benefit from this discussion.

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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