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

  • MHB
  • Thread starter mathmari
  • Start date
In summary, the conversation discusses the kernel of a map between two groups, namely $GL_3(\mathbb{Q})$ and $GL_2(\mathbb{Q})$. The kernel is shown to be isomorphic to $\mathbb{Q}^2$ with componentwise addition. An isomorphism is proven by showing that the map between the two groups is a bijective homomorphism. The conversation also discusses the group operation of the kernel, which is matrix multiplication.
  • #1
mathmari
Gold Member
MHB
5,049
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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.
 
  • #3
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)
 
  • #4
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 \(\displaystyle \begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\) in the form \(\displaystyle \begin{pmatrix}
1&0&a \\ 0&1&b \\ 0&0&1
\end{pmatrix}\).
 
  • #5
Opalg said:
That is correct. But the matrix multiplication might look more transparent if you wrote matrices such as \(\displaystyle \begin{pmatrix}
\begin{matrix}
I_2
\end{matrix} & \begin{matrix}
a\\
b\\
\end{matrix} \\ \begin{matrix}
0 & 0 \\
\end{matrix}&1
\end{pmatrix}\) in the form \(\displaystyle \begin{pmatrix}
1&0&a \\ 0&1&b \\ 0&0&1
\end{pmatrix}\).

Ok! Thanks a lot! (Smile)
 

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

What is the definition of isomorphism?

Isomorphism is a mathematical concept that describes a one-to-one correspondence between two structures, meaning that they have the same shape or structure.

How can you show that two structures are isomorphic?

The most common method is to find a bijective mapping between the elements of the two structures. This means that each element in one structure is paired with a unique element in the other structure, and vice versa. If such a mapping exists, the two structures are isomorphic.

What are the benefits of proving isomorphism between two structures?

Isomorphism allows us to translate problems from one structure to another, making it easier to solve them. It also helps us understand the relationship between different structures and identify similarities and differences between them.

Can two structures be isomorphic if they have different elements?

No, in order for two structures to be isomorphic, they must have the same number of elements and the same relationships between those elements. If the elements are different, they cannot be mapped onto each other in a one-to-one correspondence.

Are all isomorphic structures identical?

No, isomorphic structures may have different names or labels for their elements, but the underlying structure or pattern is the same. This means that while they may look different, they are essentially the same and can be transformed into each other through a bijective mapping.

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