Group Isomorphism/Examples/Order 2 Matrices with 1 Real Variable

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Example of Group Isomorphism

Let $S$ be the set defined as:

$S := \set {\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}: t \in \R}$

Consider the algebraic structure $\struct {S, \times}$, where $\times$ is used to denote (conventional) matrix multiplication.


Then $\struct {S, \times}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.


Proof

Group Axiom $\text G 0$: Closure

Let $t_1, t_2 \in \R$.

We have that:

\(\ds \begin{bmatrix} 1 & t_1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t_2 \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 \times 1 + t_1 \times 0 & 1 \times t_2 + t_1 \times 1 \\ 0 \times 1 + 1 \times 0 & 0 \times t_2 + 1 \times 1 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} 1 & t_1 + t_2 \\ 0 & 1 \end{bmatrix}\)

As $t_1 + t_2 \in \R$, it follows that $\struct {S, \times}$ is closed.


Group Axiom $\text G 1$: Associativity

By Matrix Multiplication is Associative it follows that $\struct {S, \times}$ is a semigroup.


Group Axiom $\text G 2$: Existence of Identity Element

Then from the above:

\(\ds \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}\)
\(\ds \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}\)

demonstrating that $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}$.


Group Axiom $\text G 3$: Existence of Inverse Element

Then, also from above:

\(\ds \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 & t - t \\ 0 & 1 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\ds \begin{bmatrix} 1 & -t \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 & t - t \\ 0 & 1 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

demonstrating that $\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}$ has an inverse $\begin{bmatrix} 1 & -t \\ 0 & 1 \end{bmatrix}$.


So all the group axioms are fulfilled, and $\struct {S, \times}$ is seen to be a group.

$\Box$


Isomorphism

Let $\phi: \struct {\R, +} \to \struct {S, \times}$ be the mapping defined as:

$\forall x \in \R: \map \phi x = \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}$


We have that:

\(\ds \map \phi x \map \phi y\) \(=\) \(\ds \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} 1 & x + y \\ 0 & 1 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \map \phi {x + y}\)

demonstrating that $\phi$ is a group homomorphism.


Then:

\(\ds \map \phi x\) \(=\) \(\ds \map \phi y\)
\(\ds \leadsto \ \ \) \(\ds \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}\) \(=\) \(\ds \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix}\)
\(\ds x\) \(=\) \(\ds y\)

demonstrating that $\phi$ is an injection.


Finally:

\(\ds \mathbf x\) \(\in\) \(\ds S\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in \R: \, \) \(\ds \mathbf x\) \(=\) \(\ds \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}\)

demonstrating that $\phi$ is a surjection.


Thus $\phi$ is a group homomorphism which is both injective and surjective.

That is, $\phi$ is by definition a group isomorphism.

Hence the result.

$\blacksquare$


Sources

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