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

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
Let $t_1, t_2 \in \R$.

We have that:

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

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

Then from the above:

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

Then, also from above:

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.

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:

demonstrating that $\phi$ is a group homomorphism.

Then:

demonstrating that $\phi$ is an injection.

Finally:

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.