Symmetric Groups of Same Order are Isomorphic/Proof 2

Proof
Let us define a bijection:
 * $\alpha: T_1 \to T_2$

Let $\theta: \struct {\map \Gamma {T_1}, \circ} \to \struct {\map \Gamma {T_2}, \circ}$ be defined as:


 * $\forall f \in \struct {\map \Gamma {T_1}, \circ}: \map \theta f = \alpha \circ f \circ \alpha^{-1}$

Let $f, g \in \map \Gamma {T_1}$.

We have:

That is, $\theta$ is a group homomorphism.

Let $f, g \in \map \Gamma {T_1}$ such that $\map \theta f = \map \theta g$.

Then:

Thus it is seen that $\theta$ is an injection.

Let $g \in \map \Gamma {T_2}$ be arbitrary.

Let $\alpha^{-1} \circ g \circ \alpha = f$

Then:

and so $g$ is the image of $f$ under $\theta$.

As $g$ is arbitrary, it follows that $\theta$ is a surjection.

As $\theta$ is an injection and a surjection, it follows that $\theta$ is a bijection by definition.

So $\theta$ is a bijective group homomorphism, and so a group isomorphism by definition.