Symmetric Groups of Same Order are Isomorphic

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Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $T_1$ and $T_2$ be sets whose cardinality $\card {T_1}$ and $\card {T_2}$ are both $n$.

Let $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ be the symmetric group on $S$ and $T$ respectively.


Then $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ are isomorphic.


Proof 1

Consider the symmetric group on $n$ letters $S_n$.

From Symmetric Group on n Letters is Isomorphic to Symmetric Group we have that:

$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $S_n$
$\struct {\map \Gamma {T_2}, \circ}$ is isomorphic to $S_n$

and hence from Isomorphism is Equivalence Relation:

$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $\struct {\map \Gamma {T_2}, \circ}$.

$\blacksquare$


Proof 2

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:

\(\ds \map \theta f \circ \map \theta g\) \(=\) \(\ds \paren {\alpha \circ f \circ \alpha^{-1} } \circ \paren {\alpha \circ g \circ \alpha^{-1} }\)
\(\ds \) \(=\) \(\ds \alpha \circ f \circ \paren {\alpha^{-1} \circ \alpha} \circ g \circ \alpha^{-1}\)
\(\ds \) \(=\) \(\ds \alpha \circ \paren {f \circ g} \circ \alpha^{-1}\)
\(\ds \) \(=\) \(\ds \map \theta {f \circ g}\)

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


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

Then:

\(\ds \map \theta f\) \(=\) \(\ds \map \theta g\)
\(\ds \leadsto \ \ \) \(\ds \alpha \circ f \circ \alpha^{-1}\) \(=\) \(\ds \alpha \circ g \circ \alpha^{-1}\) Definition of $\theta$
\(\ds \leadsto \ \ \) \(\ds \alpha^{-1} \circ \paren {\alpha \circ f \circ \alpha^{-1} } \circ \alpha\) \(=\) \(\ds \alpha^{-1} \circ \paren {\alpha \circ g \circ \alpha^{-1} } \circ \alpha\)
\(\ds \leadsto \ \ \) \(\ds \paren {\alpha^{-1} \circ \alpha} \circ f \circ \paren {\alpha^{-1} \circ \alpha}\) \(=\) \(\ds \paren {\alpha^{-1} \circ \alpha} \circ g \circ \paren {\alpha^{-1} \circ \alpha}\)
\(\ds \leadsto \ \ \) \(\ds f\) \(=\) \(\ds g\)

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:

\(\ds \alpha^{-1} \circ g \circ \alpha\) \(=\) \(\ds f\)
\(\ds \leadsto \ \ \) \(\ds \alpha \circ \paren {\alpha^{-1} \circ g \circ \alpha} \circ \alpha^{-1}\) \(=\) \(\ds \alpha \circ f \circ \alpha^{-1}\)
\(\ds \leadsto \ \ \) \(\ds g\) \(=\) \(\ds \alpha \circ f \circ \alpha^{-1}\)
\(\ds \) \(=\) \(\ds \map \theta f\)

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.

$\blacksquare$


Sources

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