Symmetric Groups of Same Order are Isomorphic
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:
\(\ds \map \theta f \circ \map \theta g\) | \(=\) | \(\ds \paren {\alpha \circ f \circ \alpha^{-1} } \circ \paren {\alpha \circ g \circ \alpha^{-1} }\) | Definition of $\theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \circ f \circ \paren {\alpha^{-1} \circ \alpha} \circ g \circ \alpha^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \circ \paren {f \circ g} \circ \alpha^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \theta {f \circ g}\) | Definition of $\theta$ |
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 $\theta$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^{-1} \circ \paren {\alpha \circ f \circ \alpha^{-1} } \circ \alpha\) | \(=\) | \(\ds \alpha^{-1} \circ \paren {\alpha \circ g \circ \alpha^{-1} } \circ \alpha\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\alpha^{-1} \circ \alpha} \circ f \circ \paren {\alpha^{-1} \circ \alpha}\) | \(=\) | \(\ds \paren {\alpha^{-1} \circ \alpha} \circ g \circ \paren {\alpha^{-1} \circ \alpha}\) | Group Axiom $\text G 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds f\) | \(=\) | \(\ds g\) |
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:
\(\ds \alpha^{-1} \circ g \circ \alpha\) | \(=\) | \(\ds f\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha \circ \paren {\alpha^{-1} \circ g \circ \alpha} \circ \alpha^{-1}\) | \(=\) | \(\ds \alpha \circ f \circ \alpha^{-1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds g\) | \(=\) | \(\ds \alpha \circ f \circ \alpha^{-1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \theta f\) | Definition of $\theta$ |
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
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.6$