Symmetric Groups on Sets of the Same Cardinality are Isomorphic

Theorem

Let \(G_1 = \mathrm{Sym}(X_1)\) and \(G_2 = \mathrm{Sym}(X_2)\) be symmetric groups on sets \(X_1\) and \(X_2\) such that \(|X_1| = |X_2|\). Then \(G_2 \cong G_2\).

Proof

Since \(|X_1| = |X_2|\), we have a bijection \(\phi : X_1 \to X_2\). As such, we define our isomorphism \(\psi : G_1 \to G_2\) by

\[ \psi(f) = \phi \circ f \circ \phi^{-1}.\]

That is, we define a permutation of \(X_2\) by mapping into \(X_1\), permuting within \(X_1\), and then mapping back.

To show that this is a homomorphism, note that

\[\begin{align*} \psi(f \circ g) &= \phi \circ f \circ g \circ \phi^{-1} \\ &= \phi \circ f \circ \mathrm{id} \circ g \circ \phi^{-1} \\ &= \phi \circ f \circ \phi^{-1} \circ \phi \circ g \circ \phi^{-1} \\ &= \psi(f) \circ \psi(g). \\ \end{align*}\]

The fact that \(\psi\) is a bijection follows from the fact that it is a composition of three bijections.