Proving that a function is an isomorphism of groups

Question

Let A and B be non-empty sets and f : A → B be a bijection. Consider the map $\phi$ : $S_A$ → $S_B$ that sends $\sigma$ to ${f} \circ {\sigma} \circ {f^{-1}}$. Show that $\phi$ is an isomorphism of groups.

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Answer 1

Since $f$ is a bijection, $|A|=|B|$ so $|S_A|=|S_B|$. If $\varphi(\sigma)=\varphi(\sigma')$, then $$f\circ\sigma\circ f^{-1}=f\circ\sigma'\circ f^{-1}, $$ so applying $f^{-1}$ on the left and $f$ on the right yields $\sigma=\sigma'$. If $\tau\in B$, then $f^{-1}\circ\tau\circ f\in A$ and $$\tau=\varphi(f^{-1}\circ\tau\circ f), $$ and hence $\varphi$ is a bijection. Now given $\sigma,\sigma'\in A$, we have $$\varphi(\sigma\sigma') = f\circ(\sigma\sigma')\circ f^{-1}=(f\circ\sigma\circ f^{-1})\circ(f\circ\sigma'\circ f^{-1})=\varphi(\sigma\sigma'), $$ so that $\varphi$ is an isomorphism.