The order of $bab^{-1}$ is equal to the order of $a$ for any $a,b \in G$ Let $G$ be a group. Prove that the order of $bab^{-1}$ is equal to the order of $a$, $\forall a,b \in G$

**Key Idea** $\ $ Isomorphisms preserve all "group-theoretic" properties, which includes the order of an element $\,g,\,$ since this equals the order (cardinality) of the cyclic group generated by $\,g.\,$ But an isomorphic image of a group has the same order (cardinality).

Yours is the special case of a _conjugation_ isomorphism $\ g\mapsto bgb^{-1},\, $ with inverse $\ g\mapsto b^{-1}gb.$