The order of $bab^{-1}$ is equal to the order of $a$ for any $a,b \in G$

Question

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

possible duplicate of Order of conjugate of an element given the order of this element — Bill Dubuque, Jan 18 '14 at 19:17

score 7 · Answer 1 · answered Jan 18 '14 at 18:35

7

Proof 1: $\phi(x)=bxb^{-1}$ (for a fixed $b$) is a group automorphism. So, order of $a$= order of $bab^{-1}$.

Proof 2: $(bab^{-1})^n=ba^nb^{-1}$ for all natural number $n$. This shows the above result.

answered Jan 18 '14 at 18:35

voldemort

13,182

score 1 · Accepted Answer · answered Jan 18 '14 at 19:18

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 $\ g\mapsto b^{-1}gb.$

The order of $bab^{-1}$ is equal to the order of $a$ for any $a,b \in G$

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