In Pinter's "A Book of Abstract Algebra", Chapter 10 Exercise C4 asks for the reader to prove:

${\rm ord}(bab^{-1})={\rm ord}(a)$

After playing around a little bit...and making use of the fact that $(bab^{-1})^n = ba^nb^{-1}$, I was able to prove the following implications:

  1. ${\rm ord}(bab^{-1})=q \implies a^q =e$

  2. ${\rm ord}(a) = p \implies (bab^{-1})^p=e$

I have proven that both of these statements are true. For convenience, I can reformulate these statements as follows:

  1. if ${\rm ord}(bab^{-1})=q$, then ${\rm ord}(a) \leq q$
  2. if ${\rm ord}(a)=p$, then ${\rm ord}(bab^{-1}) \leq p$

Now I am a little troubled as to how to proceed from here.

May I combine these if-then statements in the following way?

If ${\rm ord}(bab^{-1})=q$ and ${\rm ord}(a)=p$, then $p\leq q$ and $q\leq p$

Which can only be true if $p=q$...which means ${\rm ord}(bab^{-1})={\rm ord}(a)$

It is this last step where I combine the implications into one statement that has me a little worried. Is this logically precise?


  • 2
    $\begingroup$ yes, if $ \text{ord}(a)\le \text{ord}(bab^{-1})$ and $\text{ord}(bab^{-1})\le\text{ord}(a)$ then $\text{ord}(a)=\text{ord}(bab^{-1})$ $\endgroup$ – J. W. Tanner Nov 19 '19 at 1:03
  • 3
    $\begingroup$ Key Idea $ $ Isomorphisms preserve all "group-theoretic" properties, including 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.\ $ Ditto for cyclic permutations. $\endgroup$ – Bill Dubuque Nov 19 '19 at 1:14

As it stands, your proof is valid. Normally we wouldn't phrase it as "If $\text{ord}(bab^{-1}) = q$ and..." Instead we would say "let $\text{ord}(bab^{-1}) = q$ and $\text{ord}(a) = p$. Then [insert deductions] $p=q$." This is a minor quibble.

Perhaps a more direct way to solve this problem is to note that $(bab^{-1})^p = e$ if and only if $a^p = e$.


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