# Trouble finishing up the proof that ${\rm ord}(bab^{-1})={\rm ord}(a)$.

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?

Cheers.

• yes, if $\text{ord}(a)\le \text{ord}(bab^{-1})$ and $\text{ord}(bab^{-1})\le\text{ord}(a)$ then $\text{ord}(a)=\text{ord}(bab^{-1})$ – J. W. Tanner Nov 19 '19 at 1:03
• 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. – 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 $$a^p = e$$.