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If $a, b$ are in group and $ab$ has finite order $n$, why does $ba$ have order $n$ as well?

Since $(ab)^n=e$, I get $(b)(ab)^n(a)= ba$. This means that $(ba)^{n+1}=ba$, and $(ba)^n=e$. But, I don't see how this follows if the group is not abelian.

How does the left side become $(ba)^{n+1}$ without being given that group is abelian?

apnorton
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user213770
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    Hint: conjugate – Bill Dubuque Feb 07 '15 at 21:24
  • You can break that into steps: $$(ab)^n=e$$ Then you multiply on the left by $b$: $$b(ab)^n=b$$ Then you multiply on the right by $a$: $$b(ab)^n a = ba.$$ You're not using that it's abelian in your proof. If you're careful about orders, you see that everything does, indeed, work out. – Milo Brandt Feb 07 '15 at 21:36

5 Answers5

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Hint:

$$ba=a^{-1}(ab)a$$

Conjugate elements.

Timbuc
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$ab$ and $ba$ have the same order, since they are conjugate elements: $$ b(ab)b^{-1} = ba, \tag{1} $$ giving: $$ \left(b(ab)b^{-1}\right)^k = b(ab)^k b^{-1} = (ba)^{k} \tag{2}$$ from which $o(ab)=o(ba)$.

Jack D'Aurizio
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Hint: for example if $(ab)^{2}=1$, then:

$$ (ba)^{3}=b(abab)a=ba\rightarrow (ba)^{2}=1 $$

Bombyx mori
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You don't need the group to be abelian to show that $b (ab)^n a = (ba)^{n + 1}$. Just write out the $a$'s and $b$'s on both sides to see that they're the same (try it for $n = 1$ and $n = 2$ for example).

Hew Wolff
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Hint $\ $ Cyclic permutation of products arise by conjugation $\ x\mapsto a^{-1} x a,\ $ e.g.

$$ bcd\cdots za\, =\, a^{-1}(abcd\cdots z) a$$

Conjugation is an isomorphism so preserves all "group-theoretic" properties, which includes the order of an element $\,g.\,$ Indeed, the order of $\,g\,$ is the order (cardinality) of the cyclic group generated by $\,g.\,$ But an isomorphic image of a group has the same order (cardinality).

Bill Dubuque
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