If for some group with a binary operation, (G, *), we have ord $a$ := $a^n = 1$ for some positive integer $n$, would it be correct to argue that:

Case 1 (finite order): ord $ab$ = ord $ba$:

$(ab)^n = a^n b^n = 1 \iff b^n a^n b^n = b^n \iff b^n a^n (b^n (b^{-1})^n)=b^n (b^{-1})^n \iff $ $b^n a^n = (ba)^n= 1$

Case 2 (infinite order): ord $ab$ = ord $ba$:

ord $ab \to \infty$ as $n \to \infty$, and therefore ord $ab$ = ord $ba$

Do you think my proof is reliable?

  • 3
    $\begingroup$ If the group is not abelian the formula $(ab)^n=a^nb^n$ does not necessarily hold. I'm afraid that ruins your calculation at its first step. Another thing worth emphasizing is that in the definition of order it is crucial that $n$ should be the smallest positive integer with the property $a^n=1$. Last but not least. Hint: try conjugation. $\endgroup$ Sep 21 '15 at 5:52

In general, $(ab)^n = a^n b^n$ is not true in groups.

For a finite order, I advise you to write

$$(ab)^n = abababab\dots ab = a(bababa\dots ba)a^{-1}$$

  • $\begingroup$ Would it be correct to argue, for the case where ord $a$ is infinite, that since $(ab)^n \to \infty$ and $(ba)^n \to \infty$ as $n \to \infty$, then ord $ab$ = ord $ba$? $\endgroup$
    – sequence
    Sep 21 '15 at 7:20
  • 1
    $\begingroup$ @sequence The definition of "$x$ has infinite order" is "$x$ has infinite order if $x^n \neq e$ for all integers $n$." Therefore, you need to prove that $(ba)^n\neq e$. $\endgroup$
    – 5xum
    Sep 21 '15 at 7:23
  • $\begingroup$ Then, if I'm correct, the same approach as above is also good for infinite order. $\endgroup$
    – sequence
    Sep 21 '15 at 7:54

Let $G$ be a group. The element order of $g\in G$ is defined as the group order of the cyclic subgroup

$\langle g\rangle=\{g^n|n\in\mathbb Z\}$.

We have $b\langle ab\rangle=\{b(ab)^n|n\in\mathbb Z\}=\{(ba)^nb|n\in\mathbb Z\}=\langle ba\rangle b$.

Remember that for a subgroup $H\subseteq G$, all cosets $gH$ and $Hg$ have the same order as $H$.

We conclude $\operatorname{ord}(ab)=|\langle ab\rangle|=|b\langle ab\rangle|=|\langle ba\rangle b|=|\langle ba\rangle|=ord(ba)$.

(Note that with this proof, we need not distinguish between the finite and the infinite case.)

Alternatively, if you don't like using cosets, there's another easy proof for both cases at once:

This proof is more suitable if your lecture uses the definition $\operatorname{ord}(g)=\min\{n\in\mathbb N|g^n=1\}$.

$ab$ and $ba$ are conjugates, since $a^{-1}(ab)a=ba$.

Consequently, we have $(ba)^n=(a^{-1}(ab)a)^n=a^{-1}(ab)^na\ \forall\ n\in\mathbb Z$.

We conclude $(ba)^n=1\ \Leftrightarrow\ a^{-1}(ab)^na=1\ \Leftrightarrow\ a(a^{-1}(ab)^na)a^{-1}=aa^{-1}\ \Leftrightarrow\ (ab)^n=1$.

Therefore $\operatorname{ord}(ab)$ and $\operatorname{ord}(ba)$ must coincide.


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