# Is it true that the order of $ab$ is always equal to the order of $ba$?

How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?

For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?

• Why on earth are you assuming that the order of $ab$ is $2$? – Chris Eagle Dec 8 '12 at 11:14
• This question is related to math.stackexchange.com/questions/225942/… – user26857 Apr 22 '13 at 21:33
• See this question for answers that go more to the essence of the matter (conjugation). – Bill Dubuque Dec 20 '16 at 15:36

Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.

• Calling this «hand-waving» is quite misguided! – Mariano Suárez-Álvarez Apr 22 '16 at 7:33

Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.

Another hint is greyed out below (hover over with a mouse to display it):

Notice that $(ba)^{n+1} = b(ab)^na$.

• simple proof +1 – viru Apr 29 '18 at 5:19

If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.

• I think the OP should note that the orders of $a$ and $b$ are both finite. – mrs Nov 15 '12 at 20:58
• @BabakSorouh Why? The order of ab may be finite while those of a and b are infinite. – Did Nov 15 '12 at 21:11
• @GeoffreyCritzer Try $b=a^{-1}$. – Did Jun 5 '15 at 10:33
• @GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$. – Did Jun 5 '15 at 10:44
• @GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question. – Did Jun 5 '15 at 11:37

By associativity, $(ab)^p=a(ba)^{p-1}b$ for $p\geqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $p\geqslant 1$, $$(ab)^p=e\Leftrightarrow (ba)^p=e.$$

Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n= abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k $a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k

(1)

$$(ab)^n = e$$

$$\Rightarrow$$

$$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$$.

(2)

$$(ba)^n = e$$

$$\Rightarrow$$

$$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$$.

## protected by Zev ChonolesApr 22 '16 at 6:54

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