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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?

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

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    $\begingroup$ Calling this «hand-waving» is quite misguided! $\endgroup$ – Mariano Suárez-Álvarez Apr 22 '16 at 7:33
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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 $(ba)^{n+1} = b(ab)^na$.

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  • $\begingroup$ simple proof +1 $\endgroup$ – viru Apr 29 '18 at 5:19
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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.

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  • $\begingroup$ I think the OP should note that the orders of $a$ and $b$ are both finite. $\endgroup$ – mrs Nov 15 '12 at 20:58
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    $\begingroup$ @BabakSorouh Why? The order of ab may be finite while those of a and b are infinite. $\endgroup$ – Did Nov 15 '12 at 21:11
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    $\begingroup$ @GeoffreyCritzer Try $b=a^{-1}$. $\endgroup$ – Did Jun 5 '15 at 10:33
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    $\begingroup$ @GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$. $\endgroup$ – Did Jun 5 '15 at 10:44
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    $\begingroup$ @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. $\endgroup$ – Did Jun 5 '15 at 11:37
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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 $p\geqslant 1$, $$(ab)^p=e\Leftrightarrow (ba)^p=e.$$

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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

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(1)

$(ab)^n = e$

$\Rightarrow$

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

(2)

$(ba)^n = e$

$\Rightarrow$

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

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protected by Zev Chonoles Apr 22 '16 at 6:54

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