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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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    Why on earth are you assuming that the order of $ab$ is $2$?2012-12-08
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    This question is related to http://math.stackexchange.com/questions/225942/a-b-in-g-has-finite-order-then-is-the-order-of-ab-ba-a-1b-1-with2013-04-22
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    See [this question](http://math.stackexchange.com/questions/1086198/deduce-lvert-ab-rvert-lvert-ba-rvert-for-all-a-b-in-g-where-g-is-a-gro/1086267#1086267) for answers that go more to the essence of the matter (conjugation).2016-12-20

6 Answers 6

35

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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    Calling this «hand-waving» is quite misguided!2016-04-22
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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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    simple proof +12018-04-29
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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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    I think the OP should note that the orders of $a$ and $b$ are both finite.2012-11-15
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    @BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.2012-11-15
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    @Did. I would like to know of a specific example of a group with elements a and b with infinite order but the order of a*b is finite. I am not disputing what you are stating. I am a beginning student of algebra and I need " a good stock of examples". I can see that if a and b are inverses of each other then a*b=e, but I was hoping for an example where a does not equal b inverse.2015-06-05
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    @GeoffreyCritzer Try $b=a^{-1}$.2015-06-05
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    @GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$.2015-06-05
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    @Did So b transforms (x,y) to (y,x-1). Right?. |a| is infinite, |b| is infinite yet |b*a| =2. Also I think in the group of nonzero complex numbers under multiplication, a = i/2 and b=2i where i is the imaginary unit would be another example. Right?2015-06-05
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    In these proofs we assumed |ab| is finite. Is it always true that if |ab| is infinite then |ba| is infinite as well.2015-06-05
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    @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.2015-06-05
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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)

$(ba)^n = e$

$\Rightarrow$

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