19

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?

Siyanda
  • 2,549
  • 10
  • 35
  • 49

7 Answers7

48

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.

Dan Shved
  • 15,862
  • 39
  • 55
39

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

Clive Newstead
  • 64,925
23

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.

Did
  • 279,727
  • I think the OP should note that the orders of $a$ and $b$ are both finite. – Mikasa Nov 15 '12 at 20:58
  • 8
    @BabakSorouh Why? The order of ab may be finite while those of a and b are infinite. – Did Nov 15 '12 at 21:11
  • @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 ab 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 ab=e, but I was hoping for an example where a does not equal b inverse. – Geoffrey Critzer Jun 05 '15 at 10:28
  • 2
    @GeoffreyCritzer Try $b=a^{-1}$. – Did Jun 05 '15 at 10:33
  • 1
    @GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$. – Did Jun 05 '15 at 10:44
  • @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? – Geoffrey Critzer Jun 05 '15 at 11:27
  • In these proofs we assumed |ab| is finite. Is it always true that if |ab| is infinite then |ba| is infinite as well. – Geoffrey Critzer Jun 05 '15 at 11:34
  • 1
    @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 05 '15 at 11:37
  • How did you get $(ab)^{n}a = a(ba)^{n}$ – Neel Sandell Dec 02 '21 at 12:23
5

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 $p\geqslant 1$, $$(ab)^p=e\Leftrightarrow (ba)^p=e.$$

Davide Giraudo
  • 172,925
2

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

Curious student
  • 6,583
2

(1)

$(ab)^n = e$

$\Rightarrow$

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

(2)

$(ba)^n = e$

$\Rightarrow$

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

tchappy ha
  • 8,690
  • does this work: Let $k$ be the order of $xy$ then $(xy)^k=x(yx)^{k-1}y=e$ $\Rightarrow$ $(yx)^{k-1}=x^{-1}y^{-1}=(yx)^{-1}$ thus the order of $yx$ divides the order of $xy$? and by symmetry were done? – homosapien Dec 21 '22 at 21:59
0

Let $x,g\in G$. By induction, $(g^{-1}xg)^i=g^{-1}x^ig$ for all $i\in \Bbb{N}$. So $(g^{-1}xg)^i=e\Leftrightarrow x^i=e$. Which implies $|x|=| g^{-1}xg|$. Put $x=ab$ and $g=a$. Thus $|ab|=|a^{-1}aba|=|ba|$.

user264745
  • 4,143