Proving |ab|=|ba| in a Group G

[Jupiter]

I have to show that in a group G |ab|=|ba| for all a,b in G.

I don't have a clue how to proceed. I assume |ab|=n. I feel that using associativity, identity and inverse should be important. But I don't see how these tools will get me to |ba|=n.
I've considered proof by contradiciton, but that doesn't seem to be useful here.

Can someone please give me a hint?

 

[Hurkyl]

Well, I guess my first question is what you mean by |x|.

 

[Jupiter]

Sorry, I figured |ab| would be understood to mean the order of ab; that is, the least positive number n such that (ab)^n=e. I'm new in the study of algebra, so I do not know what notation is most widely recognized, or in what contexts.

 

[Hurkyl]

I thought that might be what you mean, but I wasn't sure; I've only seen that notation used for the order of a group, not the individual elements.


Anyways, try this:

Start with the equation $(ab)^n=e$. What can you do to this equation to turn the left hand side into $(ba)^n$?

Oh, and don't forget this fun bit of number theory: $a | b \wedge b | a \implies a = b$ (if both a and b are positive integers)

 

[phoenixthoth]

here's one way i think it could work:
prove $$a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}$$ (by induction, eg).

then $$\left( ba\right) ^{\left| ba\right| }=e\rightarrow \left( ab\right) ^{\left| ba\right| }=a\left( ba\right) ^{\left| ba\right| }a^{-1}=e\rightarrow \left| ab\right| |\left| ba\right|$$.

then prove $$\left( ba\right) ^{n}=a^{-1}\left( ab\right) ^{n}a$$ to get that $$\left| ba\right| |\left| ab\right|$$.

 

[NateTG]

Phoenixtoth: Your first line of LaTeX is not generally true. Although for your argument you could go with $$a(e)a^{-1}=e$$ which is obviously true.
Jupiter:
Perhaps you should look at some simple cases:

If $$|ab|=1$$ you should have little trouble.

Once you've figured that one out, try $$|ab|=2$$. You should be able to work it out from there.

I think you'll have to handle the $$|ab|=\infty$$ case seperately, and that it will involve some type of proof by contradiction.

 

[phoenixthoth]

nateTG,
if it's not true, can you find a counter example?

here's my proof by induction. can you spot an error?
$$a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}$$ is clear for n=0 if empty products are defined to equal e.

assume now that $$a\left( ba\right) ^{n}a^{-1}=\left( ab\right) ^{n}$$ for n>0. multiply the left hand side by $$a^{-1}$$ and the right hand side by $$a$$ to get this:
$$\left( ba\right) ^{n}=a^{-1}\left( ab\right) ^{n}a$$. then
$$a\left( ba\right) ^{n+1}a^{-1}=a\left( ba\right) ^{n}\left( ba\right) a^{-1}=a\left( a^{-1}\left( ab\right) ^{n}a\right) \left( ba\right) a^{-1}=\left( ab\right) ^{n+1}$$.

 

[NateTG]

Sorry. Dyslexia strikes again. I thought you had:
$$a(ab)^na^{-1}=(ab)^n$$ which is not generally true.