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I am confused since the definition of order says that $(ab)^n=e$, therefore does just showing that ab=ba solves both parts or is it two different solutions?

Assume $(ab)^n=e$, following the definition of order since n is a finite number.

$(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$

$=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$

$=a(ba)^{n-1} b$

From this, we get that $(ba)^{n-1}=a^{-1} b^{-1}=(ba)^{-1}$, therefore we have $(ba)^n=e$. Also proving that $ab=ba$.

Would this answer both of the questions above since then it show that both $ba^n=e$ and that the order of $ba$ would be $n$?

Hadi
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Heath
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  • Welcome to Mathematics Stack Exchange. $ab=ba$ does not always hold unless $G$ is Abelian – J. W. Tanner May 05 '20 at 02:13
  • "showing that $ab=ba$" But... no one said that the group was abelian. You can't show that $ab=ba$ because that could very well be false. – JMoravitz May 05 '20 at 02:14
  • Would I be able to prove that ab=ba with the information I am given? – Heath May 05 '20 at 02:15
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    proving $ab=ba$ seems too tough or rather impossible with the given info, try to write $(ab)^n= (ab)(ab)...(ab)=a(ba)(ba)...(ba)b$ – Mathronaut May 05 '20 at 02:16
  • This is what I have came up with so far::

    Assume (ab)^n=e, following the definition of order since n is a finite number. (ab)^n=(ab)(ab)(ab)………(ab)^(n-1) (ab)^n =a(ba)(ba)(ba)……(ba)b =a(ba)^(n-1) b

     – Heath May 05 '20 at 02:23
  • Maybe try entering that into your post rather than the comments. The formatting makes it very difficult to read. While you're at it, you might have a look at this MathJax Tutorial to make it even easier to read by properly formatting your equations with MathJax and $\LaTeX$. – JMoravitz May 05 '20 at 02:25
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    You concluded at the end "also proving $ab=ba$" This cannot be correct, because as mentioned, there exist non-abelian groups. What makes you believe that $ab=ba$ from what you have written? Because we need to correct this grave misconception before it causes more trouble down the road. – JMoravitz May 05 '20 at 02:33

1 Answers1

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You have $e = (ab)^n=a(ba)^{n-1}b$. From this you can multiply $a^{-1}$ from the left and $b^{-1}$ from the right to get $a^{-1}b^{-1}=(ba)^{n-1}\implies (ba)^{-1}=(ba)^{n-1}$ which gives you $(ba)^{n}=e$. From this you can conclude what the question is asking.

anumosh
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  • This shows the order of $ba$ divides $n$. Now suppose the order of $ba$ is $k\lt n$. Then $(ab)^k=e$, by interchanging $a$ and $b$, a contradiction. –  May 05 '20 at 03:44