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How can we prove that in a group, $ab$ and $ba$ have the same period?

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    possible duplicate of [Is it true that the order of $ab$ is always equal to the order of $ba$?](http://math.stackexchange.com/questions/238212/is-it-true-that-the-order-of-ab-is-always-equal-to-the-order-of-ba)2012-12-08

3 Answers 3

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Let $y\in G$. It is easy to verify that the map $\phi_y:G\rightarrow G$ that sends $x$ to $y^{-1}xy$ is a homomorhism. Thus, $|y^{-1}xy|=|\phi_y(x)|$ divides $|x|$. Similarly, $|x|=|\phi_{y^{-1}}(y^{-1}xy)|$ divides $|y^{-1}xy|$.

Hence, $|x|=|y^{-1}xy|$. Thus $|ab|=|a^{-1}aba|=|ba|$.

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    OK look the other answer is very good. it answers your question in a short way. Check it out2012-12-07
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If $(ab)^n=e$, $(ba)^n=b(ab)^nb^{-1}=e$. Conversely, if $(ba)^n=e$, then $(ab)^n=e$.

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    $(ba)^n=b(ab)^nb^{-1}=e$ I don't understand this bit.2012-12-07
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Firstly, we must not ignore the case that one element (say $ab$) has infinite period. Though that happens to be easy to handle: If $ba$ has finite period, say $(ba)^n=e$ then $(ab)^{n+1}=a(ba)^nb=ab$, a contradiction. Therefore, if one has infinite period, so does the other.

Now, we may assume both $ab$ and $ba$ have finite periods $m,n$ respectively. Without loss of generality, $m\leq n$. Then $(ba)^{m+1}=b(ab)^ma=ba$, whence $e=(ba)^{-1}(ba)^{m+1}=(ba)^m$ as desired. (Try to figure out why it was necessary to use the hypothesis $m\leq n$ in this paragraph.)