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<record version="2" id="11802">
 <title>order of products</title>
 <name>OrderOfProducts</name>
 <created>2009-05-25 16:26:53</created>
 <modified>2009-05-25 20:50:23</modified>
 <type>Theorem</type>
<parent id="2871">order (of a group)</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="20A05"/>
 </classification>
 <related>
	<object name="InverseFormingInProportionToGroupOperation"/>
 </related>
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 <content>If $a$ and $b$ are elements of a group, then both $ab$ and $ba$ have always the same order.\\

\emph{Proof.}\, Let $e$ be the indentity element of the group.\, For\, $n &gt; 1$,\, we have the 
\PMlinkname{equivalent}{Equivalent3} conditions
$$e \;=\; (ab)^n \;=\; \underbrace{(ab)(ab)\cdots(ab)}_{n} \;=\; a(ba)^{n-1}b,$$
$$a^{-1}b^{-1} \;=\; (ba)^{n-1},$$
$$(ba)^{-1} \;=\; (ba)^{n-1},$$
$$e \;=\; (ba)^n.$$
As for the infinite order, it makes the conditions false.\\


\textbf{Note.}\, More generally, all elements of any conjugacy class have the same order.</content>
</record>
