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Show that $\tau_1\tau_2$ of order $2$ or $3$, given that $\tau_1$ and $\tau_2$ are distinct transpositions.

I know that every cycle can be factored as a product of transpositions. But I'm not sure how to prove the order of $2$ or $3$.

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Write $\tau_1 = (a,b)$, $\tau_2=(r,s)$.

We have two cases:

  1. $\{a,b\}\cap\{r,s\} = \emptyset$. What happens to the product then?

  2. $\{a,b\}\cap\{r,s\}\neq\emptyset$. Then the two share exactly one element (Why?) so we can write $\tau_1 = (a,b)$, $\tau_2=(a,s)$. What is the product then?

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    @user12691: Since the two transpositions are *distinct*, they cannot both exchange the same two elements, so we cannot have $\{a,b\}=\{r,s\}$. As to what the element in common is, I don't know, and neither do you; it doesn't matter. Whichever it is they have in common, call it $a$.2011-12-15