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Consider the following permutations $x$ and $y$ in $S_6$:

$x=(1 \, 3 \, 5)(2 \, 4)$ and $y=(2 \, 3 \, 4 \, 5)$

Express $xy$ as a product of disjoint cycles.

My attempt: I first got $xy = (3 \, 5 \, 2 \, 1 \, 4)$ but realized that this is in $S_6$. I resolved and obtained $xy = (3 \, 5 \, 2 \, 1 \, 4 \, 6)$. But the answer says $xy = (1 \, 3 \, 2 \, 5 \, 4)$. What am I doing wrong?

Thank you guys!

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    $\begingroup$ The multiplication in your book must be done from right to left , as it is usually done. You did it the other way around, though... $\endgroup$ – DonAntonio Apr 2 '13 at 11:36
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The permutation $xy$ is just the application of $y$ and then $x$. So, you will see that $1 \rightarrow 1 \rightarrow 3$, $2 \rightarrow 3 \rightarrow 5$ and so on. You should eventually get the correct answer in the textbook.

I'm not sure how you got $(3\ 5\ 2\ 1\ 4\ 6)$ since neither $x$ nor $y$ moves 6.

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  • $\begingroup$ Apparently the OP multiplies cycles from left to right, against the huge majority of mathematicians' agreement. Yet there are some authors who do it that way. $\endgroup$ – DonAntonio Apr 2 '13 at 11:34
  • $\begingroup$ @DonAntonio A lot of group theorists do it that way. $\endgroup$ – Alexander Gruber Apr 2 '13 at 23:21
  • $\begingroup$ Some, a lot...I know only one who did that way (left to right), and nobody else, as that'd be tantamount to take the composition of functions $\,f\circ g\,$ from left to right, which imo is logical only if one writes functions of $\,x\,$ as $\,(x)f\,$ instead of the much more usual $\,f(x)\,$...and there aren't already many that follow this practice compared with those that don't follow it, imo. $\endgroup$ – DonAntonio Apr 2 '13 at 23:42
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There is a convention, that if a number does not appear in the disjoint cycle decomposition, then this means it is fixed. In other words, cycles of length $1$ are omitted.

So $$ x=(1 \, 3 \, 5)(2 \, 4) = (1 \, 3 \, 5)(2 \, 4) (6), \qquad y=(2 \, 3 \, 4 \, 5) = (1) (2 \, 3 \, 4 \, 5) (6). $$

With this convention, the result $$ xy = (3 \, 5 \, 2 \, 1 \, 4) = (3 \, 5 \, 2 \, 1 \, 4) (6) $$ can be regarded as an element of $S_6$. Actually, of any $S_{n}$, for $n \ge 5$.

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