Product of disjoint cycles question.

Question

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!

Answer 1

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.

Answer 2

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 $n \ge 5$.