I'm working on the following problem: Let $a = (1, 5, 3)(2, 4)$ and $b = (2, 4, 5)$ be permutations in $S_6$. Calculate the product $ba$. Answer with a product of disjoint cycles. $ba $
This is what I got, but it is wrong: $(2 4)(3 1 5)$
How can you think?
Unfortunately, the answer to this kind of question depends on whether you write a permutation on the left or the right of its argument - you'll need to check your textbook or course notes for which convention you are supposed to use. E.g., if $\theta = (1\,2\,3)(2\,4)$, then $\theta 2 = 4$, while $2\theta = 3$. How did I get that? Think of feeding $2$ through $\theta$ either from the right or from the left: working from the right, $(2\,4)$ maps $2$ to $4$ and $(1\,2\,3)$ leaves $4$ unchanged, while working from the left $(1\,2\,3)$ maps $2$ to $3$ and then $(2\,4)$ leaves $3$ unchanged. To get the disjoint cycle representation (having decided whether you are working from the left or the right), you fix a starting point, say $1$ and then keep passing it through $\theta$ until you get back to $1$. E.g., working form the right you get tyhat $\theta$ sends $1$ to $2$, $2$ to $4$, $4$ to $3$ and then $3$ back to $1$. So $\theta = (1\,2\,4\,3)$. In general, you might have missed one of the letters, so you'd carry on with the next letter that you haven't visited. (Apologies for the terminology: "letter" is the standard term for the things that permutations permute, but we usually write them as numbers.)
In your example, you have $a = (1\,5\,3)(2\,4)$ $b = (2\,4\,5)$. Using the above reasoning I get that:
$\quad ba = (1\,2\,5\,3)$ (working right to left, dropping the trivial cycle $(4)$ that the process above gives).
$\quad ba = (1\,5\,4\,3)$ (working left to right, again dropping the trivial cycle $(2)$).