Product of Disjoint Cycles

I put them all into 2 line forms,

(12345). (12345). (12345)
(52341). (24135). (21345)

Then starting from the right, I put the middle perm, underneath the right one with the rows matching ie

(12345)
(21345)

(21345)
(42135)

Cancelled out the two middle rows to get the
(12345)
(42135)

Then put the left perm underneath

(12345)
(42135)

(42135)
(42531)

Canceled the middle rows again

(12345)
(42531)

Then turned this into the one line cycle.

I'm sure there must be an easier way, but I understand this!!

Thank you!

A much easier method is to "chase the numbers". That is, find where 1 goes to, then find where that goes to, and so on until you get back to 1. Then choose the smallest number not yet accounted for, and do the same.

For instance, say you have (123)(145)(25)(34).

What happens to 1? Well working from right to left we see that
$1 \xrightarrow{(34)} 1 \xrightarrow{(25)} 1 \xrightarrow{(145)} 4 \xrightarrow{(123)} 4$
so 1 goes to 4. What happens to 4? Similarly we get
$4 \to 3 \to 3 \to 3 \to 1$
so 4 goes back to 1 and this completes a cycle.

We haven't yet accounted for 2. We find $2 \to 2 \to 5 \to 1 \to 2$, so 2 is fixed.

Next up is 3. We find $3 \to 4 \to 4 \to 5 \to 5$, so 3 goes to 5. And then $5 \to 5 \to 2 \to 2 \to 3$, so 5 goes back to 3 and we get another cycle.

So (123)(145)(25)(34) = (14)(2)(35), or we could just write it as (14)(35).

You don't even need to write anything down to this steps, you can do the bit with all the arrows in your head and just fill in the disjoint cycles as you go along.

One question for u though, as it's asked for the product does it matter that I haven't got one cycle with all 5 elements in? And that I get the 2 on its own?