just to add to ariels excellent writeup, I will exemplify the permutation multiplication in the cycle form.
Generally for permutation multiplication, we use the 2 row representation. However, the cycle representation can also be used.
Consider permutations f = (1 2 4 5 6) and g = (2 6 3 4 5). Their composition g.f can be found by multiplying the permutations right to left [note: some authors do it the other way around, but I find right to left more intutive since that corresponds to the way that we compose functions. i.e., if f and g are functions, then g.f = g (f (a)) ]
So,
g.f = (2 6 3 4 5) (1 2 4 5 6)
start with 1. locate 1 in the first set (starting from the right); if found, replace it by its permute; continue searching the replacement in subseqent cycles.
so 1->2 in the first cycle, and 2->6 in the next cycle, so overall 1->6. we write 6 as the first term of the product
(6
now we repeat the procedure for 6. 6->1 in the first cycle, and remains unaffected by the second cycle, so overall 6->1.
(6 1
we have reached the element we started out with initially (1), so we close this cycle and start a new one
(6 1) (
pick the next element 2. 2->4 and 4->5. so
(6 1) (5
continuing,
5->6 and 6->3;
3->3 and 3->4;
4->5 and 5->2;
(6 1) (5 3 4 2)
since all the elements are finished, we have our product.
Other things too are easier in the cycle representation. For example, to compute the inverse, just list the elements in the reverse order. i.e., (1 2 4) ^ -1 = (4 2 1)
Also, note that (1 2 3 4) = (1 4) (1 3) (1 2)