permutation multiplication

[theverymooon]

hello, I am trying to figure out these permutations as a product of disjoint cycles....this should be easy and i have the answers for my questions but i don't see why they are the answers. i was wondering if anyone had some insight
a. (1235)(413) = (15)(234)
what i would think it would be is to start with 1, then starting from the right, as my teacher as said(although it could be done from the right)
1 -->3 then 3-->5 so to begin (15
then 5-->1 then 1-->3 so (153
then 3-->5 and 5 is fixed so (153)
now we have 2-->3 and 3-->4 so (24)
so the final answer would be (153)(24), why is this not right?

b. (13256)(23)(46512)=(124)(35)
1-->2, 2-->3 so (13 .....???????? but supposedly 1-->2?

i am going off of an example we had in class
(16458)(23)(7)(143)(275)(68)=(1536)(7842)

so look at 1 starting from right
1-->4, 4-->5 so (15
5-->2, 2-->3 so( 153
3-->1, 1-->6 so (1536
then 6-->8, 8-->1 so close (1536)
now start with 7
7-->5, 5-->8 so (78
8-->6, 6-->4 so (784
then 4-->5, 5-->2 so (7845)

could anyone help me understand where i am going wrong?

 

[daon]

Multiplication of permutations, in the books I've studied, are done from Right to Left. It is important to agree on this "small" choice, otherwise calculations are different.

Lets compare order with: (12345)(23)

From right to left: 2->3->4 .... so 2->4
From left to right: 2->3->2 .... so this product FIXES 2.

Right to Left makes more sense since "multiplication" here is actually function composition.

So using R->L as the multiplication direction, we have

1->3->5 (15
5->1 (15)
2->3 (15)(23
3->4- (15)(234
4->1->2 (15)(234)