You can use the following rules (cycles act from the left on sequences and the notation for their action is defined so that these rules hold)
* A cycle: (a,b,c...,y,z) can be written as a product of 2-cycles via:
(a,b,c,...,y,z) = (ab)(bc)(c .) ... (. y)(yz)
* The terms in a cycle can be cycled and it's the same cycle:
(a,b,c,...y,z) = (b,c,...y,z,a) = (c,...y,z,a,b) = ... = (z,a,b,c,...,y)
( in particular (ab)=(ba) )
* Disjoint cycles commute:
(abcd)(pqr) = (pqr)(abcd) where none of p,q,r equal any of a,b,c,d.
Add to this that 2-cycles square to the identity (ab)(ab)=1 and you can always simplify permutation products.
There is a short cut you can derive from the above base rules. You can always split off a transpose (2-cycle) from either end of a general cycle (or recombine going the other way)
(a,b,c,...x,y)(y,z) = (abc...xyz) = (ab)(bc...xyz).
When combining a 2-cycle with a general cycle you must make sure here that the 2-cycle and general cycle have only one term in common and cycle both so it is positioned on the adjacent ends of each.
This is correct: (ab)(bcd) = (abcd)
but do not use: (ab)(bcad) != (abcad) as cycles with repeated terms are forbidden!
In such a case you want to factor and recombine cycle and combine again:
(ab)(bcad) = (ab)(bc)(cad) = (abc)(cad) = (bca)(cad) = (bc)(ca)(ca)(ad)=(bc)(ad).
More generally:
** you can split a cycle into a product of two cycles with one common term or append two cycles with one common term:
(abcd)(defgh...) = (abcdefgh...) With some practice you can get quite efficient at this permutation arithmetic. You can always continue until any common terms have been combined and you have a product of disjoint cycles.
Note the action of a cycle notated to obey these rules is for example:
(123)[abcd]
=(12)(23)[abcd]
=(12)[acbd] {(23) swapped 2nd and 3rd elements.}
=[cabd] {then (12) swapped first and 2nd elements.}
The net result is that the cycle (123) maps positions: 1-->2-->3-->1
Similarly for general cycles (abcd...z) maps a-->b-->c-->...-->z-->a.
(This is the notation you indicted.)
If the notion being used is the reverse of this then use a dot or star to distinguish the two notations
(123)* = (321)
and then just reverse to this notation solve and reverse back.
Example. Simplify:
P=(24)(1245)(35)(24)
=(24)(1245)(24)(35) {commuting disjoint cycles}
=(42)(2451)(24)(35) {cycle to get 2's adjacent ***}
=(42)(24)(451)(24)(35) {split off 24 transpose from 4-cycle since 2 terms in common.}
=(451)(24)(35) {square 2-cycle to identity}
=(514)(42)(35) {cycle to get 4's adjacent}
=(5142)(35) {combine cycles since only one common term}
=(1425)(53) {cycle to get 5's adjacent}
=(14253) {combine since only one common term.}
Done!
P = (14253)
*** you could have cycled to get 4's adjacent but since two terms are in common and adjacent you want to pull off both together. If two are in common but not adjacent that's fine. It will work out either way pulling out one at a time.
Some exercises:
Work out that (ax)(abcd)(ax) = (xbcd)
Work out that (ax)(abxyz)(xa) = (xbayz) (adjoint action of a transpose)
Work out that (abcd...z)(z...dcba) = 1 (reversed cycles are inverses)
Work out: (abc)(xaybzc)(cba)=? (adjoint action of a 3-cycle)
Work out: (123)(123) = (321)
Have fun.
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