Every permutation can be written as a cycle or a product of disjoint cycles. This follows by considering the effect of a sequence of permutations on each member of the setEach element i will end up after a sequence of permutations in some other positionmeaning that somewhere is the simplified result, the sequence …, i, j, … must occur. There cannot be somewhere else a sequence of the form …, j, k, … or …, k, j, …since if this occurred, we would be able to simplify it.

Disjoint cycles commute. Ifandand the cycles have no element in common, then

Sinceandare disjoint we may writewhere the's are the symbols in neitheror

Supposeis permuted bythensincefixes all theterms.

Similarlysofor eachpermuted byA similar argument is made for terms permuted byand of course, bothandleave theterms fixed. Hence

The order of a permutation of disjoint cycles is the least common multiple of the lengths of the cycles. To prove this, observe that a cycle of lengthhas orderSuppose then thatand are disjoint cycles of lengthandand letbe the least common multiple ofand andare the identity permutation, and sinceandcommuteThen the order ofmust divideSuppose the order ofisthen sobut ifandare disjoint, then so areandso both must be equal toand bothanddividehence thatdividesalso and The argument can be extended to any number of cycles in an obvious way.

Every permutation can be written in cycle form as a 2 cycle or a product of 2 cycles.

Note thatand

Permutations are odd or even. A permutation is even if it can be written as an even number of 2 cycles and odd if it can be written as an odd number of 2 cycles. Since every permutation can be written as a product of two cycles, the result follows.

The set of even permutations offorms a subgroup ofTo prove it, write each elementas a product of twocycles.

Definethen the kernel ofis the set of all even permutations.

The product of even permutations is even, as is the identity, and the inverse of an even permutation, so that the set of all even permutations is a subgroup.