# Theorems of Cyclic Permutations

Theorem 1: The product of disjoint cycles is commutative.

Proof: Let $f$ and $g$ be any two disjoint cycles, i.e. there is no element common in two when they are expressed in one row notation. Therefore, the elements permuted by $f$ are invariant under $g$ and the elements permuted by $g$ are invariant under $f$.
Hence $f \circ g = g \circ f$ the product of disjoint cycles is commutative.

Theorem 2: Every permutation can be expressed as a composite of disjoint cycles.

Proof: Let the given permutation $f$ be denoted by the usual two row symbol within a bracket. Let $a$ be any element in the first row and $b$ the element in the second row exactly beneath $a$, i.e. $f\left( a \right) = b$. Similarly, let $f\left( b \right) = c$. Continuing this process, an element 1 may be found in the upper row such that its $f$ image is $a$. Then $a,b,c,…1$ is one circular permutation. If there are additional elements $a’,b’,$, etc., in the original permutation $f$, follow the above process to obtain another cycle $\left( {a’,b’,c’,…,1′} \right)$. Even now, if some element or elements are left in the original permutation this procedure can be repeated to the extent that all the elements of $f$ are exhausted. In this way the original permutation can be put as the product of disjoint cycles.

Theorem 3: Every permutation can be expressed as a product of transpositions.

Proof: To prove the above result, we shall first show that every cycle can be expressed as a composite of transpositions. Let us consider a cycle $\left( {{a_1},{a_2},…,{a_n}} \right)$ then
$\left( {{a_1},{a_2},…,{a_n}} \right) = \left( {{a_1}\,{a_n}} \right)\left( {{a_1}\,{a_{n – 1}}} \right)…\left( {{a_1}\,{a_2}} \right)$

We have already proved that every permutation can be expressed as a composition of disjoint cycles. Therefore in the light of the two results stated above, every permutation can be expressed as a product of transpositions.