# Proof: the product of disjoint cycles of a permutation is unique

Theorem 1, p. 82, in the book “A book of abstract algebra” by Charles C. Pinter states that “Every permutation is either the identity, a single cycle, or a product of disjoint cycles”. The theorem is proved in the book. After the proof, it is stated that it is easy to see that the product of cycles is unique, except for the order of factors. I have seen online proofs of this claim of uniqueness based on the concept of orbit. In what follows, an alternative proof is offered without relying on orbits.

Assume that there exists a permutation $\pi$ which can be written as a product of disjoint cycles in two ways. So, there are two collections, each consisting of disjoint cycles, $C=\left\{c_1,c_2,\dots,c_n\right\}$ and $S=\left\{s_1,s_2,\dots,s_m\right\}$ such that $C\ne S$ and $\pi=\prod_{i=1}^{n}c_i=\prod_{i=1}^{m}s_i$.

$C\ne S\Rightarrow C\not\subseteq S~\mbox{or}~S\not\subseteq C$. Without loss of generality, assume that $C\not\subseteq S$. Thus, $(\exists~\mbox{cycle}~c_i \in C)(\forall~\mbox{cycles}~s_k\in S) c_i\ne s_k$.

Let $a_q$ be some element in cycle $c_i$. Since $a_q\in c_i$ and the cycles in $C$ are disjoint, $a_q$ is permuted by $\pi$. Therefore, $\exists~\mbox{cycle}~s_l\in S$ such that $a_q\in s_l$.

The cycles $c_i$ and $s_l$ are disjoint from the rest of cycles in $C$ and $S$, respectively. Thus, $c_i(a_q)=s_l(a_q)=\pi(a_q)$.

Since $c_i\ne s_l$, there exists an element $a_p$ that breaks the equality arising from the successive implementation of permutation $\pi$ in each of the two cycles $c_i$ and $s_l$ starting from $a_q$. Thereby, there exists an element $a_p$ in $c_i$ and in $s_l$ with $c_i(a_p)\ne s_l(a_p)$. This is a contradiction, since the permutation $\pi$ is a function, so $a_p$ should be mapped to a unique element $\pi(a_p)$.