📂Abstract Algebra

Orbits, Cycles, and Permutations in Abstract Algebra

Definitions ¹

1. The equivalence classes of $\sim$ are called the Orbits of $\sigma$.
2. A permutation that has at most one orbit with more than one element is called a Cycle.
3. Among the orbits a cycle has, the orbit with the largest cardinality is called the Length of the cycle.
4. A cycle with length $2$ is called a Transposition.
5. Orbits corresponding to a cycle that do not share elements are called Disjoint.

Explanation

It’s normal to not understand just by definitions, let’s look into some actual examples.

Orbit

Considering the permutation $$\sigma = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 & 8 & 6 & 7 & 4 & 1 & 5 & 2 \end{bmatrix}$$ from $S_{8}$. This expression represents $$1 \to 3 \to 6 \to 1 \\ 2 \to 8 \to 2 \\ 4 \to 7 \to 5 \to 4$$ . Therefore, the equivalence relation $\sim$ determines the following three equivalence classes. $$\left\{ 1, 3, 6 \right\} \\ \left\{ 2, 8 \right\} \\ \left\{ 4 , 5 , 7 \right\}$$

Cycle

Considering the permutation $$\mu_{1} = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 5 & 1 & 4 \end{bmatrix}$$ from $S_{5}$. This permutation is represented as $1 \to 3 \to 5 \to 4 \to 1$, excluding the unchanged $2$, it can also be represented simply as $(1,3,5,4)$. It is important to note that while using this representation, order matters so that $(1,3,5,4) = (3,5,4,1)$ but not $(1,3,5,4) \ne (1,5,3,4)$. Also, considering $$\mu_{2} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix}$$ , since $(1,2)$ does not even represent the presence of $3$, it should be clearly stated in $S_{3}$ that it is $(1,2)$.

Length

There are only two orbits of the cycle $$\mu_{1} = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 5 & 1 & 4 \end{bmatrix}$$ , which are $$\left\{ 1,3,4,5 \right\} \\ \left\{ 2 \right\}$$ . Therefore, since $| \left\{ 1,3,4,5 \right\} | = 4$ and $| \left\{ 2 \right\} | =1$, the length of $\mu_{1}$ becomes $4$.

Transposition

The cycle $$\mu_{2} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix} = (1,2)$$ has a length of $2$, so it’s a transposition. In simple terms, it’s a cycle that just swaps two elements. Generally, $$(1,2, \cdots , n) = (1, n) (1, n-1 ) \cdots (1,3) (1,2)$$ can be represented. If one wishes to base it on $3$, $$(1,2, \cdots , n) = (3, 4, \cdots , n , 1, 2 ) = (3 , 2) (3, 1) \cdots (3,4)$$ can be changed accordingly. This is a very useful property to know.

Disjoint

Considering $$\sigma = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 & 8 & 6 & 7 & 4 & 1 & 5 & 2 \end{bmatrix} = (1,3,6) (2,8) (4,7,5)$$ , the three cycles $(1,3,6)$, $(2,8)$, and $(4,7,5)$ have corresponding orbits that do not share elements, making them disjoint. What can be understood from this representation is that it’s perfectly fine to be represented as $$(1,3,6) (2,8) (4,7,5) = (4,7,5) (2,8) (1,3,6)$$ . Permutations can be represented as the product of cycles, and if such products are considered the same, then the orbits are uniquely determined.

Theorem

• [1]: Every permutation of a finite symmetric group with more than one element can be represented as the product of transpositions.
• [2]: Every permutation of a finite symmetric group can be represented as the product of disjoint cycles.