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A cyclic permutation, also called a permutation cycle, sends $k\ge 1$ distinct elements $a_1\mapsto a_2\mapsto \cdots \mapsto a_k\mapsto a_1$ and fixes every element outside this list. Its notation is $(a_1{a}_2\dots a_k)$. For example, $(1342)$ is the permutation with image list $(3,1,4,2)$: all four labels lie on one directed loop. In contrast,has two separate loops, exchanging $1$ with $3$ and $2$ with $4$. It is a permutation composed of two transpositions, rather than a single cyclic permutation.

Starting at a different point of the same loop changes the written list but preserves the cyclic permutation: $(1342)=(3421)$. Reversing the direction generally changes the permutation; it gives the inverse function, here $(1243)$. A transposition is a cyclic permutation of length two; a one-element permutation cycle records a fixed point. In a cycle decomposition of a permutation, one-element permutation cycles are usually omitted, with the underlying set understood.

The beauty of this representation is that repeated rearrangements become motion around independent loops. Applying $(1342)$ four times returns every label to its start, while applying $(13)(24)$ twice already does so. Disjoint permutation cycles commute under function composition, since each moves only labels the others fix. On $n\ge 1$ labels, there are $(n-1)!$ cyclic permutations whose single loop contains every label: fix $1$ as the first written entry and order the other $n-1$ labels freely, giving a factorial. This is also the count of cyclic orderings; choosing a starting position would otherwise count each loop $n$ times. Rotations are identified here, while reversing direction is allowed to give a different cyclic ordering.

Proof outline for decomposition. Starting at any label in a finite set, repeatedly apply the permutation. The pigeonhole principle forces a repeated label. If the first repetition were at a label other than the starting one, its two predecessors would be different labels with the same image, contradicting the injective function property. This produces a permutation cycle returning to the start. Start again at a label not yet visited; the new permutation cycle cannot meet an earlier one, since the bijection gives every label a unique predecessor. Continue until every label belongs to a loop. The resulting cycle decomposition of a permutation is unique up to the order of the disjoint permutation cycles and the starting point written in each one. ^[ref]

A group action on a nonempty set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $g\cdot x=y$. There is exactly one group orbit: every position is reachable from every other. Requiring $X\ne \varnothing$ avoids calling the group action on the empty set transitive merely because the quantified statement has no points to test.^[ref]

Here a left group action of a group $G$ means a function $G\times X\to X$, written $(g,x)\mapsto g\cdot x$, satisfyingwhere $e$ is the identity element. Products move the point by $h$ first, then by $g$. Transitivity says that an orbit map $G\to X$, $g\mapsto g\cdot x$, is surjective for every $x\in X$; it suffices to check one chosen point.^[ref][ref]

For example, the symmetric group $S_3$ acts on $X=\{1,2,3\}$ by $\sigma \cdot i=\sigma (i)$. This is a transitive group action, but it is not a free group action: the stabilizer subgroup of $1$ is $\{e,(23)\}$. Both the transposition $(12)$ and the cyclic permutation $(123)$ send $1$ to $2$. In contrast, the cyclic group $C_3=\mathbb{Z}/3\mathbb{Z}$ acts on itself by $k\cdot x=k+x(\mathrm{mod}3)$, using modular arithmetic. The group element sending $x$ to $y$ is uniquely $k=y-x(\mathrm{mod}3)$. These two group actions reach the same number of points, but only the second has unique transport between them.

The stabilizer subgroup $G_x=\{h\in G:h\cdot x=x\}$ measures the ambiguity of transport. If $g_0\cdot x=y$, thenThe first set is a left coset, and the second identity says that stabilizer subgroups in one group orbit are conjugate subgroups. Briefly, $g\cdot x=g_0\cdot x$ exactly when $g_0^{-1}g\in G_x$. Thus, for a finite group in a transitive group action, the orbit-stabilizer theorem gives $|G|=|X||G_x|$, and there are exactly $|G_x|$ transports for each pair of points. The example counts are $6=3\cdot 2$ for $S_3$ and $3=3\cdot 1$ for $C_3$. This separates the number of reachable positions from the number of ways to reach each one.^[ref]

A transposition is a permutation that exchanges two elements and fixes every other element. Every transposition has order two.

A transposition is a cyclic permutation of length two.