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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]

Cyclic permutation of the coordinates in the preceding commutator calculation gives the remaining orbital angular momentum commutation relations:Reversing the order of a commutator reverses its sign.

The argument order in $p$ is a cyclic permutation. Undo it by rotating in the other direction. For $n\ge 1$, choose the lambda termUse fresh bound variables throughout. Substituting $qz$ for the first argument of $p$ and then performing beta reduction givesIndeed, the arguments received by $qz$ are $y_1=x_2,\dots ,y_{n-1}=x_n,y_n=x_1$, so its last argument moves back to the front. Apply eta conversion successively, starting with the innermost lambda abstraction. At each step the bound variable being removed is absent from the preceding expression. Thus the last expression is related to $z$ by beta-eta equivalence, and consequentlyFor the induction needed in the next part, also compute the reverse composition of lambda terms:So this $q$ is a two-sided inverse. The empty argument list at $n=0$ means $p=\lambda z.z$; choose $q=I$. For $n=1$, both displayed rotations give $\lambda z{y}_1.z{y}_1$, which is also related to $I$ by eta conversion. These boundary cases require no exceptional inverse construction.

Use structural induction on the construction of $p\in \mathcal{R}$, with the stronger induction claim that $p$ is an invertible lambda term: there is a lambda term $q$ satisfying bothThe base constructor has exactly the two-sided inverse proved in part (c). Strengthening the claim is useful because the inductive construction needs inverses of its constituent lambda terms in both orders.

For the inductive constructor, let $t_i$ be the two-sided inverse supplied for $s_i$. Thus, for a fresh variable $u$, beta-eta equivalence givesFor $n\ge 1$, defineThis lambda term reverses the cyclic permutation and applies the appropriate inverse to each argument. First use beta reduction to expand one order of composition of lambda terms:For the other order, beta reduction givesThese computations close the stronger induction. For $n=1$, each formula has the single argument involving $s_1$ and $t_1$; for $n=0$, the constructor is $I$ and its inverse is $I$. Every construction is finite, so the inverse is a finite lambda term. There is no requirement that this inverse itself belong to $\mathcal{R}$.

Save the last entry, shift the others right in descending index order, and put the saved value at the left. Descending order prevents overwriting an unread entry.For one pass, after the shifts at offsets greater than t, those positions contain the original preceding entries; offsets through t are still available, and x retains the original last entry. The loop variant t decreases to zero. The outer loop invariant is that the segment has been rotated right by step places relative to its initial state. Repeating a one-step cyclic permutation $k$ times therefore gives the required array rotation. The method uses one extra T variable and $O(kn)$ time, with $O(1)$ auxiliary space. The early return avoids all accesses for an identity rotation.

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.