Permutation notation

Created by Watchduck

This article examines different notations for the composition of permutations with each other and with vectors.

Permutations are bijections from a set to itself, and the set does not need to have an order.
But they can also be described as operations that move things from places to other places — which is the natural mental image when the permutation is e.g. a rotation of a cube.
In the end it does not matter what kind of mental image is used to understand what a permutation is or does, as long as there is no ambiguity what result a formula will yield.
When a permutation $\pi$ is interpreted as moving objects from places to other places, there are two ways to describe it.

Visualisations of $P$ as an active permutation

Arrow diagram of $P$
Matrix representations of $P$
	Left (prefix) $M{\bigl (}\pi (i),i{\bigr )}=1$	Right (postfix) $M{\bigl (}i,\pi (i){\bigr )}=1$
Map
Cycle
There are two ways to assign a matrix to a permutation, corresponding to the two ways of multiplying permutations. There are two ways to draw arrows in the chosen matrix, one similar to two-line and the other to cycle notation. (The former is used in the blue boxes 14 and 15, the latter in the rest of the article.)

The usual way is as an active permutation or map or substitution:

$\pi$ moves an object from place $i$ to place $\pi (i)$. In the arrow diagram the one-line notation denotes where the arrows go.

The result of applying $\pi (i)$ on a vector $(x_{1},\dots ,x_{n})$ is $(x_{\pi ^{-1}(1)},\dots ,x_{\pi ^{-1}(n)})$.

But $\pi$ may also be represented by the result of applying it to the natural order. This is called a passive permutation or (re)arrangement or ordering:

$\pi$ replaces an object in position $i$ by that in position $\pi (i)$. In the arrow diagram the one-line notation denotes where the arrows come from.

In this case the result of applying $\pi (i)$ on a vector $(x_{1},\dots ,x_{n})$ is $(x_{\pi (1)},\dots ,x_{\pi (n)})$.

Confusing these two interpretations will lead to confusing permutations with their inverses.
Active and passive transformations seem to be a related concept.

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Composition of permutations is associative, but not commutative.

With prefix notation or left action $\pi \sigma (x)=\pi {\bigl (}\sigma (x){\bigr )}$.

$\pi \sigma$ can be thought of as $\pi ~{\scriptstyle after}~\sigma$, and $\pi x$ as $\pi ~{\scriptstyle of}~x$

This notation corresponds to the usual way of writing function composition.

With postfix notation or right action $\pi \sigma (x)=x\pi \sigma =\sigma {\bigl (}\pi (x){\bigr )}$.

$\pi \sigma$ can be thought of as $\pi ~{\scriptstyle before}~\sigma$, and $x\pi$ as (image of) $x~{\scriptstyle under}~\pi$.

This notation is common in group theory.

(In the Python examples in this article (P*Q)(2) = 2^P^Q = Q(P(2)). The result is R(2) = 4).

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In its graphics this article shows all possible interpretations, including the passive ones.
But to avoid confusion the accompanying calculations use only active right notation, which is also used by SymPy and Wolfram.

Notations

$\alpha ={\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}},~\beta ={\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}},~v=(v_{1},v_{2},v_{3}),~~~~~~~~~~~\alpha ^{-1}={\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}}$

Active

If permuting $v$ by $\alpha$ gives $(v_{\alpha ^{-1}(1)},v_{\alpha ^{-1}(2)},v_{\alpha ^{-1}(3)})=(v_{2},v_{3},v_{1})$ the permutation is active.

Active left

AL

If $\alpha \beta ={\begin{pmatrix}2&1&3\\1&3&2\end{pmatrix}}{\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}}={\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}}$ the notation should be seen as active left. * $\alpha \beta$ means $\alpha ~{\scriptstyle after}~\beta$ and can be written $\alpha (\beta )$.      $\alpha (1)$ can be written $\alpha 1$. Permuting $v$ by $\alpha$ may be written in different ways: $~~v_{\alpha ^{-1}}~~=~~v(\alpha ^{-1})~~=~~\alpha .v~~=~~v\alpha ^{-1}$ The notation with the dot is just an abbreviation one may define. With two permutations it means: $\alpha .\beta ~~=~~(\alpha \beta ^{-1})^{-1}~~=~~\beta \alpha ^{-1}$

Active right

AR

If $\alpha \beta ={\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}}{\begin{pmatrix}3&1&2\\3&2&1\end{pmatrix}}={\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}}$ the notation should be seen as active right. * $\alpha \beta$ means $\alpha ~{\scriptstyle before}~\beta$ and can be written $\beta (\alpha )$.      $\alpha (1)$ can be written $1\alpha$. Permuting $v$ by $\alpha$ may be written in different ways: $~~v_{\alpha ^{-1}}~~=~~v(\alpha ^{-1})~~=~~v.\alpha ~~=~~\alpha ^{-1}v$      (Compare box 12a.) The abbreviation with the dot corresponds to the function permute in Wolfram. With two permutations it means: $\alpha .\beta ~~=~~(\alpha ^{-1}\beta )^{-1}~~=~~\beta ^{-1}\alpha$

Passive

If permuting $v$ by $\alpha$ gives $(v_{\alpha (1)},v_{\alpha (2)},v_{\alpha (3)})=(v_{3},v_{1},v_{2})$ the permutation is passive.

Passive left should be seen as a misunderstanding of active right, and passive right should be seen as a misunderstanding of active left. ^*

*   (Unless permuting a vector proofs otherwise.)

Examples with 5-element permutations

Permutations can be combined with:

• other permutations (boxes 1 to 11)
• vectors of the same length with arbitrary elements (green boxes 12 and 16)
• vectors of arbitrary length with elements from the same range,
  i.e. maps with the same codomain as the permutation, but bijectivity not required (blue boxes 14 and 15)
• scalars from the same range (red boxes 13 and 17, compare blue box 14)

The variables used in the main examples are:

• the permutations $P={\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}=(0132)$ and $Q={\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}=(04)(13)$
• the vector $V=(V_{0},\dots ,V_{4})$ (in the graphics simply $0\dots 4$)
• the vector $E=(0,0,4,2,3,2)$

In the Wolfram calculations the 1-based equivalents are used.

Permutations of four elements are identified by their index numbers in RevCoLex order, i.e. as the $n$-th finite permutation.
$P=(0132)$ is denoted ${\mathit {10}}$ where it appears among other permutations of only four elements. (See box 19, compare with box 1.)
(Equivalents for the permutations of five elements would be $Q={\mathit {119}}$, $R={\mathit {109}}$ and $S={\mathit {83}}$ , but only the letters are used.)

Box 0 Initialisation for the computations

Python: Permutations in SymPy should be interpreted as AR (the documentation states that "p*q means first apply p, then q"). But the results do not discourage the interpretation as PL. The simple function times gives the same results. The result of using it on the non-bijective map $E$ suggests the interpretation as AR, but that of permuting the vector $V$ suggests the interpretation as PL. from sympy.combinatorics import Permutation def times(positions, sequence): return [sequence[i] for i in positions] def inv(perm): length = len(perm) if sorted(perm) != list(range(length)): print('Inverse not possible!') else: inverse = [0] * length for key, val in enumerate(perm): inverse[val] = key return inverse p = [1, 3, 0, 2, 4] q = [4, 3, 2, 1, 0] r = times(p, q) P = Permutation(p) Q = Permutation(q) R = P * Q v = ['0', '1', '2', '3', '4'] e = [0, 0, 4, 2, 3, 2] Wolfram: Permutations in Mathematica are 1-based: p = {2,4,1,3,5} q = {5,4,3,2,1} v = {v1,v2,v3,v4,v5} e = {1,1,5,3,4,3}

Active vs. passive

A simple rule to avoid passive permutations: If the rows of a permutation matrix are labelled, the cycles must go clockwise (AR). If the columns are labelled, the cycles must go anticlockwise (AL).

Box 1 Ambiguity of $P\cdot Q=R$
Passive (PL):     $P~{\scriptstyle after}~Q~=~R$		Active (AR):     $P~{\scriptstyle before}~Q~=~R$
If it is only required, that $P\cdot Q=R$, and no one intends to permute vectors, there is no need to choose an interpretation. In other terms: It only matters that the 1s in the permutation matrices are at ${\bigl (}i,\pi (i){\bigr )}$, and the question if the arrows go clockwise (AR) or anticlockwise (PL) is irrelevant. If on the other hand it is only required that a permutation permutes the vector $V$ into $Y$, it does dot matter if it is labelled $R$ or $R^{-1}$. In other terms: As long as the arrow goes from 0 to 3, it does not matter if it is labelled 3 (AR) or 0 (PR). While the permutations on the left and right have the same labels and do different things, the permutations below do exactly the same as those above them, but are labelled differently. (Compare box 19, showing the same for 4-element permutations.) Box 1a Ambiguity of $P^{-1}\cdot Q^{-1}=R^{-1}$ Active (AL):     $P^{-1}~{\scriptstyle after}~Q^{-1}~=~R^{-1}$ Passive (PR):     $P^{-1}~{\scriptstyle before}~Q^{-1}~=~R^{-1}$ In these matrices the 1s are in the positions ${\bigl (}\pi (i),i{\bigr )}$. They are exactly as those above, except that rows instead of columns are labelled.

Left vs. right

Box 2 Ambiguity of $P\cdot Q$ with active permutations
Left (AL):     $P~{\scriptstyle after}~Q~=~S$		Right (AR):     $P~{\scriptstyle before}~Q~=~R$

Box 3 Ambiguity of $P\cdot Q$ with passive permutations
Left (PL):     $P~{\scriptstyle after}~Q~=~R$		Right (PR):     $P~{\scriptstyle before}~Q~=~S$

The following examples show with arrow diagrams what the composed permutations actually do, and how this can be interpreted as a product in two different ways, corresponding to two different matrix multiplications.

The two arrangements of matrices are symmetric to each other, including the direction of the arrows in the matrices.
In other terms: The matrix image on the left and on the right show exactly the same thing in a different way.
The arrows in both arrangements of matrices are the same as in the arrow diagram to their left.

In prefix notation (left action) $\alpha \cdot \beta$ means $\alpha ~{\scriptstyle after}~\beta$. In postfix notation (right action) $\beta \cdot \alpha$ means $\beta ~{\scriptstyle before}~\alpha$.
Obviously $\alpha ~{\scriptstyle after}~\beta$ and $\beta ~{\scriptstyle before}~\alpha$ are just different ways to say the same thing.

Active

Box 4 Formulas for $R$ and $S$ in two-line and cycle notation
AL:     $\alpha \beta =\alpha ~{\scriptstyle after}~\beta$	AR:     $\beta \alpha =\beta ~{\scriptstyle before}~\alpha$
${\begin{aligned}QP&~~=~~{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}\\&~~=~~{\begin{pmatrix}1&3&0&2&4\\3&1&4&2&0\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\3&1&4&2&0\end{pmatrix}}~~=~~R\end{aligned}}$	${\begin{aligned}PQ&~~=~~{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}{\begin{pmatrix}1&3&0&2&4\\3&1&4&2&0\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\3&1&4&2&0\end{pmatrix}}~~=~~R\end{aligned}}$
${\begin{aligned}PQ&~~=~~{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}\\&~~=~~{\begin{pmatrix}4&3&2&1&0\\4&2&0&3&1\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\4&2&0&3&1\end{pmatrix}}~~=~~S\end{aligned}}$	${\begin{aligned}QP&~~=~~{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4\\1&3&0&2&4\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\4&3&2&1&0\end{pmatrix}}{\begin{pmatrix}4&3&2&1&0\\4&2&0&3&1\end{pmatrix}}\\&~~=~~{\begin{pmatrix}0&1&2&3&4\\4&2&0&3&1\end{pmatrix}}~~=~~S\end{aligned}}$
The top left row is reordered like the bottom right, so the bottom left row becomes the result.	The top right row is reordered like the bottom left, so the bottom right row becomes the result.
$QP~=~(04)(13)\cdot (0132)~=~(0324)~=~R$	$PQ~=~(0132)\cdot (04)(13)~=~(0324)~=~R$
$PQ~=~(0132)\cdot (04)(13)~=~(0412)~=~S$	$QP~=~(04)(13)\cdot (0132)~=~(0412)~=~S$

Box 5 Computation (AR only)

Box 5a Python >>> Permutation.print_cyclic = False >>> P Permutation[1, 3, 0, 2, 4] >>> Q Permutation[4, 3, 2, 1, 0] >>> ~P Permutation[2, 0, 3, 1, 4] >>> ~Q Permutation[4, 3, 2, 1, 0] >>> ~Q * ~P Permutation[4, 1, 3, 0, 2] >>> P * Q Permutation[3, 1, 4, 2, 0] >>> Q * P Permutation[4, 2, 0, 3, 1] >>> Permutation.print_cyclic = True >>> P Permutation(4)(0, 1, 3, 2) >>> Q Permutation(0, 4)(1, 3) >>> ~P Permutation(4)(0, 2, 3, 1) >>> ~Q Permutation(0, 4)(1, 3) >>> ~Q * ~P Permutation(0, 4, 2, 3) >>> P * Q Permutation(0, 3, 2, 4) >>> Q * P Permutation(0, 4, 1, 2)     >>> p [1, 3, 0, 2, 4] >>> q [4, 3, 2, 1, 0] >>> inv(p) [2, 0, 3, 1, 4] >>> inv(q) [4, 3, 2, 1, 0] >>> times(inv(q), inv(p)) [4, 1, 3, 0, 2] >>> times(p, q) [3, 1, 4, 2, 0] >>> times(q, p) [4, 2, 0, 3, 1] Box 5b Wolfram I: p O: {2,4,1,3,5} I: q O: {5,4,3,2,1} I: PermutationCycles[p] O: Cycles[{{1,2,4,3}}] I: PermutationCycles[q] O: Cycles[{{1,5},{2,4}}] I: PermutationProduct[InversePermutation[q], InversePermutation[p]] (* R^-1 *) O: {5,2,4,1,3} I: PermutationProduct[p, q] (* R *) O: {4,2,5,3,1} I: PermutationProduct[q, p] (* S *) O: {5,3,1,4,2}

Box 6 $R^{-1}$
First doing $Q^{-1}$ and then doing $P^{-1}$ is the same as doing $R^{-1}$. AL:     $P^{-1}~{\scriptstyle after}~Q^{-1}~=~R^{-1}$ AR:     $Q^{-1}~{\scriptstyle before}~P^{-1}~=~R^{-1}$

Box 7 $R$
First doing $P$ and then doing $Q$ is the same as doing $R$. AL:     $Q~{\scriptstyle after}~P~=~R$ AR:     $P~{\scriptstyle before}~Q~=~R$

Box 8 $S$
First doing $Q$ and then doing $P$ is the same as doing $S$. AL:     $P~{\scriptstyle after}~Q~=~S$ AR:     $Q~{\scriptstyle before}~P~=~S$

Passive

Box 9 $R^{-1}$
First doing $P^{-1}$ and then doing $Q^{-1}$ is the same as doing $R^{-1}$. PL:     $Q^{-1}~{\scriptstyle after}~P^{-1}~=~R^{-1}$ PR:     $P^{-1}~{\scriptstyle before}~Q^{-1}~=~R^{-1}$

Box 10 $R$
First doing $Q$ and then doing $P$ is the same as doing $R$. PL:     $P~{\scriptstyle after}~Q~=~R$ PR:     $Q~{\scriptstyle before}~P~=~R$

Box 11 $S$
First doing $P$ and then doing $Q$ is the same as doing $S$. PL:     $Q~{\scriptstyle after}~P~=~S$ PR:     $P~{\scriptstyle before}~Q~=~S$

Active

Box 12 Permuting a vector

Passive equivalent Box 12a Formulas and computation ${\begin{aligned}&P^{-1}V&~=~&V.P&~=~&V_{P^{-1}}&~=~&(V_{P^{-1}(0)},\dots ,V_{P^{-1}(4)})&~=~&(V_{2},V_{0},V_{3},V_{1},V_{4})&&&~=~&X\\&Q^{-1}X&~=~&X.Q&~=~&X_{Q^{-1}}&~=~&(X_{Q^{-1}(0)},\dots ,X_{Q^{-1}(4)})&~=~&(X_{4},X_{3},X_{2},X_{1},X_{0})&~=~&(V_{4},V_{1},V_{3},V_{0},V_{2})&~=~&Y\\\\&R^{-1}V&~=~&V.R&~=~&V_{R^{-1}}&~=~&(V_{R^{-1}(0)},\dots ,V_{R^{-1}(4)})&~=~&(V_{4},V_{1},V_{3},V_{0},V_{2})&&&~=~&Y\end{aligned}}$ Python: >>> x = times(inv(p), v) >>> x ['2', '0', '3', '1', '4'] >>> times(inv(q), x) # Y ['4', '1', '3', '0', '2'] >>> times(inv(r), v) # Y ['4', '1', '3', '0', '2'] Wolfram:   I: x = Permute[v, p] O: {v3,v1,v4,v2,v5} I: Permute[x, q] (* Y *) O: {v5,v2,v4,v1,v3} I: Permute[v, r] (* Y *) O: {v5,v2,v4,v1,v3}

Box 13 Concatenation with a scalar as permuting a vector
Passive equivalent Box 13a Computation Python: >>> times([0], p) [1] >>> times([1], q) [3] >>> times([0], r) [3] Wolfram: I: PermutationReplace[1, p] O: 2 I: PermutationReplace[2, q] O: 4 I: PermutationReplace[1, r] O: 4

Box 14 Concatenation with a scalar as concatenation with a constant

Box 15 Concatenation with a non-bijective map

$E=(0,0,4,2,3,2)$ ${\begin{aligned}RE&~~=~~{\Bigl (}R{\bigl (}E(0){\bigr )},\dots ,R{\bigl (}E(5){\bigr )}{\Bigr )}\\&~~=~~{\bigl (}R(0),R(0),R(4),R(2),R(3),R(2){\bigr )}\\&~~=~~(3,3,0,4,2,4)~~=~~G\end{aligned}}$ ${\begin{aligned}ER&~~=~~(0ER,\dots ,5ER)\\&~~=~~(0R,0R,4R,2R,3R,2R)\\&~~=~~(3,3,0,4,2,4)~~=~~G\end{aligned}}$ Box 15a Computation Python: >>> f = times(e, p) >>> f [1, 1, 4, 0, 2, 0] >>> times(f, q) # G [3, 3, 0, 4, 2, 4] >>> times(e, r) # G [3, 3, 0, 4, 2, 4] Wolfram:   I: f = PermutationReplace[e, p] O: {2,2,5,1,3,1} I: PermutationReplace[f, q] (* G *) O: {4,4,1,5,3,5} I: PermutationReplace[e, r] (* G *) O: {4,4,1,5,3,5}

Passive

Box 16 Permuting a vector

Active equivalent Box 16a Formulas and computation ${\begin{aligned}&QV&~~=~~&(V_{Q(0)},\dots ,V_{Q(4)})&~~=~~&(V_{4},V_{3},V_{2},V_{1},V_{0})&&&~~=~~&A\\&PA&~~=~~&(A_{P(0)},\dots ,A_{P(4)})&~~=~~&(A_{1},A_{3},A_{0},A_{2},A_{4})&~~=~~&(V_{3},V_{1},V_{4},V_{2},V_{0})&~~=~~&B\\\\&RV&~~=~~&(V_{R(0)},\dots ,V_{R(4)})&~~=~~&(V_{3},V_{1},V_{4},V_{2},V_{0})&&&~~=~~&B\end{aligned}}$ Python: >>> a = times(q, v) >>> a ['4', '3', '2', '1', '0'] >>> times(p, a) # B ['3', '1', '4', '2', '0'] >>> times(r, v) # B ['3', '1', '4', '2', '0'] Wolfram:   I: a = Permute[v, InversePermutation[q]] O: {v5,v4,v3,v2,v1} I: Permute[a, InversePermutation[p]] (* B *) O: {v4,v2,v5,v3,v1} I: Permute[v, InversePermutation[r]] (* B *) O: {v4,v2,v5,v3,v1}

Box 17 Concatenation with a scalar as permuting a vector
Active equivalent Box 17a Computation Python: >>> times([3], inv(q)) [1] >>> times([1], inv(p)) [0] >>> times([3], inv(r)) [0] Wolfram: I: PermutationReplace[4,InversePermutation[q]] O: 2 I: PermutationReplace[2,InversePermutation[p]] O: 1 I: PermutationReplace[4,InversePermutation[r]] O: 1

Examples with permutations of the square

The dihedral group Dih₄ is the group of symmetries of the square.

Box 18 Cayley graphs
The three Cayley graphs below show how each element of the group can be displayed as a product of two different elements. The formulas are in postfix (AR) notation, i.e. an arrow representing $\beta$ goes from $\alpha$ to $\alpha \beta$. An undirected edge is shorthand for arrows in both directions.
$p={\mathit {13}}$   (orange arrow),     $t={\mathit {2}}$   (green edge)	$a={\mathit {10}}$   (red arrow),     $b={\mathit {16}}$   (blue edge)	$b={\mathit {16}}$   (blue edge),     $c={\mathit {21}}$   (pink edge)
orange arrow from ${\mathit {21}}$ to ${\mathit {7}}$: ${\mathit {21}}\cdot {\mathit {13}}={\mathit {7}}$ green edge between ${\mathit {21}}$ and ${\mathit {23}}$: ${\mathit {21}}\cdot {\mathit {2}}={\mathit {23}}$   and   ${\mathit {23}}\cdot {\mathit {2}}={\mathit {21}}$	red arrow from ${\mathit {7}}$ to ${\mathit {21}}$: ${\mathit {7}}\cdot {\mathit {10}}={\mathit {21}}$ blue edge between ${\mathit {7}}$ and ${\mathit {23}}$: ${\mathit {7}}\cdot {\mathit {16}}={\mathit {23}}$   and   ${\mathit {23}}\cdot {\mathit {16}}={\mathit {7}}$	blue edge between ${\mathit {13}}$ and ${\mathit {2}}$: ${\mathit {13}}\cdot {\mathit {16}}={\mathit {2}}$   and   ${\mathit {2}}\cdot {\mathit {16}}={\mathit {13}}$ pink edge between ${\mathit {2}}$ and ${\mathit {23}}$: ${\mathit {2}}\cdot {\mathit {21}}={\mathit {23}}$   and   ${\mathit {23}}\cdot {\mathit {21}}={\mathit {2}}$

Active vs. passive

PL:     ${\mathit {10}}$ is rotation to the left, ${\mathit {13}}$ to the right	AR:     ${\mathit {10}}=a$ is rotation to the right, ${\mathit {13}}=p$ to the left	Cycle graph (formulas on the left corresponding to Cayley graph in the middle, thus AR)
		Cayley table (AR or PL)

Box 19 Ambiguity of ${\mathit {10}}\cdot {\mathit {2}}={\mathit {16}}$
Passive (PL):     ${\mathit {10}}~{\scriptstyle after}~{\mathit {2}}={\mathit {16}}$	Active (AR):     ${\mathit {10}}~{\scriptstyle before}~{\mathit {2}}={\mathit {16}}$
The inverse of ${\mathit {10}}$ is ${\mathit {13}}$, while ${\mathit {2}}$ and ${\mathit {16}}$ are self-conjugate. The Cayley table above contains ${\mathit {10}}\cdot {\mathit {2}}={\mathit {16}}$, so it fits the row above, and it contains ${\mathit {2}}\cdot {\mathit {13}}={\mathit {16}}$, so its transpose would fit the row below. As above (box 1) the permutations on the left and right have the same labels and do different things, while the permutations below do exactly the same as those above them, but are labelled differently. Box 19a Ambiguity of ${\mathit {10}}^{-1}\cdot {\mathit {2}}^{-1}={\mathit {16}}^{-1}$ which is ${\mathit {13}}\cdot {\mathit {2}}={\mathit {16}}$ Active (AL):     ${\mathit {13}}~{\scriptstyle after}~{\mathit {2}}={\mathit {16}}$ Passive (PR):     ${\mathit {13}}~{\scriptstyle before}~{\mathit {2}}={\mathit {16}}$

Left vs. right

Active

Box 20 ${\mathit {7}}$
AL:     ${\mathit {10}}~{\scriptstyle after}~{\mathit {2}}={\mathit {7}}$ AR:     ${\mathit {2}}~{\scriptstyle before}~{\mathit {10}}={\mathit {7}}$

Box 21 ${\mathit {16}}$
AL:     ${\mathit {2}}~{\scriptstyle after}~{\mathit {10}}={\mathit {16}}$ AR:     ${\mathit {10}}~{\scriptstyle before}~{\mathit {2}}={\mathit {16}}$

Passive

Box 22 ${\mathit {7}}$
PL:     ${\mathit {2}}~{\scriptstyle after}~{\mathit {10}}={\mathit {7}}$ PR:     ${\mathit {10}}~{\scriptstyle before}~{\mathit {2}}={\mathit {7}}$

Box 23 ${\mathit {16}}$
PL:     ${\mathit {10}}~{\scriptstyle after}~{\mathit {2}}={\mathit {16}}$ PR:     ${\mathit {2}}~{\scriptstyle before}~{\mathit {10}}={\mathit {16}}$

Languages

Wolfram

$P$ as permutation matrix and arrow diagram in Wolfram Alpha

Wolfram Alpha displays permutations in a way that resembles PL representation in this article,
which – if only permutation concatenation is concerned – is the same as AR. (See boxes 1 and 19.)

$P$ is shown as ${\begin{pmatrix}1&2&3&4&5\\2&4&1&3&5\end{pmatrix}}=(1243)$, but in the arrow diagram and the permutation matrix the arrow from 1 goes to 3 — not to 2.

Wolfram does not accept the element 0, so the examples are converted to 1-based permutations:

P = (1243)(5)   = 2 4 1 3 5  http://www.wolframalpha.com/input/?i=permutation+(1+2+4+3)(5)
Q = (15)(24)(3) = 5 4 3 2 1  http://www.wolframalpha.com/input/?i=permutation+(1+5)(2+4)(3)
R = (1435)(2)   = 4 2 5 3 1  http://www.wolframalpha.com/input/?i=permutation+(1+4+3+5)(2)
S = (1523)(4)   = 5 3 1 4 2  http://www.wolframalpha.com/input/?i=permutation+(1+5+2+3)(4)

A blogpost from 2011 shows that back then the list notation with curly braces was the inverse of the one-line notation.
But in the calculations shown in this article (done in Mathematica Online in December 2016) the permutation p = {2,4,1,3,5} corresponds to Cycles[{{1,2,4,3}}].
Apparently in the meantime a notation people found confusing was dropped in favour of a more mainstream one.

This article uses three binary Wolfram functions:

PermutationProduct[a, b, c] gives the product of permutations a, b, c.     (box 5b)

PermutationReplace[expr, perm] replaces each part in expr by its image under the permutation perm.     (boxes 13a, 15a, 17a)

Permute[l, p] permutes list l according to permutation p.     (boxes 12a, 16a)

More examples

I: p O: {2,4,1,3,5} I: q O: {5,4,3,2,1} I: PermutationProduct[p, q] (* R *) O: {4,2,5,3,1} I: PermutationProduct[q, p] (* S *) O: {5,3,1,4,2} For two permutations PermutationReplace gives the same result as PermutationProduct: I: PermutationReplace[p, q] (* R *) O: {4,2,5,3,1} I: PermutationReplace[q, p] (* S *) O: {5,3,1,4,2} Permute[p, q] Permute[q, p] Permute[p, q] is PermutationProduct[q, p] but Permute[q, p] is not PermutationProduct[p, q]: I: Permute[p, q] O: {5,3,1,4,2} I: Permute[q, p] O: {3,5,2,4,1} Permute[a, b] with permutations a, b behaves like InversePermutation[PermutationProduct[InversePermutation[a], b]] = PermutationProduct[InversePermutation[b], a]. Confusingly the result of Permute[q, p] changes to {4,2,5,3,1} when the Combinatorica package is used (Needs["Combinatorica`"]), i.e. Permute[a, b] with permutations a, b now behaves like PermutationProduct[b, a]. The result of Permute[p, q] is {5,3,1,4,2} in both cases because q is self-conjugate — thus PermutationProduct[q, p] = PermutationProduct[InversePermutation[q], p].

SymPy

http://docs.sympy.org/dev/modules/combinatorics/permutations.html

The composite of two permutations p*q means first apply p, then q, so i^(p*q) = (i^p)^q which is i^p^q according to Python precedence rules.

SymPy has two operators for permutations: Multiplication (*) and what could be called exponentiation (^). The latter seems to be intended only to access elements of a permutation, but can be used to combine two combinations. The result of a^b is the same as ~b*a*b (see Python script), which is a permutation in the same conjugacy class.

Permutations ${\mathit {0}}\dots {\mathit {23}}$ Example:    ${\mathit {15}}=(031)$

$(a,b)~~=~~a\cdot b$ Example:    ${\mathit {6}}\cdot {\mathit {15}}~~=~~{\mathit {9}}$

$(a,b)~~=~~a^{b}~~=~~b^{-1}\cdot a\cdot b$       (details on file page) Example:    ${\mathit {6^{15}}}~~=~~{\mathit {15}}^{-1}\cdot {\mathit {6}}\cdot {\mathit {15}}~~=~~{\mathit {20}}\cdot {\mathit {6}}\cdot {\mathit {15}}~~=~~{\mathit {2}}$

Negative results for ~p(i)

Inverse permutations behave strangely when accessed from the right. ${\mathit {15}}^{-1}$ should behave exactly like ${\mathit {20}}$. While i^~p15 behaves as expected, ~p15(i) gives negative results: p15 = Permutation([3, 0, 2, 1]), p20 = Permutation([1, 3, 2, 0]) The two permutations are inverses: p20 == ~p15 = True [i^~p15 for i in range(4)] = [1, 3, 2, 0] [~p15(i) for i in range(4)] = [-4, -1, -3, -2] These numbers would work to access the values of a list with four elements in Python - e.g. a[-1] = a[3] is the last element of a. [-4, -1, -3, -2] converted accordingly is [0, 3, 1, 2], which is permutation ${\mathit {8}}$. Doing the same for all permutations of four elements (see Python script) shows the following result: Where the expected permutation is $a$, the converted permutation is $b$ with $a\cdot b={\mathit {23}}$. Here the expected permutations is ${\mathit {20}}$, and the converted permutation is ${\mathit {8}}$, because ${\mathit {20}}\cdot {\mathit {8}}={\mathit {23}}$.

Sources

There appear to be no sources that actually use the passive interpretation of permutations. So the sources below are active left and active right.

Left

Aigner 2007

We read a product always from right to left, thus for

$\sigma ={\begin{pmatrix}1&2&3&4&5&6\\2&3&4&1&6&5\end{pmatrix}}$,   $\tau ={\begin{pmatrix}1&2&3&4&5&6\\1&3&4&5&2&6\end{pmatrix}}$,

we have

$\tau \sigma ={\begin{pmatrix}1&2&3&4&5&6\\3&4&5&1&6&2\end{pmatrix}}$   and   $\sigma \tau ={\begin{pmatrix}1&2&3&4&5&6\\2&4&1&6&3&5\end{pmatrix}}$.

We call $\sigma =\sigma (1)\sigma (2)\dots \sigma (n)$ the word representation of $\sigma$.

Aigner, Martin (2007). A course in enumeration. Berlin New York: Springer. ISBN 3642072534.  — Word Representation (p. 27 ff)

Right

Knuth 1973

In TAoCP Knuth states that in his book permutations are always multiplied from left to right, and gives the following example:

$\alpha \beta ={\begin{pmatrix}0&1&2&3&4&5\\2&5&0&1&4&3\end{pmatrix}}{\begin{pmatrix}0&1&2&3&4&5\\5&4&3&2&1&0\end{pmatrix}}={\begin{pmatrix}0&1&2&3&4&5\\2&5&0&1&4&3\end{pmatrix}}{\begin{pmatrix}2&5&0&1&4&3\\3&0&5&4&1&2\end{pmatrix}}={\begin{pmatrix}0&1&2&3&4&5\\3&0&5&4&1&2\end{pmatrix}}$

$\alpha \beta =(02)(153)\cdot (05)(14)(23)=(0341)(25)$

SymPy
>>> from sympy.combinatorics import Permutation >>> a = Permutation(0, 2)(1, 5, 3) >>> b = Permutation(0, 5)(1, 4)(2, 3) >>> a * b Permutation(0, 3, 4, 1)(2, 5) >>> b * a Permutation(0, 3)(1, 4, 5, 2) >>> Permutation.print_cyclic = False >>> a Permutation[2, 5, 0, 1, 4, 3] >>> b Permutation[5, 4, 3, 2, 1, 0] >>> a * b Permutation[3, 0, 5, 4, 1, 2] >>> b * a Permutation[3, 4, 1, 0, 5, 2]

The left permutation sees the argument first:

Notice that the image of 1 under $\alpha \beta$ is $1(\alpha \beta )=(1\alpha )\beta =5\beta =0$, etc.

SymPy: (1^a)^b = 5^b = 0

This is not to be confused with right to left multiplication of permutations, i.e. the way functions are composed:

[...] traditional functional notation, in which one writes $\alpha (1)=5$, makes it natural to think that $\alpha \beta (1)$ should mean $\alpha (\beta (1))=\alpha (4)=4$.

SymPy: a(b(1)) = a(4) = 4

Knuth, Donald (1973). The art of computer programming. Reading, Mass: Addison-Wesley Pub. Co. ISBN 0201896850.  — Chapter 7.2.1.2. Generating all permutations

Cameron 1994

Cameron decribes the active form of a permutation as a one-to-one mapping from $X=\{x_{1},\dots ,x_{n}\}$ to itself, and the passive form as the ordered n-tuple ${\bigl (}\pi (x_{1}),\dots ,\pi (x_{n}){\bigr )}$.

I wrote $\pi (x)$ for the result of applying the function $\pi$ to the element $x$.
[...] In order that the result of applying first $\pi _{1}$ and then $\pi _{2}$ can be called $\pi _{1}\pi _{2}$, it is more natural to denote the image of $x$ under $\pi$ as $x\pi$.
[Footnote:] We say that permutations act on the right if they compose according to this rule.

What Cameron calls the passive permutation is simply the one-line or word representation of an (active) permutation, not a passive permutation in the sense of this article.

Cameron, Peter J. (1994), Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, ISBN 0-521-45761-0 — Chapter 3.5 Permutations (p. 29)

Grimaldi 2004

Illustrations of $\pi _{1}$ and $r_{1}$ as in the book, illustration of $\pi _{1}r_{1}$ added

Grimaldi defines $\pi _{1}={\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}}$ as the left rotation of a triangle and $r_{1}={\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}}$ as the reflection that leaves the right base angle in place.
Their product $\pi _{1}r_{1}={\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}}=r_{3}$ is the reflection about the vertical axis.

Grimaldi, Ralph (2004). Discrete and combinatorial mathematics : an applied introduction. Reading, Mass: Addison-Wesley Longman. ISBN 0321211030.  — Example 16.7 (p.749)