PermutationReplace

Apply a permutation by VALUE rather than by position: a point $i$ becomes $\sigma(i)$, wherever it shows up. On a plain point or a list this renames entries in place; on Cycles it conjugates — $\sigma \cdot c \cdot \sigma^{-1}$ — since conjugating a cycle is exactly relabelling the points it names.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguagePermutationReplace✓↗

PermutationReplace(i, perm)$\sigma(i)$, or $i$ itself past perm's own support

PermutationReplace(list, perm)list with every entry $i$ replaced by $\sigma(i)$

PermutationReplace(Cycles(...), perm)the cycles conjugated by perm — each point renamed, same cycle structure

Typed signatures

Type	Package	Applies
(expression<Cycles> | integer | list<integer>, expression<Cycles> | list<integer>) -> expression<Cycles> | integer | list<integer>	groupalgebra	always

Domain: Permutations

Details

• Permute moves items by POSITION; PermutationReplace renames them by VALUE — Permute(List(a, b, c), perm) and PermutationReplace(List(1, 2, 3), perm) answer different questions
• perm in either notation — Cycles or a one-line word
• A point beyond perm's own support is fixed and passes through unchanged
• A conjugated Cycles prints in ITS canonical form — smallest point first, singletons dropped, cycles ordered by that first point — not whatever rotation the per-point relabelling produced; this is the same presentation PermutationCycles and GroupElements use

Examples

Basic #

$(1\,2\,3)$ sends $2 \to 3$

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In$\mathrm{PermutationReplace}(2, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack))$Out$3$

each VALUE is replaced, not the position it sits in — 4 is past the cycle's support, so it is fixed

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In$\mathrm{PermutationReplace}(\bigl\lbrack1, 2, 3, 4\bigr\rbrack, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack2, 3, 1, 4\bigr\rbrack$

on Cycles, PermutationReplace conjugates: every point the cycle names is relabelled by the second permutation

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In$\mathrm{PermutationReplace}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack\bigr\rbrack))$Out$\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 2\bigr\rbrack\bigr\rbrack)$

Applies the cycle to a point: 4 goes to 6

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In$\mathrm{PermutationReplace}(4, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 3, 4, 6\bigr\rbrack\bigr\rbrack))$Out$6$

The identity leaves a point alone

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In$\mathrm{PermutationReplace}(4, \mathrm{Cycles}(\bigl\lbrack\bigr\rbrack))$Out$4$

Applying one permutation to another composes them

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In$\mathrm{PermutationReplace}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 4\bigr\rbrack, \bigl\lbrack2, 5, 6, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 3, 5, 1, 4, 6\bigr\rbrack\bigr\rbrack))$Out$\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 5, 3\bigr\rbrack, \bigl\lbrack4, 6\bigr\rbrack\bigr\rbrack)$

Scope #

5 moves to 3, the next point in its cycle

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In$\mathrm{PermutationReplace}(5, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5, 3\bigr\rbrack, \bigl\lbrack2, 7\bigr\rbrack\bigr\rbrack))$Out$3$

The cycle wraps around: 3 returns to 1

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In$\mathrm{PermutationReplace}(3, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5, 3\bigr\rbrack, \bigl\lbrack2, 7\bigr\rbrack\bigr\rbrack))$Out$1$

4 is in no cycle, so it is fixed

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In$\mathrm{PermutationReplace}(4, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5, 3\bigr\rbrack, \bigl\lbrack2, 7\bigr\rbrack\bigr\rbrack))$Out$4$

Replaces every entry of a list through the permutation

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In$\mathrm{PermutationReplace}(\bigl\lbrack1, 2, 3, 4, 5, 6\bigr\rbrack, \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5, 3\bigr\rbrack, \bigl\lbrack2, 7\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack5, 7, 1, 4, 3, 6\bigr\rbrack$

Properties #

the result prints the way Cycles always does when it's computed rather than typed: smallest point first, singletons dropped — not whatever rotation the per-point relabelling happened to produce

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In$\mathrm{PermutationReplace}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 2\bigr\rbrack\bigr\rbrack))$Out$\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3\bigr\rbrack\bigr\rbrack)$

conjugating by $p$ then by $q$ is conjugating by their composite: two steps land where one big permutation would

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In$\mathrm{PermutationReplace}(\mathrm{PermutationReplace}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack\bigr\rbrack)), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack))$Out$\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 2\bigr\rbrack\bigr\rbrack)$

About

Examples	12
Checked against	Wolfram Language 12 agree
Crosswalk	Wolfram Language

See also: Permute, Cycles, PermutationList, InversePermutation