AlternatingGroup

The alternating group $A_n$: the even permutations of $\{1, \dots, n\}$, of order $n!/2$. A pure carrier like SymmetricGroup — GroupOrder, GroupElements and GroupGenerators read it directly, without materialising a Cayley table.

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

AlternatingGroup(n)the alternating group on $n$ points

SageMathAlternatingGroup(n)!↗

Typed signatures

Type	Package	Applies
(integer) -> expression<AlternatingGroup>	groupalgebra	always

Domain: Permutations

Details

• GroupOrder(AlternatingGroup(n)) is $n!/2$ without materialising all $n!/2$ permutations — $A_1$ and $A_2$ are both trivial, order 1
• GroupGenerators(AlternatingGroup(n)) is the 3-cycle $(1\,2\,3)$ together with the longest even cycle available: $(1\,2\,\dots\,n)$ when $n$ is odd, $(2\,3\,\dots\,n)$ when $n$ is even — an $n$-cycle is an even permutation exactly when $n$ is odd
• GroupElements(AlternatingGroup(n)), like PermutationGroup's, prints elements as Cycles and takes an optional Part-style position selector

Examples

Basic #

$|A_5| = 5!/2 = 60$

live ↗

In$\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(5))$Out$60$

for odd $n$, an $n$-cycle is even — $(1\,2\,3\,4\,5)$ decomposes into 4 transpositions

live ↗

In$\mathrm{GroupGenerators}(\mathrm{AlternatingGroup}(5))$Out$\bigl\lbrack\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3, 4, 5\bigr\rbrack\bigr\rbrack)\bigr\rbrack$

for even $n$, the full $n$-cycle would be odd, so the second generator is one point shorter: $(2\,3\,4)$

live ↗

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

at $n = 3$ the two generators coincide, so there is only one

live ↗

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

$A_3$ is cyclic of order 3 — the identity and the two 3-cycles

live ↗

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

Order is $10!/2$, half of the symmetric group

live ↗

In$\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(10))$Out$1\,814\,400$

A 3-cycle and an 9-cycle generate $A_{10}$

live ↗

In$\mathrm{GroupGenerators}(\mathrm{AlternatingGroup}(10))$Out$\bigl\lbrack\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 3, 4, 5, 6, 7, 8, 9, 10\bigr\rbrack\bigr\rbrack)\bigr\rbrack$

All twelve even permutations of four points

live ↗

In$\mathrm{GroupElements}(\mathrm{AlternatingGroup}(4))$Out$\bigl\lbrack\mathrm{Cycles}(\bigl\lbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 3, 4\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 4, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack, \bigl\lbrack3, 4\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 4\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 2\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 4\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3\bigr\rbrack, \bigl\lbrack2, 4\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 4, 2\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 4, 3\bigr\rbrack\bigr\rbrack), \mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 4\bigr\rbrack, \bigl\lbrack2, 3\bigr\rbrack\bigr\rbrack)\bigr\rbrack$

Scope #

$A_1$ is the trivial group — $1!/2$ isn't even an integer, so this is a base case, not the formula

live ↗

In$\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(1))$Out$1$

$A_2$ is trivial too — $S_2$'s only even permutation is the identity

live ↗

In$\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(2))$Out$1$

The trivial group has only the identity

live ↗

In$\mathrm{GroupElements}(\mathrm{AlternatingGroup}(1))$Out$\bigl\lbrack\mathrm{Cycles}(\bigl\lbrack\bigr\rbrack)\bigr\rbrack$

$A_2$ is also trivial: no even permutation but the identity

live ↗

In$\mathrm{GroupElements}(\mathrm{AlternatingGroup}(2))$Out$\bigl\lbrack\mathrm{Cycles}(\bigl\lbrack\bigr\rbrack)\bigr\rbrack$

About

Examples	12
Checked against	Wolfram Language 12 agreeSageMath 10 agree2 differ
Crosswalk	Wolfram LanguageSageMath

Implementation

mappedsagepackages/oracle/src/mappings.ts

See also: SymmetricGroup, GroupOrder, GroupElements, GroupGenerators, PermutationGroup, Permute