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

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 #

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

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

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: $(2\,3\,4)$

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

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

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

In

\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(10))

Out

1\,814\,400

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

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

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

In

\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(1))

Out

1

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

In

\mathrm{GroupOrder}(\mathrm{AlternatingGroup}(2))

Out

1

The trivial group has only the identity

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

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 agree
Crosswalk	Wolfram Language

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