Alternating group

This article is about a basic definition in group theory. The article text may, however, contain advanced material.
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Definition

For a finite set

Let $S$ be a finite set. The alternating group on $S$ is defined in the following equivalent ways:

1. It is the group of all even permutations on $S$ under composition. An even permutation is a permutation whose cycle decomposition has an even number of cycles of even size. Specifically, the alternating group on $S$ is the subgroup of the symmetric group on $S$ comprising the even permutations.
2. It is the kernel of the sign homomorphism from the symmetric group on $S$ to the group $\pm 1$.

For $S$ having size zero or one, the alternating group on $S$ equals the whole symmetric group on $S$. For $S$ having size at least two, the alternating group on $S$ is the unique subgroup of index two in the symmetric group on $S$.

The alternating group on a set of size $n$ is denoted $A_{n}$ and is termed the alternating group of degree $n$.

For an infinite set

Let $S$ be an infinite set. The finitary alternating group on $S$ is defined in the following equivalent ways:

1. It is the group of all even permutations on $S$ under composition. Note that the concept of even permutation requires the permutation to be finitary, i.e., finite support, or in other words, to only move finitely many elements.
2. It is the kernel of the sign homomorphism on the finitary symmetric group on $S$.

There is no infinitary equivalent, i.e., for an infinite set, there is no "infinitary alternating group".

Facts

• The alternating group on a set of size five or more is simple. Also, the finitary alternating group on an infinite set is simple. For full proof, refer: A5 is simple, alternating groups are simple
• Projective special linear group equals alternating group in only finitely many cases

Arithmetic functions

Here, $n$ is the degree of the alternating group, i.e., the size of the set it acts on.

For all the statements involving $n\geq 5$, we use the fact that A5 is simple and alternating groups are simple for degree at least five.

Function	Value	Explanation
order	$n!/2$ for $n\geq 2$, $1$ for $n=0,1$	It has index two in the symmetric group of degree $n$.
exponent	$\operatorname {lcm} \{1,2,\dots ,n-2,n\}$ (skipping over $n-1$) if $n$ is odd, $\operatorname {lcm} \{1,2,\dots ,n-1\}$ if $n$ is even.	A $k$-cycle can occur as a cycle in the cycle decomposition of an even permutation if either $k$ is odd with $k\leq n$ (in which case we can use that cycle itself as the even permutation) or $k$ is even with $k+2\leq n$ (in which case we can take the even permutation as that cycle times a disjoint 2-cycle). The exponent is the lcm of possible cycle sizes that can occur, so the above method gives that the lcm is over all the odd numbers less than or equal to $n$, and all the even numbers less than or equal to $n-2$. Working separately the cases of odd and even $n$ gives the conclusion.
nilpotency class	$1$ for $n\leq 3$, undefined for $n\geq 4$	abelian group for $n\leq 3$, not a nilpotent group for $n\geq 4$.
derived length	$1$ for $n\leq 3$, $2$ for $n=4$, undefined for $n\geq 5$	abelian for $n\leq 3$, simple for $n\geq 5$.
Frattini length	$0$ for $n\leq 2$, $1$ for $n\geq 3$	Frattini-free group: intersection of maximal subgroups is trivial.
Fitting length	$1$ for $n\leq 3$, $2$ for $n=4$, undefined for $n\geq 5$	abelian for $n\leq 3$, simple for $n\geq 5$.
minimum size of generating set	2
subgroup rank	At most $n/2$
max-length	?
chief length	$0$ for $n\leq 2$, $1$ for $n=3,n\geq 5$, $2$ for $n=4$.
composition length	$0$ for $n\leq 2$, $1$ for $n=3,n\geq 5$, $3$ for $n=4$.

Group properties

Property	Satisfied	Explanation
Abelian group	Yes for $n\leq 3$, no for $n\geq 4$	$(1,2,3)$ and $(1,2)(3,4)$ don't commute.
Nilpotent group	Yes for $n\leq 3$, no for $n\geq 4$	$A_{4}$ is centerless.
Solvable group	Yes for $n\leq 4$, no for $n\geq 5$	alternating groups are simple for degree five or more.
Supersolvable group	Yes for $n\leq 3$, no for $n\geq 4$
Simple group	Yes for $n=3$ and $n\geq 5$, no for $n=4$.
Rational group	No for all $n$.
Ambivalent group	Yes for $n=1,2,5,6,10,14$, no otherwise	See classification of ambivalent alternating groups.