# Linear representation theory of alternating group:A4

View linear representation theory of particular groups | View other specific information about alternating group:A4

This article discusses the linear representation theory of the alternating group of degree four, a group of order 12. For convenience, the underlying set is $\{ 1,2,3,4 \}$, and permutations are written using the cycle decomposition notation.

See alternating group:A4 and subgroup structure of alternating group:A4 for background information on the group structure.

## Summary

Item Value
Degrees of irreducible representations over a splitting field 1,1,1,3
maximum: 3, lcm: 3, number: 4
Schur index values of irreducible representations 1,1,1,1
Smallest ring of realization of all representations (characteristic zero) $\mathbb{Z}[e^{2\pi i/3})] \cong \mathbb{Z}[x]/(x^2 + x + 1)$
Smallest field of realization of all representations (characteristic zero) $\mathbb{Q}(e^{2\pi i/3}) \cong \mathbb{Q}[x]/(x^2 + x + 1)$
Criterion for a field to be a splitting field Any field of characteristic not 2 or 3 that contains a primitive cube root of unity, i.e., the polynomial $x^2 + x + 1$ splits.
For a finite field, this is equivalent to saying that the size the field is odd and congruent to 1 modulo 3 (or equivalently, congruent to 1 modulo 6).
Degrees of irreducible representations over a non-splitting field 1,2,3
maximum: 3, lcm: 6, number: 3
Smallest size splitting field Field:F7, the field with 7 elements.

## Family contexts

Family name Parameter values General discussion of linear representation theory of family
alternating group degree $n = 4$ linear representation theory of alternating groups
projective special linear group of degree two over a finite field field:F3, i.e., the group is $PSL(2,3)$ linear representation theory of projective special linear group of degree two over a finite field
COMPARE AND CONTRAST: View linear representation theory of groups of order 12 to compare and contrast the linear representation theory with other groups of order 12.

## Representations

### Summary information

Below is summary information on irreducible representations that are absolutely irreducible, i.e., they remain irreducible in any bigger field, and in particular are irreducible in a splitting field. We assume that the characteristic of the field is not 2 or 3, except in the last two columns, where we consider what happens in characteristic 2 and characteristic 3.

Name of representation type Number of representations of this type Degree Schur index Criterion for field Kernel Quotient by kernel (on which it descends to a faithful representation) Characteristic 2 Characteristic 3
trivial 1 1 1 any whole group trivial group works works
one-dimensional nontrivial 2 1 1 contains a primitive cube root of unity V4 in A4 cyclic group:Z3 works works, same as trivial
three-dimensional irreducible 1 3 1 any trivial subgroup, i.e., the representation is faithful alternating group:A4 indecomposable but not irreducible works

Below are representations that are irreducible over a non-splitting field but split up over a splitting field.

Name of representation type Number of representations of this type Degree Criterion for field What happens over a splitting field? Kernel Quotient by kernel (on which it descends to a faithful representation)
two-dimensional, kernel of order four 1 2 does not contain a primitive cube root of unity splits into the two one-dimensional nontrivial representations V4 in A4 cyclic group:Z3

### Trivial representation

The trivial representation works over all fields. It is a one-dimensional representation that sends every element of the group to the $1 \times 1$ identity matrix $( 1 )$.

### Two one-dimensional representations with kernel of order four

The alternating group of degree four has a unique proper nontrivial normal subgroup, namely V4 in A4. This is a subgroup of order four isomorphic to the Klein four-group, and equals the derived subgroup. It is explicitly given by: $\! K := \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$.

There are two one-dimensional representations with kernel $K$ over any field that has characteristic not equal to $3$ and has primitive cuberoots of unity.. These correspond to the two one-dimensional representations of the quotient group, which is cyclic of order three (see linear representation theory of cyclic group:Z3 for details).

If we denote by $\omega$ a primitive cube root of unity, the two representations are explicitly as follows (note that since the representations are one-dimensional, we simply write them as numbers rather than as matrices):

Coset of $K$ (V4 in A4) Image of every element of this coset under the first one-dimensional representation Image of this coset under the second one-dimensional representation $\{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$ 1 1 $\{ (1,2,3), (2,4,3), (1,4,2), (1,3,4) \}$ $\omega$ $\omega^2$ $\{ (1,3,2), (2,3,4), (1,2,4), (1,4,3) \}$ $\omega^2$ $\omega$

In the case of characteristic zero, $\omega = e^{2\pi i/3} = \cos(2\pi/3) + i\sin(2\pi/3)$ where $i$ is a square root of $-1$.

### Three-dimensional irreducible representation

There is a three-dimensional representation that works over any field. Here is one way of describing this representation. Consider the action of the alternating group on a four-dimensional vector space, by permuting the basis vectors through its action on a set of size four. This action has an invariant subspace of codimension one: the subspace comprising vectors whose coordinates add to zero. This gives a three-dimensional vector space on which the alternating group acts, and this is an irreducible representation.

Here is a description of the values of this representation at each element:

Here is what happens with various characteristics of fields:

Case for the field What happens in this case
characteristic not equal to 2 The representation is irreducible
characteristic equal to 2 The representation is indecomposable but not irreducible. In fact, it induces an injective homomorphism from $A_4$ to GL(3,2) and, with a change of basis, the image of this mapping lies completely inside a parabolic subgroup which is isomorphic to symmetric group:S4.

### Two-dimensional representation: irreducible in the non-splitting case

For any field $F$, there is a two-dimensional representation over $F$ with kernel V4 in A4, given as follows:

Coset of V4 in A4 Matrix that all elements in this coset go to Trace Minimal polynomial $\{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ 2 $x - 1$ $\{ (1,2,3), (2,4,3), (1,4,2), (1,3,4) \}$ $\begin{pmatrix} 0 & -1 \\ 1 & -1 \\\end{pmatrix}$ -1 $x^2 + x + 1$ $\{ (1,3,2), (2,3,4), (1,2,4), (1,4,3) \}$ $\begin{pmatrix} -1 & 1 \\ -1 & 0 \\\end{pmatrix}$ -1 $x^2 + x + 1$

For more on the representation of cyclic group:Z3 that gives rise to this representation, see linear representation theory of cyclic group:Z3.

Here are the three cases for the field $F$, and what happens in each case:

Case for the field $F$ What happens in this case $F$ does not have characteristic $3$ does not contain a primitive cube root of unity The representation is irreducible. However, it is not absolutely irreducible, it decomposes as a direct sum in a quadratic extension containing a primitive cube root of unity (See next point). $F$ does not have characteristic $3$ and contains a primitive cube root of unity The representation decomposes as a direct sum of the two one-dimensional nontrivial representations with kernel V4 in A4
Characteristic of $F$ is $3$ The representation is indecomposable but not irreducible, because there is an invariant one-dimensional subspace.

## Degrees of irreducible representations

### Summary

FACTS TO CHECK AGAINST FOR DEGREES OF IRREDUCIBLE REPRESENTATIONS OVER SPLITTING FIELD:
Divisibility facts: degree of irreducible representation divides group order | degree of irreducible representation divides index of abelian normal subgroup
Size bounds: order of inner automorphism group bounds square of degree of irreducible representation| degree of irreducible representation is bounded by index of abelian subgroup| maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup
Cumulative facts: sum of squares of degrees of irreducible representations equals order of group | number of irreducible representations equals number of conjugacy classes | number of one-dimensional representations equals order of abelianization

Described below for a field of characteristic not $2$ or $3$ (i.e., the non-modular case):

Type of field Splitting field? Condition on polynomial Condition on $q$ for finite field of size $q$ Example values of $q$ (size of field) Degrees of irreducible representations
Contains a primitive cube root of unity Yes $x^2 + x + 1$ splits $3$ divides $q - 1$ 7 (field:F7), 13, 19, 25, 31, 37, 43, 49, 61 1,1,1,3
Does not contain a primitive cube root of unity No $x^2 + x + 1$ does not split $3$ does not divide $q - 1$ 5, 11, 17, 23, 29, 41, 47, 53, 59 1,2,3

## Character table

### Character table over a splitting field

FACTS TO CHECK AGAINST (for characters of irreducible linear representations over a splitting field):
Orthogonality relations: Character orthogonality theorem | Column orthogonality theorem
Separation results (basically says rows independent, columns independent): Splitting implies characters form a basis for space of class functions|Character determines representation in characteristic zero
Numerical facts: Characters are cyclotomic integers | Size-degree-weighted characters are algebraic integers
Character value facts: Irreducible character of degree greater than one takes value zero on some conjugacy class| Conjugacy class of more than average size has character value zero for some irreducible character | Zero-or-scalar lemma

Let $\omega$ be a primitive cube root of unity. The character table over a splitting field is as follows:

Representation/conjugacy class representative $()$ (size 1) $(1,2)(3,4)$ (size 3) $(1,2,3)$ (size 4) $(1,3,2)$ (size 4)
trivial representation 1 1 1 1
first nontrivial one-dimensional representation 1 1 $\omega$ $\omega^2$
second nontrivial one-dimensional representation 1 1 $\omega^2$ $\omega$
three-dimensional irreducible representation 3 -1 0 0

Note that this character table is interpreted differently depending on what the splitting field is and which of the primitive cube roots we choose to be $\omega$. Switching the roles of $\omega$ and $\omega^2$ in the above table simply permutes the two nontrivial one-dimensional representations and has no effect on the overall character table.

In characteristic zero (for instance, over $\mathbb{C}$), $\omega$ can be taken as $e^{2\pi i/3}$ or $\cos(2\pi/3) + i\sin(2\pi/3)$, which is $(-1 + i\sqrt{3})/2$. $\omega^2$ is the other primitive cube root of unity, and is given as $e^{-2\pi i/3}$ or $\cos(2\pi/3) - i\sin(2\pi/3)$ or $(-1 - i\sqrt{3})/2$. Here are the characters multiplied by conjugacy class size and divided by the degree of the representation. Note that size-degree-weighted characters are algebraic integers:

Representation/conjugacy class representative $()$ $(1,2)(3,4)$ $(1,2,3)$ $(1,3,2)$
trivial representation 1 3 4 4
first nontrivial one-dimensional representation 1 3 $4e^{2\pi i/3}$ $4e^{-2\pi i/3}$
second nontrivial one-dimensional representation 1 3 $4e^{-2\pi i/3}$ $4e^{2\pi i/3}$
three-dimensional irreducible representation 1 -1 0 0

### Character table over a non-splitting field

For a field that is not a splitting field for the group, there are only three equivalence classes of irreducible representations. But also, the number of Galois conjugacy classes is three. Specifically, the two conjugacy classes of 3-cycles become a single conjugacy class. Here is the character table:

Representation/conjugacy class representative $()$ $(1,2)(3,4)$ $(1,2,3)$
trivial representation 1 1 1
irreducible two-dimensional representation 2 2 -1
irreducible three-dimensional representation 3 -1 0

Over a finite field, the character values are interpreted as integers modulo the field characteristic; over an infinite field, they are interpreted as rational numbers and hence field elements.

If doing character theory over the real numbers, we know that the number of irreducible representations over reals equals number of equivalence classes under real conjugacy and for the rational numbers, we know that the . The above is the character table both over the rationals and over the reals.

## Realization information

### Smallest ring of realization

Representation Smallest ring of realization Smallest set of elements that can be used as matrix entries for the ring
trivial representation $\mathbb{Z}$ -- ring of integers $\{ 1 \}$
first nontrivial irreducible representation $\mathbb{Z}[e^{2\pi i/3}]$ $\{ 1, e^{2\pi i/3}, e^{-2\pi i/3} \}$
second nontrivial irreducible representation $\mathbb{Z}[e^{2\pi i/3}]$ $\{ 1, e^{2\pi i/3}, e^{-2\pi i/3} \}$
irreducible three-dimensional representation $\mathbb{Z}$ $\{ 1, 0, -1 \}$
irreducible two-dimensional representation over fields not containing a primitive cuberoot of unity $\mathbb{Z}$ $\{ 1,0,-1 \}$