# Element structure of alternating group:A5

View element structure of particular groups | View other specific information about alternating group:A5

## Summary

Item Value
order of the whole group (total number of elements) 60
prime factorization $2^2 \cdot 3^1 \cdot 5^1= 4 \cdot 3 \cdot 5$
See order computation for more
conjugacy class sizes 1,12,12,15,20
maximum: 20, number: 5, sum (equals order of group): 60, lcm: 60
See conjugacy class structure for more.
number of conjugacy classes 5
See element structure of alternating group:A5#Number of conjugacy classes
order statistics 1 of order 1, 15 of order 2, 20 of order 3, 24 of order 5
maximum: 5, lcm (exponent of the whole group): 30

## Family contexts

Family name Parameter values General discussion of element structure of family
alternating group 5 element structure of alternating groups
projective general linear group of degree two over a finite field field:F4 element structure of projective general linear group of degree two over a finite field
projective special linear group of degree two over a finite field field:F5 element structure of projective special linear group of degree two over a finite field
COMPARE AND CONTRAST: View element structure of groups of order 60 to compare and contrast the element structure with other groups of order 60.

## Elements

### Order computation

The alternating group of degree five has order 60, with prime factorization $60 = 2^2 \cdot 3^1 \cdot 5 = 4 \cdot 3 \cdot 5$. Below are listed various methods that can be used to compute the order, all of which should give the answer 60:

Family Parameter values Formula for order of a group in the family Proof or justification of formula Evaluation at parameter values Full interpretation of conjugacy class structure Orders for members of this family
alternating group $A_n$ of degree $n$ degree $n = 5$ $n!/2$ See alternating group, element structure of alternating groups $5!/2 = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1/2 = 60$ #Interpretation as alternating group For $A_n, n = 3,4,5,\dots$: 3,12,60,360,2520,20160,...
projective special linear group of degree two over a finite field of size $q$ $q = 5$, i.e., field:F5, so the group is $PSL(2,5)$ $\frac{q^3 - q)}{\operatorname{gcd}(2,q - 1)}$ which becomes $\frac{q^3 - q}{2}$, same as $\frac{q(q - 1)(q + 1)}{2}$ for $q$ odd
$q^3 - q$, same as $q(q - 1)(q + 1)$ for $q$ a power of 2
See order formulas for linear groups of degree two, order formulas for linear groups, and projective special linear group of degree two $\frac{5^3 - 5}{2} = \frac{125 - 5}{2} = 60$
Factored version: $5(5 - 1)(5 + 1)/2 = 5(4)(6)/2 = 60$
#Interpretation as projective special linear group of degree two For $PSL(2,q), q = 3,5,7,9,\dots$ (odd case): 12,60,168,360,...
special linear group of degree two over a finite field of size $q$ $q = 4$, i.e., field:F4, so the group is $SL(2,4)$ $q^3 - q$, same as $q(q - 1)(q + 1)$ See order formulas for linear groups of degree two, order formulas for linear groups, and special linear group of degree two $q^3 - q = 4^3 - 4 = 60$
In factored form: $q(q - 1)(q + 1) = 4(3)(5) = 60$
#Interpretation as special linear group of degree two over field:F4 For $SL(2,q), q = 2,3,4,5,7,8,9,\dots$: 6,24,60,120,336,504,720,...
von Dyck group with parameters $(p,q,r)$ $(p,q,r) = (2,3,5)$ (note that the order of the parameters is irrelevant, though we usually arrange them in ascending or descending order depending on the convention being followed). $\frac{2}{1/p + 1/q + 1/r - 1}$ See element structure of von Dyck groups $\frac{2}{1/2 + 1/3 + 1/5 - 1} = \frac{2}{1/30} = 60$ #Interpretation as von Dyck group (omitted here due to multiple parameters)

## Conjugacy class structure

FACTS TO CHECK AGAINST FOR CONJUGACY CLASS SIZES AND STRUCTURE:
Divisibility facts: size of conjugacy class divides order of group | size of conjugacy class divides index of center | size of conjugacy class equals index of centralizer
Bounding facts: size of conjugacy class is bounded by order of derived subgroup
Counting facts: number of conjugacy classes equals number of irreducible representations | class equation of a group

There is a total of 5 conjugacy classes, of which 3 are unsplit from symmetric group:S5, and 2 are a split pair arising from a single conjugacy class in $S_5$. The conjugacy class sizes are 1, 12, 12, 15, 20.

### Interpretation as alternating group

FACTS TO CHECK AGAINST SPECIFICALLY FOR SYMMETRIC GROUPS AND ALTERNATING GROUPS:
Conjugacy class parametrization: cycle type determines conjugacy class (in symmetric group)
Conjugacy class sizes: conjugacy class size formula in symmetric group
Other facts: even permutation (definition) -- the alternating group is the set of even permutations | splitting criterion for conjugacy classes in the alternating group (from symmetric group)| criterion for element of alternating group to be real

For a symmetric group, cycle type determines conjugacy class. The statement is almost true for the alternating group, except for the fact that some conjugacy classes of even permutations in the symmetric group split into two in the alternating group, as per the splitting criterion for conjugacy classes in the alternating group, which says that a conjugacy class of even permutations splits in the alternating group if and only if it is the product of odd cycles of distinct length.

Here are the unsplit conjugacy classes:

Partition Verbal description of cycle type Representative element of the cycle type All elements of the cycle type Size of conjugacy class Formula for size Element order
1 + 1 + 1 + 1 + 1 five fixed points $()$ -- the identity element $()$ 1 $\! \frac{5!}{(1)^5(5!)}$ 1
3 + 1 + 1 one 3-cycle, two fixed points $(1,2,3)$ [SHOW MORE] 20 $\! \frac{5!}{(3)(1)^2(2!)}$ 3
2 + 2 + 1 double transposition: two 2-cycles, one fixed point $(1,2)(3,4)$ [SHOW MORE] 15 $\! \frac{5!}{(2)^2(2!)(1)}$ 2

Here is the split pair of conjugacy classes:

Partition Verbal description of cycle type Combined size of conjugacy classes Formula for combined size Size of each half Representative of first half Representative of second half Real? Rational? Element order
5 one 5-cycle 24 $\! \frac{5!}{5}$ 12 $(1,2,3,4,5)$ $(1,3,5,2,4)$ Yes No 5

### Interpretation as projective special linear group of degree two

We consider the group as $PSL(2,q)$ with $q = 5$. We use the letter $q$ to denote the generic case of $q \equiv 1 \pmod 4$.

Nature of conjugacy class upstairs in $SL_2$ Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class (generic $q$ that is 1 mod 4) Size of conjugacy class ($q = 5$) Number of such conjugacy classes (generic $q$ that is 1 mod 4) Number of such conjugacy classes ($q = 5$) Total number of elements (generic $q$ that is 1 mod 4) Total number of elements ($q = 5$) Matrix representatives upstairs (one per conjugacy class) Representatives as permutations
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar $\{ 1,1 \}$ or $\{ -1,-1\}$, both correspond to the same element $(x - a)^2$ where $a \in \{ -1,1 \}$ $x - a$ where $a \in \{ -1,1\}$ 1 1 1 1 1 1 $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $()$
Not diagonal, has Jordan block of size two $1$ (multiplicity 2) or $-1$ (multiplicity 2). Each conjugacy class has one representative of each type. $(x - a)^2$ where $a \in \{ -1,1 \}$ Same as characteristic polynomial $(q^2 - 1)/2$ 12 2 2 $q^2 - 1$ 24 $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$, $\begin{pmatrix} 1 & 2 \\ 0 & 1 \\\end{pmatrix}$ $(1,2,3,4,5)$, $(1,3,5,2,4)$
Diagonlizable over $\mathbb{F}_q$ with diagonal entries squaring to $-1$ $\{ 2,3 \}$ $x^2 + 1$ $x^2 + 1$ $q(q + 1)/2$ 15 1 1 $q(q + 1)/2$ 15 $\begin{pmatrix} 2 & 0 \\ 0 & 3 \\\end{pmatrix}$ $(1,2)(3,4)$
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of field:F25 of norm 1. Each pair identified with its negative pair. $x^2 - x + 1$, $x^2 + x + 1$, get identified. Same as characteristic polynomial $q(q - 1)$ 20 $(q - 1)/4$ 1 $q(q - 1)^2/4$ 20 $\begin{pmatrix} 0 & -1 \\ 1 & -1 \\\end{pmatrix}$ $(1,2,3)$
Diagonalizable over $\mathbb{F}_q$ with distinct (and hence mutually inverse) diagonal entries, whose square is not $-1$ None for this field -- -- $q(q + 1)$ 30 $(q - 5)/4$ 0 $q(q + 1)(q - 5)/4$ 0 -- --
Total NA NA NA NA NA $(q + 5)/2$ 5 $(q^3 - q)/2$ 60 NA NA

### Interpretation as special linear group of degree two over field:F4

Nature of conjugacy class Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class Number of such conjugacy classes Total number of elements Semisimple? Diagonalizable over $\mathbb{F}_q$? Splits in $SL_2$ relative to $GL_2$? Representative matrices (one for each conjugacy class) Representative element as permutation
Diagonalizable over field:F4 with distinct (and hence mutually inverse) diagonal entries $\lambda, 1/\lambda$ where $\lambda \in \mathbb{F}_4 \setminus \{ 0,1 \}$ $x^2 + x + 1$ $x^2 + x + 1$ 20 1 20 Yes Yes No $\begin{pmatrix}0 & 1 \\ 1 & 1 \\\end{pmatrix}$ $(1,2,3)$
Diagonalizable over field:F16, not over field:F4. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of field:F16 of norm 1 $x^2 - ax + 1$, $a \ne 0, 1$. Same as characteristic polynomial 12 2 24 Yes No No PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $(1,2,3,4,5)$, $(1,3,5,2,4)$
Diagonalizable over field:F4 with equal diagonal entries, hence a scalar. $1, 1$ $x^2 + 1$ $x + 1$ 1 1 1 Yes Yes No $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ $()$
Not diagonal, has Jordan block of size two $1$ (multiplicity 2) $x^2 + 1$ $x^2 + 1$ 15 1 15 No No No $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ $(1,2)(3,4)$
Total NA NA NA NA 5 60 45 NA NA

## Conjugacy class structure: additional information

### Number of conjugacy classes

The alternating group of degree five has 5 conjugacy classes. Below are listed various methods that can be used to compute the number of conjugacy classes, all of which should give the answer 5:

Family Parameter values Formula for number of conjugacy classes of a group in the family Proof or justification of formula Evaluation at parameter values Full interpretation of conjugacy class structure Conjugacy class sizes for groups in the family
alternating group $A_n$ of degree $n$ $n = 5$, i.e., the group $A_5$ (Number of pairs of non-self-conjugate partitions of $n$) + 2(Number of self-conjugate partitions of $n$) See element structure of alternating groups $3 + 2(1) = 5$ #Interpretation as alternating group For $A_n, n = 3,4,5,6,\dots$: 3,4,5,7,9,14,18,24,...
projective special linear group of degree two over a finite field of size $q$ $q = 5$, i.e., field:F5, so the group is $PSL(2,5)$ $(q + 5)/2$ for $q$ odd
$q + 1$ for $q$ a power of 2
See element structure of projective special linear group of degree two over a finite field, number of conjugacy classes in projective special linear group of fixed degree over a finite field is PORC function of field size $(q + 5)/2$ simplifies to $(5 + 5)/2 = 5$ #Interpretation as projective special linear group of degree two For $PSL(2,q), q = 3,5,7,9,\dots$ (odd $q$): 4,5,6,7,...
special linear group of degree two over a finite field of size $q$ $q = 4$, i.e., field:F4, so the group is $SL(2,4)$ $q + 4$ for $q$ odd
$q + 1$ for $q$ a power of 2
See element structure of special linear group of degree two, number of conjugacy classes in projective special linear group of fixed degree over a finite field is PORC function of field size $q + 1 = 4 + 1 = 5$ #Interpretation as special linear group of degree two over field:F4 For $SL(2,q), q = 2,4,8,16,\dots$ (powers of 2): 3,5,9,17,...