# Element structure of alternating group:A6

View element structure of particular groups | View other specific information about alternating group:A6

For convenience, we take the set acted upon as $\{ 1,2,3,4,5,6 \}$.

## Elements

### Order computation

The alternating group of degree six has order 360, with prime factorization $360 = 2^3 \cdot 3^2 \cdot 5^1 = 8 \cdot 9 \cdot 5$. Below are listed various methods that can be used to compute the order, all of which should give the answer 360:

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
alternating group $A_n$ of degree $n$ degree $n = 6$ $n!/2$ See alternating group, element structure of alternating groups $6!/2 = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1/2 = 360$ #Interpretation as alternating group
projective special linear group of degree two over a finite field of size $q$ $q = 9$, i.e., field:F9, so the group is $PSL(2,9)$ $(q^3 - q)/2 = q(q - 1)(q + 1)/2$ for $q$ odd
$q^3 - q = q(q - 1)(q + 1)$ 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 $(9^3 - 9)/2 = (729 - 9)/2 = 360$
Factored version: $9(9 - 1)(9 + 1)/2 = 9(8)(10)/2 = 360$
#Interpretation as projective special linear group of degree two

## 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 7 conjugacy classes, of which 5 are unsplit from symmetric group:S6, and 2 are a split pair arising from a single conjugacy class in $S_6$. The conjugacy class sizes are 1, 40, 45, 90, 40, 72, 72.

### 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 Size of conjugacy class Formula for size Element order
1 + 1 + 1 + 1 + 1 + 1 six fixed points $()$ -- the identity element 1 $\! \frac{6!}{(1)^6(6!)}$ 1
3 + 1 + 1 + 1 one 3-cycle, three fixed points $(1,2,3)$ 40 $\! \frac{6!}{(3)(1)^3(3!)}$ 3
2 + 2 + 1 + 1 double transposition: two 2-cycles, two fixed points $(1,2)(3,4)$ 45 $\frac{6!}{(2)^2(2!)(1)^2(2!)}$ 2
4 + 2 one 4-cycle, one 2-cycle $(1,2,3,4)(5,6)$ 90 $\! \frac{6!}{(4)(2)}$ 4
3 + 3 two 3-cycles $(1,2,3)(4,5,6)$ 40 $\! \frac{6!}{(3)^2(2!)}$ 3

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 + 1 one 5-cycle, one fixed point 144 $\! \frac{6!}{(5)(1)}$ 72 $(1,2,3,4,5)$ $(1,3,5,2,4)$ Yes No 5

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

We consider the group as $PSL(2,q)$, $q = 9$. 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 ($q = 9$) Number of such conjugacy classes (generic $q$ that is 1 mod 4) Number of such conjugacy classes ($q = 9$) Total number of elements (generic $q$ that is 1 mod 4) Total number of elements ($q = 9$) 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 $a \in \{ -1,1 \}$ $x - a$ where $a \in \{ -1,1\}$ $(q^2 - 1)/2$ 40 2 2 $q^2 - 1$ 80 $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$ $(1,2,3)$ and $(1,2,3)(4,5,6)$
Diagonalizable over $\mathbb{F}_q$ with diagonal entries squaring to $-1$ $i, -i$ where $i^2 = -1$ $x^2 + 1$ $x^2 + 1$ $q(q + 1)/2$ 45 1 1 $q(q + 1)/2$ 45 PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $(1,2)(3,4)$
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of $\mathbb{F}_{q^2}$ of norm 1. Each pair identified with its negative pair. $x^2 - ax + 1$, irreducible; note that $x^2 - ax + 1$'s pair and $x^2 + ax +1$'s pair get identified. Same as characteristic polynomial $q(q - 1)$ 72 $(q - 1)/4$ 2 $q(q - 1)^2/4$ 144 PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $(1,2,3,4,5)$ and $(1,3,5,2,4)$
Diagonalizable over $\mathbb{F}_q$ with distinct (and hence mutually inverse) diagonal entries, whose square is not $-1$ $\lambda, 1/\lambda$ where $\lambda \in \mathbb{F}_q \setminus \{ 0,1,-1,i,-i \}$ where $i,-i$ are square roots of $-1$. Note that the representative pairs $\{ \lambda, 1/\lambda \}$ and $\{ -\lambda,-1/\lambda \}$ get identified. $x^2 - (\lambda + 1/\lambda)x + 1$, again with identification. $x^2 - (\lambda + 1/\lambda)x + 1$, again with identification. $q(q + 1)$ 90 $(q - 5)/4$ 1 $q(q + 1)(q - 5)/4$ 90 PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] $(1,2,3,4)(5,6)$
Total NA NA NA NA NA $(q + 5)/2$ 7 $(q^3 - q)/2$ 360 NA NA

## Conjugacy class structure: additional information

### Number of conjugacy classes

The alternating group of degree six has 7 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 7:

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
alternating group $A_n$ of degree $n$ $n = 6$, i.e., the group $A_6$ (Number of pairs of non-self-conjugate partitions of $n$) + 2(Number of self-conjugate partitions of $n$) See element structure of alternating groups $5 + 2(1) = 7$ #Interpretation as alternating group
projective special linear group of degree two over a finite field of size $q$ $q = 9$, i.e., field:F9, so the group is $PSL(2,9)$ $(q + 5)/2$ 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 $(9 + 5)/2 = 7$ #Interpretation as projective special linear group of degree two