# Element structure of symmetric group:S5

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## Summary

Item Value
order of the whole group (total number of elements) 120
conjugacy class sizes 1,10,15,20,20,24,30
maximum: 30, number of conjugacy classes: 7, lcm: 120
order statistics 1 of order 1, 25 of order 2, 20 of order 3, 30 of order 4, 24 of order 5, 20 of order 6
maximum: 6, lcm (exponent of the whole group): 60

## Family contexts

Note that if you go to the #Conjugacy class structure section of this article, you'll find a discussion of the conjugacy class structure with each of the below family interpretations.

Family name Parameter values General discussion of element structure of family
symmetric group degree $n = 5$ element structure of symmetric groups
projective general linear group of degree two over a finite field field:F5, i.e., the group is $PGL(2,5)$ element structure of projective general linear group of degree two over a finite field
projective semilinear group of degree two over a finite field field:F4, i.e., the group is $P\Gamma L(2,4)$ element structure of projective semilinear group of degree two over a finite field
COMPARE AND CONTRAST: View element structure of groups of order 120 to compare and contrast the element structure with other groups of order 120.

## Elements

### Order computation

The symmetric group of degree five has order 120, with prime factorization $120 = 2^3 \cdot 3^1 \cdot 5^1= 8 \cdot 3 \cdot 5$. Below are listed various methods that can be used to compute the order, all of which should give the answer 120:

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
symmetric group $S_n$ of degree $n$ degree $n = 5$ $n!$ See symmetric group, element structure of symmetric groups $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ #Interpretation as symmetric group
projective general linear group of degree two over a finite field of size $q$ size $q = 5$, i.e., field:F5, so the group is $PGL(2,5)$ $q^3 - q$
In factored form: $q(q - 1)(q + 1)$
See order formulas for linear groups of degree two, order formulas for linear groups, projective general linear group of degree two $q^3 - q$ becomes $5^3 - 5 = 120$
In factored form: $q(q - 1)(q + 1)$ becomes $5(4)(6) = 120$
#Interpretation as projective general linear group of degree two
projective semilinear group of degree two over a finite field of size $q = p^r$, $p$ prime size $q = 4$, $p = 2$, $r = 2$, i.e., field:F4, so the group is $P\Gamma L(2,4)$ $r(q^3 - q)$
In factored form: $rq(q - 1)(q + 1)$
See order formulas for linear groups of degree two, order formulas for linear groups, projective general linear group of degree two $r(q^3 - q)$ becomes $2(4^3 - 4) = 120$
In factored form: $rq(q - 1)(q + 1)$ becomes $2(4)(3)(5) = 120$
#Interpretation as projective semilinear 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

### Interpretation as symmetric 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 any symmetric group, cycle type determines conjugacy class, i.e., the cycle type of a permutation (which describes the sizes of the cycles in a cycle decomposition of that permutation), determines its conjugacy class. In other words, two permutations are conjugate if and only if they have the same number of cycles of each size.

The cycle types (and hence the conjugacy classes) are parametrized by partitions of the size of the set. We describe the situation for this group:

Partition Partition in grouped form Verbal description of cycle type Representative element with the cycle type Size of conjugacy class Formula calculating size Even or odd? If even, splits? If splits, real in alternating group? Element order Formula calcuating element order
1 + 1 + 1 + 1 + 1 1 (5 times) five fixed points $()$ -- the identity element 1 $\! \frac{5!}{(1)^5(5!)}$ even; no 1 $\operatorname{lcm}\{1 \}$
2 + 1 + 1 + 1 2 (1 time), 1 (3 times) transposition: one 2-cycle, three fixed point $(1,2)$ 10 $\! \frac{5!}{[(2)^1(1!)][(1)^3(3!)]}$ or $\! \frac{5!}{(2)(1)^3(3!)}$, also $\binom{5}{2}$ in this case odd 2 $\operatorname{lcm}\{2,1 \}$
3 + 1 + 1 3 (1 time), 1 (2 times) one 3-cycle, two fixed points $(1,2,3)$ 20 $\! \frac{5!}{[(3)^1(1!)][(1)^2(2!)]}$ or $\! \frac{5!}{(3)(1)^2(2!)}$ even; no 3 $\operatorname{lcm}\{3,1\}$
2 + 2 + 1 2 (2 times), 1 (1 time) double transposition: two 2-cycles, one fixed point $(1,2)(3,4)$ 15 $\frac{5!}{[2^2(2!)][1^1(1!)]}$ or $\! \frac{5!}{2^2(2!)(1)}$ even; no 2 $\operatorname{lcm}\{2,1 \}$
4 + 1 4 (1 time), 1 (1 time) one 4-cycle, one fixed point $(1,2,3,4)$ 30 $\! \frac{5!}{[4^1(1!)][1^1(1!)]}$ or$\! \frac{5!}{(4)(1)}$ odd 4 $\operatorname{lcm}\{4,1\}$
3 + 2 3 (1 time), 2 (1 time) one 3-cycle, one 2-cycle $(1,2,3)(4,5)$ 20 $\! \frac{5!}{[3^1(1!)][2^1(1!)]}$ or $\! \frac{5!}{(3)(2)}$ odd 6 $\operatorname{lcm}\{3,2 \}$
5 5 (1 time) one 5-cycle $(1,2,3,4,5)$ 24 $\frac{5!}{5^1(1!)}$ or $\! \frac{5!}{5}$ even; yes; yes 5 $\operatorname{lcm} \{ 5 \}$
Total (7 rows, 7 being the number of unordered integer partitions of 5) -- -- -- 120 (equals order of the group) -- odd: 60 (3 classes)
even;no: 36 (3 classes)
even;yes;yes: 24 (1 class)

FACTS TO CHECK AGAINST ON FIXED POINTS AND CYCLES
Fixed points: probability distribution of number of fixed points of permutations | expected number of fixed points of permutation equals one
Number of cycles: probability distribution of number of cycles of permutations | expected number of cycles of permutation equals harmonic number of degree
Partition Number of elements in conjugacy class Order of elements Number of fixed points Number of cycles (including fixed points) Minimum number of transpositions that must be multiplied to obtain this cycle decomposition
1 + 1 + 1 + 1 + 1 1 1 5 5 0
2 + 1 + 1 + 1 10 2 3 4 1
3 + 1 + 1 20 3 2 3 2
2 + 2 + 1 15 2 1 3 2
4 + 1 30 4 1 2 3
3 + 2 20 6 0 2 3
5 24 5 0 1 4
Mean over conjugacy classes 120/7 18/7 12/7 20/7 15/7
Mean over elements 1301/60 157/40 1 137/60 25/12

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

Compare with element structure of projective general linear group of degree two over a finite field. Here, the field is field:F5, so $q = 5$.

Nature of conjugacy class upstairs in $GL_2$ Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class (generic odd $q$) Size of conjugacy class ($q = 5$) Number of such conjugacy classes (generic odd $q$) Number of such conjugacy classes ($q = 5$) Total number of elements (generic odd $q$) Total number of elements ($q = 5$) Representative as permutation (one per conjugacy class)
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar $\{ a, a \}$ where $a \in \mathbb{F}_q^\ast$ $(x - a)^2$ where $a \in \mathbb{F}_q^\ast$ $x - a$ where $a \in \mathbb{F}_q^\ast$ 1 1 1 1 1 1 $()$
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$, eigenvalues are negatives of each other. Pair of mutually negative conjugate elements of $\mathbb{F}_{q^2}$. All such pairs identified. $x^2 - \mu$, $\mu$ a nonzero non-square Same as characteristic polynomial $q(q - 1)/2$ 10 1 1 $q(q - 1)/2$ 10 $(1,2)$
Diagonalizable over $\mathbb{F}_q$ with mutually negative diagonal entries. $\{ \lambda, - \lambda \}$, all such pairs identified. $x^2 - \lambda^2$, all identified Same as characteristic polynomial $q(q + 1)/2 = (q^2 + q)/2$ 15 1 1 $q(q + 1)/2 = (q^2 + q)/2$ 15 $(1,2)(3,4)$
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$, eigenvalues are not negatives of each other. Pair of conjugate elements of $\mathbb{F}_{q^2}$. Each pair identified with anything obtained by multiplying both elements of it by an element of $\mathbb{F}_q$. $x^2 - ax + b$, $a \ne 0$, irreducible; with identification. Same as characteristic polynomial $q(q - 1)$ 20 $(q - 1)/2$ 2 $q(q -1)^2/2 = (q^3 - 2q^2 + q)/2$ 40 $(1,2,3)$ and $(1,2,3)(4,5)$
Not diagonal, has Jordan block of size two $a \in\mathbb{F}_q^\ast$ (multiplicity 2). Each conjugacy class has one representative of each type. $(x - a)^2$ Same as characteristic polynomial $q^2 - 1$ 24 1 1 $q^2 - 1$ 24 $(1,2,3,4,5)$
Diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries whose sum is not zero. $\lambda, \mu$ where $\lambda,\mu \in \mathbb{F}_q^\ast$ and $\lambda + \mu \ne 0$. The pairs $\{ \lambda, \mu \}$ and $\{ a\lambda, a\mu \}$ are identified. $x^2 - (\lambda + \mu)x + \lambda\mu$, again with identification. Same as characteristic polynomial. $q(q + 1)$ 30 $(q - 3)/2$ 1 $q(q+1)(q - 3)/2 = (q^3 - 2q^2 - 3q)/2$ 30 $(1,2,3,4)$
Total NA NA NA NA NA $q + 2$ 7 $q^3 - q$ 120 --

## Conjugacy class structure: additional information

### Number of conjugacy classes

The symmetric group of degree five 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
symmetric group $S_n$ of degree $n$ degree $n = 5$ number of unordered integer partitions of $n$ Follows from cycle type determines conjugacy class. For more, see element structure of symmetric groups number of unordered integer partitions of 5 equals 7 #Interpretation as symmetric group
projective general linear group of degree two $PGL(2,q)$ over a finite field of size $q$ $q = 5$, i.e., field:F5 Case $q$ odd: $q + 2$
Case $q$ even: $q + 1$
element structure of projective general linear group of degree two over a finite field. See also number of conjugacy classes in projective general linear group of fixed degree over a finite field is PORC function of field size Since 5 is odd, we use the odd case formula, and get $q + 2 = 5 + 2 = 7$ #Interpretation as projective general linear group of degree two