# Element structure of symmetric group:S4

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This article discusses the element structure of symmetric group:S4, the symmetric group of degree four. We denote its elements as acting on the set $\{ 1,2,3,4 \}$, written using cycle decompositions, with composition by function composition where functions act on the left.

Since this group is a complete group (i.e., every automorphism is inner and the center is trivial), the classification of elements up to conjugacy is the same as the classification up to automorphisms. Further, since cycle type determines conjugacy class for symmetric groups, the conjugacy classes are parametrized by cycle types, which in turn are parametrized by unordered integer partitions of $4$.

This page concentrates on the more group-theoretic aspects of the element structure. For the more combinatorial aspects, see combinatorics of symmetric group:S4.

## Summary

Item Value
order of the whole group (total number of elements) 24
prime factorization $2^3 \cdot 3^1 = 8 \cdot 3$
See order computation for more
For other groups of the same order, see groups of order 24
conjugacy class sizes 1,3,6,6,8
maximum: 8, number: 5, sum (equals order of group): 24, lcm: 24
See conjugacy class structure for more.
number of conjugacy classes 5
See element structure of symmetric group:S4#Number of conjugacy classes
order statistics 1 of order 1, 9 of order 2, 8 of order 3, 6 of order 4
maximum: 4, lcm (exponent of the whole group): 12

## 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 $n = 4$ element structure of symmetric groups
projective general linear group of degree two over a finite field field:F3, i.e., the group is $PGL(2,3)$ element structure of projective general linear group of degree two over a finite field
general affine group of degree two over a finite field field:F2, i.e., the group is $GA(2,2)$ element structure of general affine group of degree two over a finite field
COMPARE AND CONTRAST: View element structure of groups of order 24 to compare and contrast the element structure with other groups of order 24.

## Elements

### Multiple ways of describing permutations

Note that the matrix for the right action is obtained by taking the transpose of the matrix for the left action. For the identity element and the elements of order 2, both matrices coincide.

Cycle decomposition notation One-line notation, i.e., image of string $1,2,3,4$ Matrix (left action) Matrix (right action) Order of element (= lcm of cycle sizes)
$()$ 1234 $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 1
$(3,4)$ 1243 $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 2
$(2,3)$ 1324 $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 2
$(2,3,4)$ 1342 $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 3
$(2,4,3)$ 1423 $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 3
$(2,4)$ 1432 $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 2
$(1,2)$ 2134 $\begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 2
$(1,2)(3,4)$ 2143 $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 2
$(1,2,3)$ 2314 $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 3
$(1,2,3,4)$ 2341 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 4
$(1,2,4,3)$ 2413 $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 4
$(1,2,4)$ 2431 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 3
$(1,3,2)$ 3124 $\begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 3
$(1,3,4,2)$ 3142 $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 4
$(1,3)$ 3214 $\begin{pmatrix}0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ $\begin{pmatrix}0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{pmatrix}$ 2
$(1,3,4)$ 3241 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 3
$(1,3)(2,4)$ 3412 $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 2
$(1,3,2,4)$ 3421 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 4
$(1,4,3,2)$ 4123 $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 4
$(1,4,2)$ 4132 $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 3
$(1,4,3)$ 4213 $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{pmatrix}$ 3
$(1,4)$ 4231 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 2
$(1,4,2,3)$ 4312 $\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\\end{pmatrix}$ 4
$(1,4)(2,3)$ 4321 $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\\end{pmatrix}$ 2

### Order computation

The symmetric group of degree four has order 24, with prime factorization $24 = 2^3 \cdot 3^1 = 8 \cdot 3$. Below are listed various methods that can be used to compute the order, all of which should give the answer 24:

Family Parameter values Formula for order of a group in the family Proof or justification of formula Evaluation at parameter values (should give the answer 24) Full interpretation of conjugacy class structure
symmetric group $S_n$ of degree $n$ degree $n = 4$ $n!$ See symmetric group, element structure of symmetric groups $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ #Interpretation as symmetric group
projective general linear group of degree two over a finite field of size $q$ size $q = 3$, i.e., field:F3, so the group is $PGL(2,3)$ $q^3 - q$
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 $3^3 - 3 = 24$
In factored form: $q(q - 1)(q + 1)$ becomes $3(2)(4) = 24$
#Interpretation as projective general linear group of degree two
general affine group of degree two over a finite field of size $q$ size $q = 2$, i.e., field:F2, so the group is $GA(2,2)$ or $AGL(2,2)$ $q^2(q^2 - 1)(q^2 - q)$ See order formulas for linear groups of degree two $2^2(2^2 - 1)(2^2 - 2) = (4)(3)(2) = 24$ #Interpretation as general affine group of degree two
general semiaffine group of degree one over a finite field of size $q = p^r$, $p$ prime $q = 4, p = 2, r = 2$, i.e., field:F4, so the group is $\Gamma A(1,4)$ or $A\Gamma L(1,4)$. $rq(q - 1)$ See order formulas for linear groups of degree one, general semiaffine group of degree one $rq(q - 1) = 2(4)(3) = 24$ #Interpretation as general semiaffine group of degree one
projective special linear group of degree two over a finite discrete valuation ring with length $l$ and residue field of size $q$ $q = 2, l = 2$, the ring is $\mathbb{Z}/4\mathbb{Z}$, and the group is $PSL(2,\mathbb{Z}/4\mathbb{Z})$ $q^{3l - 2}(q - 1)(q + 1)/m$ where $m$ is the number of square roots of unity in the ring See order formulas for linear groups of degree two Here $q = 2,l = 2$ and $m= 2$, so we get $2^{3\cdot 2 - 2}(2 - 1)(2 + 1)/2 = 2^4(1)(3)/2 = 24$ #Interpretation as projective special linear group of degree two
von Dyck group with parameters $(p,q,r)$ $(p,q,r) = (2,3,4)$ (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 $\frac{2}{1/2 + 1/3 + 1/4 - 1} = \frac{2}{1/12} = 24$ #Interpretation as von Dyck group
triangle group with parameters $(p,q,r)$ $(p,q,r) = (2,3,3)$ (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). $\frac{4}{1/p + 1/q + 1/r - 1}$ See element structure of triangle groups $\frac{4}{1/2 + 1/3 + 1/3 - 1} = \frac{4}{1/6} = 24$ #Interpretation as triangle group

#### Computation of prime powers in order

The prime factorization of the order is:

$24 = 2^3 \cdot 3^1$

Family Parameter values Formula for order of a group in the family Formula for the largest power of a generic prime $\ell$ dividing the order Case $\ell = 2$ (answer should be 3) Case $\ell = 3$ (answer should be 1) Case $\ell \ge 5$ (answer should be 0)
symmetric group $S_n$ of degree $n$ degree $n = 4$ $n!$ $\sum_{k=1}^{\lfloor \log_\ell n\rfloor} \lfloor \frac{n}{\ell^k}\rfloor$ $\lfloor\frac{4}{2^1}\rfloor + \lfloor\frac{4}{2^2}\rfloor = 2 + 1 = 3$ $\lfloor \frac{4}{3^1} \rfloor= 1$ no summands, so sum is 0
projective general linear group of degree two over a finite field of size $q$ size $q = 3$, i.e., field:F3, so the group is $PGL(2,3)$ $q^3 - q$
In factored form: $q(q - 1)(q + 1)$
If $\ell$ is the underlying prime of $q$, then $\log_\ell q$.
If $\ell = 2$ and $q \equiv 1 \pmod 4$, then 1 + the exponent for the largest power of 2 dividing $q -1$.
If $\ell = 2$ and $q \equiv -1 \pmod 4$, then 1 + the exponent for the largest power of 2 dividing $q + 1$.
Otherwise, if $\ell$ divides $q - 1$ or $q + 1$ (can't divide both) the exponent for the largest power dividing that. If it divides neither, then zero.
We use the $\ell = 2$ and $q \equiv -1 \pmod 4$ case, to get (1 + exponent for largest power of 2 dividing 4) = 1 + 2 = 3 We use the case that $\ell$ is the underlying prime of $q$, and get $\log_\ell q = \log_3 3 = 1$ Since 3 - 1 = 2 and 3 + 1 = 4 are powers of 2, there are no primes involved other than 2 and 3.
general affine group of degree two over a finite field of size $q$ size $q = 2$, i.e., field:F2, so the group is $GA(2,2)$ or $AGL(2,2)$ $q^2(q^2 - 1)(q^2 - q)$ If $\ell$ is the underlying prime of $q$, then $3\log_\ell q$.
If $\ell = 2$ and $q \equiv 1 \pmod 4$, then 1 + twice the exponent for the largest power of 2 dividing $q - 1$.
If $\ell = 2$ and $q \equiv -1 \pmod 4$, then 2 + the exponent for the largest power of 2 dividing $q + 1$.
Otherwise, if $\ell$ divides $q - 1$, then twice the exponent for the largest power of $\ell$ dividing $q - 1$.
Otherwise, if $\ell$ divides $q + 1$, then the exponent for the largest power of $\ell$ dividing $q + 1$.
Otherwise, 0.
$\ell = 2$ is the underlying prime for $q = 2$, so $3 \log_\ell q = 3 \log_22 = 3(1) = 3$. $\ell$ is odd and divides $q + 1$, so the exponent for the largest power of $\ell = 3$ dividing $q + 1 = 2 + 1 = 3$, which is 1. No primes other than 2 and 3 divide 2, 2 + 1, or 2 - 1.

## 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

The conjugacy class sizes are $1,3,6,6,8$.

### 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 Elements with the cycle type Size of conjugacy class Formula for size Even or odd? If even, splits? If splits, real in alternating group? Element order Formula calculating element order
1 + 1 + 1 + 1 1 (4 times) four cycles of size one each, i.e., four fixed points $()$ -- the identity element 1 $\! \frac{4!}{(1)^4(4!)}$ even; no 1 $\operatorname{lcm}\{ 1,1,1,1 \}$
2 + 1 + 1 2 (1 time), 1 (2 times) one transposition (cycle of size two), two fixed points $(1,2)$, $(1,3)$, $(1,4)$, $(2,3)$, $(2,4)$, $(3,4)$ 6 $\! \frac{4!}{[(2)^1(1!)][(1)^2(2!)]}$, also $\binom{4}{2}$ odd 2 $\operatorname{lcm}\{2,1,1 \}$
2 + 2 2 (2 times) double transposition: two cycles of size two $(1,2)(3,4)$, $(1,3)(2,4)$, $(1,4)(2,3)$ 3 $\! \frac{4!}{(2)^2(2!)}$ even; no 2 $\operatorname{lcm}\{2,2 \}$
3 + 1 3 (1 time), 1 (1 time) one 3-cycle, one fixed point $(1,2,3)$, $(1,3,2)$, $(2,3,4)$, $(2,4,3)$, $(3,4,1)$, $(3,1,4)$, $(4,1,2)$, $(4,2,1)$ 8 $\! \frac{4!}{[(3)^1(1!)][(1)^1(1!)]}$ or $\! \frac{4!}{(3)(1)}$ even; yes; no 3 $\operatorname{lcm}\{3,1 \}$
4 4 (1 time) one 4-cycle, no fixed points $(1,2,3,4)$, $(1,2,4,3)$, $(1,3,2,4)$, $(1,3,4,2)$, $(1,4,2,3)$, $(1,4,3,2)$ 6 $\! \frac{4!}{(4)^1(1!)}$ or $\frac{4!}{4}$ odd 4 $\operatorname{lcm} \{ 4 \}$
Total (5 rows, 5 being the number of unordered integer partitions of 4) -- -- -- 24 (equals 4!, the order of the whole group) -- odd: 12 (2 classes)
even; no: 4 (2 classes)
even; yes; no: 8 (1 class)
order 1: 1 (1 class)
order 2: 9 (2 classes)
order 3: 8 (1 class)
order 4: 6 (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 4 4 0
2 + 1 + 1 6 2 2 3 1
2 + 2 3 2 0 2 2
3 + 1 8 3 1 2 2
4 6 4 0 1 3
Mean over conjugacy classes 24/5 8/5 7/5 6 8/5
Mean over elements 73/12 67/24 1 25/12 23/12

The mean over elements of the number of fixed points is $1$ for all symmetric groups on finite sets. The mean over elements of the number of cycles is $1 + 1/2 + 1/3 + \dots + 1/n$, which in this case is $1 + 1/2 + 1/3 + 1/4$.

For characters, see linear representation theory of symmetric group:S4.

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

The symmetric group $S_4$ is isomorphic to $PGL(2,3)$, i.e., the projective general linear group of degree two over field:F3. Compare with element structure of projective general linear group of degree two over a finite field.

Nature of conjugacy class upstairs in $GL(2,q)$ (here $q = 3$) Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class (generic odd $q$) Size of conjugacy class ($q = 3$) Number of such conjugacy classes (generic odd $q$) Number of such conjugacy classes ($q = 3$) Total number of elements (generic odd $q$) Total number of elements ($q = 3$) Representatives of conjugacy classes as permutations
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar $\{ a, a \}$ where $a \in \mathbb{F}_q^\ast$ $(t - a)^2$ where $a \in \mathbb{F}_q^\ast$ $t - a$ 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. $t^2 - \mu$, $\mu$ a nonzero non-square Same as characteristic polynomial $q(q - 1)/2$ 3 1 1 $q(q - 1)/2$ 3 $(1,2)(3,4)$
Diagonalizable over $\mathbb{F}_q$ with mutually negative diagonal entries. $\{ \lambda, - \lambda \}$, all such pairs identified. $t^2 - \lambda^2$, all identified Same as characteristic polynomial $q(q + 1)/2$
$= (q^2 + q)/2$
6 1 1 $q(q + 1)/2$
$= (q^2 + q)/2$
6 $(1,2)$
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 $\mathbb{F}_q$. $t^2 - at + b$, $a \ne 0$, irreducible; with identification. Same as characteristic polynomial $q(q - 1)$ 6 $(q - 1)/2$ 1 $q(q -1)^2/2$
$= (q^3 - 2q^2 + q)/2$
6 $(1,2,3,4)$
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. $(t - a)^2$ Same as characteristic polynomial $q^2 - 1$ 8 1 1 $q^2 - 1$ 8 $(1,2,3)$
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. $t^2 - (\lambda + \mu)t+ \lambda\mu$, again with identification. Same as characteristic polynomial. $q(q + 1)$ 12 $(q - 3)/2$ 0 $q(q+1)(q - 3)/2 =$
$(q^3 - 2q^2 - 3q)/2$
0 --
Total NA NA NA NA NA $q + 2$ 5 $q^3 - q$ 24 --

### Interpretation as general affine group of degree two

The symmetric group $S_4$ is isomorphic to $GA(2,2)$, i.e., the general affine group of degree two over field:F2. Compare with element structure of general affine group of degree two over a finite field. In the table below, $q = 2$. The transformation is of the form $x \mapsto Ax + v$ where $A \in GL(2,2)$ and $v \in \mathbb{F}_2^2$.

Nature of conjugacy class Eigenvalues Characteristic polynomial of $A$ Minimal polynomial of $A$ Size of conjugacy class (generic $q$ Size of conjugacy class ($q = 2$) Number of such conjugacy classes (generic $q$) Number of such conjugacy classes ($q = 2$) Total number of elements (generic $q$ Total number of elements ($q = 2$) Representative element as permutation, one for each conjugacy class
$A$ is the identity, $v = 0$ $\{ 1,1 \}$ $(t - 1)^2$ $t- 1$ 1 1 1 1 1 1 $()$
$A$ is the identity, $v \ne 0$ $\{ 1,1 \}$ $(t - 1)^2$ $t - 1$ $q^2 - 1$ 3 1 1 $q^2 - 1$ 3 $(1,2)(3,4)$
$A$ is diagonalizable over $\mathbb{F}_q$ with equal diagonal entries not equal to 1, hence a scalar. The value of $v$ does not affect the conjugacy class. $\{a,a \}$ where $a \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $(t - a)^2$ where $a \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $t - a$ where $a \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $q^2$ 4 $q - 2$ 0 $q^2(q - 2)$ 0 --
$A$ is diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. The value of $v$ does not affect the conjugacy class. Pair of conjugate elements of $\mathbb{F}_{q^2}$ $t^2 - at + b$, irreducible Same as characteristic polynomial $q^3(q - 1)$ 8 $q(q - 1)/2 = (q^2 - q)/2$ 1 $q^4(q-1)^2/2$ 8 $(1,2,3)$
$A$ has Jordan block of size two, with repeated eigenvalue equal to 1, $v$ is in the image of $A - 1$ $\{ 1, 1 \}$ $(t - 1)^2$ Same as characteristic polynomial $q(q^2 - 1)$ 6 1 1 $q(q^2 - 1)$ 6 $(1,2)$
$A$ has Jordan block of size two, with repeated eigenvalue equal to 1, $v$ is not in the image of $A - 1$ $\{ 1, 1 \}$ $(t - 1)^2$ Same as characteristic polynomial $(q^2 - 1)(q^2 - q)$ 6 1 1 $(q^2 - 1)(q^2 - q)$ 6 $(1,2,3,4)$
$A$ has Jordan block of size two, with repeated eigenvalue not equal to 1 $a$ (multiplicity two) where $a \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $(t - a)^2$ where $a \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ Same as characteristic polynomial $q^2(q^2 - 1)$ 12 $q - 2$ 0 $q^2(q^2 - 1)(q - 2)$ 0 --
$A$ diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries, one of which is 1, $v$ is in the image of $A - 1$ $1,\mu$, $\mu \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $t^2 - (\mu + 1)t + \mu$ Same as characteristic polynomial $q^2(q + 1)$ 12 $q - 2$ 0 $q^2(q+1)(q-2)$ 0 --
$A$ diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries, one of which is 1, $v$ is not in the image of $A - 1$ $1,\mu$, $\mu \in \mathbb{F}_q^\ast \setminus \{ 1 \}$ $t^2 - (\mu + 1)t + \mu$ Same as characteristic polynomial $q(q + 1)(q^2 - q)$ 12 $q - 2$ 0 $q^2(q+1)(q - 1)(q-2)$ 0 --
$A$ diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries, neither of which is 1 $\lambda, \mu$ (interchangeable) distinct elements of $\mathbb{F}_q^\ast$, neither equal to 1 $t^2 - (\lambda + \mu)t + \lambda \mu$ Same as characteristic polynomial $q^3(q+1)$ 24 $(q - 2)(q - 3)/2$ 0 $q^3(q+1)(q-2)(q-3)/2$ 0 --
Total NA NA NA NA NA $q^2 + q - 1$ 5 $q^2(q^2 - 1)(q^2 - q)$ 24

### Interpretation as general semiaffine group of degree one

Compare and contrast with element structure of general semiaffine group of degree one over a finite field

We view the group as the general semiaffine group of degree one $\Gamma A(1,q)$ with $q = p^2$. Here, $q = 4$ and $p = 2$.

Nature of conjugacy class Size of conjugacy class (generic $p$) Size of conjugacy class ($p = 2$) Number of such conjugacy classes (generic $p$) Number of such conjugacy classes ($p = 2$) Total number of elements (generic $p$) Total number of elements ($p = 2$) Representative as permutation (one per class)
identity element of group, acts as identity map 1 1 1 1 1 1 $()$
non-identity element, in additive group. Acts as translation $q - 1 = p^2 - 1$ 3 1 1 $q - 1 = p^2 - 1$ 3 $(1,2)(3,4)$
outside the additive group, but in GA(1,q) and the multiplicative part is in the prime subfield. Acts as $x \mapsto ax + b$, with $a \in \mathbb{F}_p^\ast \setminus \{ 1 \}$ $q = p^2$ 4 $p - 2$ 0 $p^2(p - 2)$ 0 --
outside the additive group, but in GA(1,q) and the multiplicative part is outside the prime subfield. Acts as $x \mapsto ax + b$, with $a \in \mathbb{F}_q^\ast \setminus \mathbb{F}_p^\ast$ $2q = 2p^2$ 8 $p(p - 1)/2$ 1 $p^3(p - 1)$ 8 $(1,2,3)$
outside GA(1,q), has order two (i.e., generates a permutable complement to the subgroup of index two that is GA(1,q)) $p(p + 1)$ 6 1 1 $p(p + 1)$ 6 $(1,2)$
outside GA(1,q), its image mod the additive group has order two but it does not itself have order two $p (p - 1)(p + 1) = p^3 - p$ 6 1 1 $p(p - 1)(p + 1) = p^3 - p$ 6 $(1,2,3,4)$
outside GA(1,q), its image mod the additive group does not have order two $p^2(p + 1)$ 12 $p - 2$ 0 $p^2(p + 1)(p - 2)$ 0 --
Total -- -- $p(p + 3)/2 = (p^2 + 3p)/2$ (equals number of conjugacy classes in the group) 5 $2p^2(p^2 - 1)$ (equals order of the whole group) 24 --

## Conjugacy class structure: additional information

### Number of conjugacy classes

The symmetric group of degree four 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
symmetric group $S_n$ of degree $n$ degree $n = 4$ 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 4 equals 5 #Interpretation as symmetric group
projective general linear group of degree two $PGL(2,q)$ over a finite field of size $q$ $q = 3$, i.e., field:F3 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 3 is odd, we use the odd case formula, and get $q + 2 = 3 + 2 = 5$ #Interpretation as projective general linear group of degree two
general affine group of degree two $GA(2,q)$ over a finite field of size $q$ $q = 2$, i.e., field:F2 $q^2 + q - 1$ element structure of general affine group of degree two over a finite field; see also number of conjugacy classes in general affine group of fixed degree over a finite field is polynomial function of field size $2^2 + 2 - 1 = 5$ #Interpretation as general affine group of degree two
general semiaffine group of degree one $\Gamma A(1,q)$ over a finite field of size $q = p^2$, $p$ prime $p = 2$, so $q = 4$, i.e., field:F4 $p(p + 3)/2$ element structure of general semiaffine group of degree one over a finite field $p(p + 3)/2 = 2(2 + 3)/2 = 5$ #Interpretation as general semiaffine group of degree one

## Cayley graphs

### With generating set all transpositions

Here, the generating set is the set of all transpositions. Since the generating set is a conjugacy class of involutions, the left and right Cayley graphs are identical. Further, we can unambiguously give a direction to each edge (away from the identity element) because there are no cycles of odd length, which follows from the fact that all elements of the generating set are odd permutations.

The following is some useful tabulated information about the Cayley graph. The edges to/from listed here are the edges for any representative element, not the total across the conjugacy class. Note that the sum of edges to/from in each row is $6$, which is the number of generators used in the generating set.

Conjugacy class Distance from identity in Cayley graph Number of elements Edges to/from $()$ Edges to/from $(1,2)$-class Edges to/from $(1,2,3)$-class Edges to/from $(1,2)(3,4)$-class Edges to/from $(1,2,3,4)$-class
$()$ 0 1 0 6 0 0 0
$(1,2)$ 1 6 1 0 4 1 0
$(1,2,3)$ 2 8 0 3 0 0 3
$(1,2)(3,4)$ 2 3 0 2 0 0 4
$(1,2,3,4)$ 3 6 0 0 4 2 0

## Bruhat ordering

The basic picture:

A fuller picture:

### Basic tabulation by length

Length Number of elements of that length Elements of that length Conjugacy class information for these elements
0 1 $()$ -- the identity element forms a single conjugacy class
1 3 $(1,2)$, $(2,3)$, $(3,4)$ 3 of 6 elements in conjugacy class of $(1,2)$
2 5 $(1,2)(3,4)$, $(1,2,3)$, $(1,3,2)$, $(2,3,4)$, $(2,4,3)$ 1 of 3 elements in conjugacy class of $(1,2)(3,4)$, 4 of 8 elements in conjugacy class of $(1,2,3)$
3 6 $(1,3)$, $(2,4)$, $(1,2,3,4)$, $(1,4,3,2)$, $(1,2,4,3)$, $(1,3,4,2)$ 2 of 6 elements in conjugacy class of $(1,2)$, 4 of 6 elements in conjugacy class of $(1,2,3,4)$
4 5 $(1,3)(2,4)$, $(1,2,4)$, $(1,4,2)$, $(1,3,4)$, $(1,4,3)$ 1 of 3 elements in conjugacy class of $(1,2)(3,4)$, 4 of 8 elements in conjugacy class of $(1,2,3)$
5 3 $(1,3,2,4)$, $(1,4,2,3)$, $(1,4)$ 2 of 6 elements in conjugacy class of $(1,2,3,4)$, 1 of 6 elements in conjugacy class of $(1,4)$
6 1 $(1,4)(2,3)$ 1 of 3 elements in conjugacy class of $(1,2)(3,4)$

### Equivalence classes up to symmetries

These are equivalence classes up to the two symmetries: flipping the $s_i$s and a left-to-right order reversal symmetry. Elements in the same equivalence class are in the same conjugacy class in the group and the corresponding points in the Bruhat ordering are in the same orbit under automorphisms of the graph of the Bruhat ordering.

Elements in equivalence class Number of elements Length Indegree Outdegree Image class under the anti-automorphism
$()$ 1 0 0 1 $(1,4)(2,3)$
$(1,2)$, $(3,4)$ 2 1 1 3 $(1,3,2,4)$, $(1,4,2,3)$
$(2,3)$ 1 1 1 4 $(1,4)$
$(1,2,3)$, $(1,3,2)$, $(2,3,4)$, $(2,4,3)$ 4 2 2 3 $(1,2,4)$, $(1,3,4)$, $(1,4,2)$, $(1,4,3)$
$(1,2)(3,4)$ 1 2 2 2 $(1,3)(2,4)$
$(1,3)$, $(2,4)$ 2 3 2 2 $(1,3)$, $(2,4)$
$(1,2,3,4)$, $(1,4,3,2)$ 2 3 2 2 $(1,2,3,4)$, $(1,4,3,2)$
$(1,2,4,3)$, $(1,3,4,2)$ 2 3 3 3 $(1,2,4,3)$, $(1,3,4,2)$
$(1,2,4)$, $(1,3,4)$, $(1,4,2)$, $(1,4,3)$ 4 4 3 2 $(1,2,3)$, $(1,3,2)$, $(2,3,4)$, $(2,4,3)$
$(1,3)(2,4)$ 1 4 2 2 $(1,2)(3,4)$
$(1,3,2,4)$, $(1,4,2,3)$ 2 5 3 1 $(1,2)$, $(3,4)$
$(1,4)$ 1 5 4 1 $(2,3)$
$(1,4)(2,3)$ 1 6 1 0 $()$