Element structure of special linear group:SL(2,5)

This article gives specific information, namely, element structure, about a particular group, namely: special linear group:SL(2,5).
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This article gives detailed information about the element structure of special linear group:SL(2,5), which is a group of order 120.

Summary

Item	Value
order of the whole group (total number of elements)	120
conjugacy class sizes	1,1,12,12,12,12,20,20,30 in grouped form: 1 (2 times), 12 (4 times), 20 (2 times), 30 (1 time) maximum: 30, number of conjugacy classes: 9, lcm: 60
order statistics	1 of order 1, 1 of order 2, 20 of order 3, 30 of order 4, 24 of order 5, 20 of order 6, 24 of order 10 maximum: 10, lcm (exponent of the whole group): 60

Elements

Order computation

The group $SL(2,5)$ 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
special linear group of degree two over a finite field of size $q$	$q = 5$, i.e., field:F5, so the group is $SL(2,5)$	$q^3 - q$, in factored form $q(q - 1)(q + 1)$	See order formulas for linear groups of degree two, order formulas for linear groups, and special linear group of degree two	$5^3 - 5 = 120$ Factored version: $5(5 - 1)(5 + 1) = 5(4)(6) = 120$	#Interpretation as special linear group of degree two
double cover of alternating group $2 \cdot A_n$ of degree $n$	degree $n = 5$, so the group is $2 \cdot A_5$	$n!$	See double cover of alternating group, element structure of double cover of alternating group	$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$	#Interpretation as double cover of alternating group
binary von Dyck group with parameters $(p,q,r)$	$(p,q,r) = (5,3,2)$ (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 binary von Dyck groups	$\frac{4}{1/5 + 1/3 + 1/2 - 1} = \frac{4}{1/30} = 120$	#Interpretation as binary von Dyck group

Conjugacy class structure

Conjugacy classes

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Interpretation as special linear group of degree two

Further information: element structure of special linear group of degree two over a finite field

In the table below, $q = 5$. Note that the information is presented for generic odd $q$ and then computed numerically for $q = 5$.

Nature of conjugacy class	Eigenvalue pairs of all conjugacy classes	Characteristic polynomials of all conjugacy classes	Minimal polynomials of all conjugacy classes	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 matrices (one per conjugacy class)
Scalar	$\{ 1, 1 \}$ or $\{ -1,-1\}$	$x^2 - 2x + 1$ or $x^2 + 2x + 1$	$x - 1$ or $x + 1$	1	1	2	2	2	2	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$ and $\begin{pmatrix} -1 & 0 \\ 0 & -1\\\end{pmatrix}$
Not diagonal, Jordan block of size two	$\{ 1, 1 \}$ or $\{ -1,-1\}$	$x^2 - 2x + 1$ or $x^2 + 2x + 1$	$x^2 - 2x + 1$ or $x^2 + 2x + 1$	$(q^2 - 1)/2$	12	4	4	$2(q^2 - 1)$	48	[SHOW MORE] $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$, $\begin{pmatrix} 1 & 2 \\ 0 & 1 \\\end{pmatrix}$, $\begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$, $\begin{pmatrix} -1 & 2 \\ 0 & -1 \\\end{pmatrix}$
Diagonalizable over $\mathbb{F}_{q^2}$ (in our case field:F25), not over $\mathbb{F}_q$ (in our case, field:F5). Must necessarily have no repeated eigenvalues.	For $q = 5$: $\{ 2 + \sqrt{3}, 2 - \sqrt{3} \}$ and $\{ -2 + \sqrt{3}, -2 - \sqrt{3} \}$, where $\sqrt{3}$ is interpreted a an element of field:F25 that squares to 3	For $q = 5$: $x^2 - x + 1$, $x^2 + x + 1$	For $q = 5$: $x^2 - x + 1$, $x^2 + x + 1$	$q(q - 1)$	20	$(q - 1)/2$	2	$q(q - 1)^2/2$	40	$\begin{pmatrix}0 & -1\\ 1 & 1\\\end{pmatrix}$, $\begin{pmatrix}0 & -1\\ 1 & -1\\\end{pmatrix}$
Diagonalizable over $\mathbb{F}_q$, i.e., field:F5, with distinct diagonal entries	For $q = 5$: $\{ 2,3 \}$	For $q = 5$: $x^2 + 1$	For $q = 5$: $x^2 + 1$	$q(q+1)$	30	$(q - 3)/2$	1	$q(q+1)(q-3)/2$	30	$\begin{pmatrix} 2 & 0 \\ 0 & 3 \\\end{pmatrix}$
Total	NA	NA	NA	NA	NA	$q + 4$	9	$q^3 - q$	120	NA

Interpretation as double cover of alternating group

Further information: element structure of double cover of alternating group

$SL(2,5)$ is isomorphic to $2 \cdot A_n,n = 5$. Recall that we have the following rules to determine splitting and orders. The rules listed below are only for partitions that already correspond to even permutations, i.e., partitions that have an even number of even parts:

Hypothesis: does the partition have at least one even part?	Hypothesis: does the partition have a repeated part? (the repeated part may be even or odd)	Conclusion: does the conjugacy class split from $S_n$ to $A_n$ in 2?	Conclusion: does the fiber in $2 \cdot A_n$ over a conjugacy class in $A_n$ split in 2?	Total number of conjugacy classes in $2 \cdot A_n$ corresponding to this partition (4 if Yes to both preceding columns, 2 if Yes to one and No to other, 1 if No to both)	Number of these conjugacy classes where order of element = lcm of parts	Number of these conjugacy classes where order of element = twice the lcm of parts
No	No	Yes	Yes	4	2	2
No	Yes	No	Yes	2	1	1
Yes	No	No	Yes	2	0	2
Yes	Yes	No	No	1	0	1

Partition	Partition in grouped form	Does the partition have at least one even part?	Does the partition have a repeated part?	Conclusion: does the conjugacy class split from $S_n$ to $A_n$ in 2?	Conclusion: does the fiber in $2 \cdot A_n$ over a conjugacy class in $A_n$ split in 2?	Total number of conjugacy classes in $2 \cdot A_n$ corresponding to this partition (4 if Yes to both preceding columns, 2 if Yes to one and No to other, 1 if No to both)	Size of each conjugacy class	Size formula (we take the size formula in $S_n$, multiply by 2, and divide by the number (1,2, or 4) two columns preceding)	Total number of elements (= twice the size of the $S_n$-conjugacy class)	Element orders	Formula for element orders
1 + 1 + 1 + 1 + 1	1 (5 times)	No	Yes	No	Yes	2	1	$\frac{2}{2} \frac{5!}{(1)^5(5!)}$	2	1 (1 class), 2 (1 class)	$\operatorname{lcm} \{ 1 \}$ (1 class) $2\operatorname{lcm} \{ 1 \}$ (1 class)
2 + 2 + 1	2 (2 times), 1 (1 time)	Yes	Yes	No	No	1	30	$\frac{2}{1} \frac{5!}{(2)^2(2!)(1)}$	30	4	$2\operatorname{lcm} \{ 2,1 \}$ (1 class)
3 + 1 + 1	3 (1 time), 1 (2 times)	No	Yes	No	Yes	2	20	$\! \frac{2}{2} \frac{5!}{(3)(1)^2(2!)}$	40	3 (1 class) 6 (1 class)	$\operatorname{lcm} \{ 3,1 \}$ (1 class) $2\operatorname{lcm} \{ 3,1 \}$ (1 class)
5	5 (1 time)	No	No	Yes	Yes	4	12	$\frac{2}{4} \frac{5!}{5}$	48	5 (2 classes), 10 (2 classes)	$\operatorname{lcm} \{ 5 \}$ (2 classes) $2 \operatorname{lcm} \{ 5 \}$ (2 classes)
Total	--	--	--	--	--	9	--	--	120	--	--

Conjugacy class structure: additional information

Number of conjugacy classes

The group has 9 conjugacy classes. This number can be computed in a variety of ways:

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
special linear group of degree two $SL(2,q)$ over a finite field of size $q$	$q = 5$, i.e., field:F5	Case $q$ odd: $q + 4$ Case $q$ even: $q + 1$	element structure of special linear group of degree two over a finite field; see also number of conjugacy classes in special 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 $q + 4 = 5 + 4 = 9$	#Interpretation as special linear group of degree two
double cover of alternating group $2 \cdot A_n$	$n = 5$, i.e., the group is double cover of alternating group:A5	(number of unordered integer partitions of $n$) + 3(number of partitions of $n$ into distinct odd parts) - (number of partitions of $n$ with a positive even number of even parts and with at least one repeated part)	See element structure of double cover of alternating group, splitting criterion for conjugacy classes in double cover of alternating group	For $n = 5$, the three numbers to calculate are respectively 7,1,1. So, we get $7 + 3(1) - 1 = 9$.	#Interpretation as double cover of alternating group
binary von Dyck group with parameters $p,q,r$ satisfying $1/p + 1/q + 1/r > 1$	$(p,q,r) = (5,3,2)$	$p + q + r - 1$		$p +q + r - 1 = 5 + 3 + 2 - 1 = 9$	#Interpretation as binary von Dyck group