# A5 (Group) - A5 (Gruppe)

The group considered in the mathematical branch of group theory${\displaystyle A_{5}}$is the 5th degree alternating group . It has 60 elements and is the smallest non-Abelian simple group and the smallest non- resolvable group . It finds a geometric realization as a group of rotations of the icosahedron .

## Definitions

The cycle (234) as an illustration

We consider the set of all bijective mappings of the 5-element set${\displaystyle \{1,2,3,4,5\}}$in itself. This forms a group with the sequential execution of images as a link ; this link is also called a product and is written as${\displaystyle \cdot }$or without any connection signs. This is the symmetrical group ${\displaystyle S_{5}}$ With ${\displaystyle 5!=120}$ Elements.

Such mappings are called permutations and the so-called cycle notation is used for them ${\displaystyle (i_{1}\ldots i_{k})}$ with different elements ${\displaystyle i_{1},\ldots ,i_{k}\in \{1,2,3,4,5\}}$. The illustration${\displaystyle (i_{1}\ldots i_{k})}$maps each element in the cycle list to the one on the right and finally the last element in the list to the first. The cycle${\displaystyle (2\,3\,4)}$So maps 2 to 3, 3 to 4 and 4 to 2 and leaves elements 1 and 5 fixed. A cycle${\displaystyle (i\,j)}$ of length 2 therefore only interchanged ${\displaystyle i}$ and ${\displaystyle j}$and leaves all other elements fixed, such mappings are called transpositions . Different cycles can describe the same permutation, for example${\displaystyle (2\,3\,4)=(3\,4\,2)}$'Uniqueness is obtained by agreeing to put the smallest number occurring in the cycle at the beginning.

Any permutation can be written as the product of cycles, even as the product of transpositions. The representation as a product of transpositions is ambiguous. See for example

${\displaystyle (1\,2\,3)=(1\,2)\cdot (2\,3)=(1\,2)\cdot (2\,3)\cdot (4\,5)\cdot (4\,5)}$

We use the usual sequence for illustrations, that is, the illustration first ${\displaystyle (2\,3)}$ applied, then ${\displaystyle (1\,2)}$. (This is not done consistently in the literature; authors who write operations and functions on the right-hand side of the elements to be mapped use exactly the opposite convention.) However, it is clear whether a straight permutation is used to represent a permutation as a product of transpositions or an odd number of transpositions is required, correspondingly the permutations are called even or odd.

Then it is clear that the product of even permutations is even again, because the numbers of transpositions used add up when combining. The even permutations therefore form a subgroup , that is the alternating group${\displaystyle A_{5}}$.

It goes without saying that there are analogous concepts for ${\displaystyle \{1,\ldots ,n\}}$ instead of ${\displaystyle \{1,2,3,4,5\}}$possible, which then leads to the alternating group A n . In this article we cover the case${\displaystyle n=5}$.

## Elementary properties

### Number of elements

Is ${\displaystyle \sigma \in S_{5}}$ some permutation so is ${\displaystyle (1\,2)\cdot \sigma }$ odd or even if and only if ${\displaystyle \sigma }$is odd or even. So there are as many even as odd permutations, and it follows that${\displaystyle A_{5}}$ Has 60 elements.

### Cycle of three

A cycle of three, that is to say, a cycle ${\displaystyle (i\,j\,k)}$ of length three, is straight, because

${\displaystyle (i\,j\,k)=(i\,j)\cdot (j\,k)}$.

A cycle of three ${\displaystyle (i\,j\,k)}$ is apparently a figure that consists of each of the elements ${\displaystyle \{i,j,k\}}$ maps to a different element of this set of three and the other two elements ${\displaystyle \{1,2,3,4,5\}}$leaves solid; there are exactly two such images, namely${\displaystyle (i\,j\,k)}$ and ${\displaystyle (i\,k\,j)}$. As it total${\displaystyle \textstyle {\binom {5}{3}}=10}$If there are such sets of three, we come to a total of 20 cycles of three. The other way around

${\displaystyle (i\,j)\cdot (j\,k)=(i\,j\,k)}$ For ${\displaystyle i,j,k}$ different in pairs
${\displaystyle (i\,j)\cdot (k\,l)=(i\,k\,j)\cdot (i\,k\,l)}$ For ${\displaystyle i,j,k,l}$ different in pairs,

every even permutation is a product of cycles of three, that is, the group ${\displaystyle A_{5}}$ is generated by the triple cycles.

### Orders

As in every group, there is exactly one element of order 1, namely the neutral element .

The elements of order 2 are obtained from transpositions, which obviously have order 2. There${\displaystyle A_{5}}$ contains only even permutations, the permutations of order 2 are exactly the products of two transpositions foreign to the element ${\displaystyle (i\,j)\cdot (k\,l)}$ with different pairs ${\displaystyle i,j,k,l\in \{1,2,3,4,5\}}$. There are 5 possibilities for a set of four${\displaystyle \{i,j,k,l\}\subset \{1,2,3,4,5\}}$ (one element does not belong in each case) and for each such set of four one can have three different elements ${\displaystyle (i\,j)\cdot (k\,l),(i\,k)\cdot (j\,l),(i\,l)\cdot (j\,k)\in A_{5}}$ of order 2, that makes a total of ${\displaystyle 5\cdot 3=15}$ Elements of order 2.

The elements of order 3 are the 20 three-cycle cycles mentioned above.

All five-cycle ${\displaystyle (i\,j\,k\,l\,m)=(i\,j\,k)\cdot (k\,l\,m)}$ are products of two cycles of three and therefore elements of the ${\displaystyle A_{5}}$ and obviously have the order 5. Since all 5 numbers are in ${\displaystyle (i\,j\,k\,l\,m)}$occur, the 1 is also included, which is put in the first position. Therefore there are exactly the five-cycle${\displaystyle (1\,j\,k\,l\,m)}$ with different pairs ${\displaystyle j,k,l,m\in \{2,3,4,5\}}$, and there is also ${\displaystyle 4!=24}$Options. There are therefore 24 elements of order 5.

With this we have determined the orders of 1 + 15 + 20 + 24 = 60 elements, so there are no elements of further orders. This gives us the following overview:

order number Typical element description
1 1 ${\displaystyle e}$ neutral element
2 15 ${\displaystyle (i\,j)\cdot (k\,l)}$ two transpositions foreign to the element
3 20 ${\displaystyle (i\,j\,k)}$ Cycle of three
5 24 ${\displaystyle (i\,j\,k\,l\,m)}$ Five-cycle

Linking table of the alternating group A 5 in color. The neutral element is black

With the alternating group${\displaystyle A_{4}}$ it is still possible to obtain the group elements and the link table from the geometric image of the rotations of a tetrahedron. The group${\displaystyle A_{5}}$occurs as a rotation group of the icosahedron (and the dodecahedron , which is dual to the icosahedron). That is why it is also called the icosahedral rotating group and alternatively denotes it with the letter${\displaystyle I}$. It is a subgroup of the full icosahedral group ${\displaystyle I_{h}}$.

Geometric assignments are hardly practicable for a group with 60 elements. With a link table with 60 x 60 positions, it would also be confusing to write numbers, letters [1] or symbols in the table . However, it is possible to represent the elements using colored squares and, accordingly, the link table, as is done, for example, in the online encyclopedia for mathematics MathWorld . [2]

It should be noted that, in general, no particular arrangement can be identified for the elements of a group. The only fixed rule is that the neutral element is the first element of every row and column (top left corner). In order for a link table to make sense without specifying the individual elements, for example as permutations , one should commit to a comprehensible rule for the order of the elements. This is possible if you choose the order according to the faculty-based number system . With a permutation generator you can generate all 120 permutations of 5 objects in an orderly order. [3] This gives the elements of the symmetrical group${\displaystyle S_{5}}$. To go to the alternating group${\displaystyle A_{5}}$you just have to delete all odd permutations. Now the 59 x 59 group multiplications have to be carried out with these elements.

Since the order has been determined according to the faculty-based number system, each permutation is assigned an ordinal number from 0 to 59. They are assigned to a hue in the HSV color space with constant color saturation and constant brightness . The size of the hue is usually specified on a color wheel with a range of values ​​from 0 to 360 (in degrees). The hues for the permutations are now distributed equidistantly according to their permutation number over the range of values ​​of the hue quantity . This gives you a set of rules for assigning a color to an element of any group.

In this order, the ordinal numbers 1 and 2 are fixed points for elements 1, 2 and 3, and 1 is the sole fixed point for elements 4 to 12. Thus the first three elements form an alternating subgroup of the type${\displaystyle A_{3}}$ and elements 1 to 12 an alternating subgroup of the type ${\displaystyle A_{4}}$. These two subsets of the alternating group${\displaystyle A_{5}}$can be clearly seen as diagonal blocks from the graphic on the link table. Furthermore, you immediately notice a grouping in blocks of 12 elements each.

## presentation

A presentation by generators and relations looks like this: The group${\displaystyle A_{5}}$ becomes through two generators ${\displaystyle x,y}$ and the relations

${\displaystyle x^{5}=y^{2}=(xy)^{3}=e}$

Are defined. That is, any group made up of two elements${\displaystyle x,y}$ is generated, which also fulfill the mentioned relations, is isomorphic to ${\displaystyle A_{5}}$.

The ${\displaystyle A_{5}}$ itself is from ${\displaystyle x=(1\,2\,3\,4\,5)}$ and ${\displaystyle y=(1\,2)\cdot (3\,4)}$generated, and these elements satisfy the specified relations. [4]

## Transitive operation on 6 elements

The group ${\displaystyle A_{5}}$has 24 elements of order 5, of which 4 together with the neutral element form a subgroup of order 5, there are therefore six subgroups of order 5, which are also the 5- Sylow groups . Since the group operates transitively on the six 5-Sylow groups by means of conjugation , because every two 5-Sylow groups are conjugated, we get in total that${\displaystyle A_{5}}$operates transitively on a variable element set. This operation is even faithful . The following reversal applies: [5]

• Every 60-element transitive permutation group on 6 elements is isomorphic to ${\displaystyle A_{5}}$.

## A 5 cannot be resolved

To any group ${\displaystyle G}$is the commutator group ${\displaystyle K^{1}(G)}$defined as that of all commutators ${\displaystyle [x,y]:=x^{-1}y^{-1}xy}$generated subgroup. One explains inductively${\displaystyle K^{n+1}(G):=K^{1}(K^{n}(G))}$and calls the group resolvable if there is one${\displaystyle n}$ gives with ${\displaystyle K^{n}(G)=\{e\}}$.

The group ${\displaystyle A_{5}}$is not resolvable. Is namely${\displaystyle (i\,j\,k)}$ a cycle of three, so be ${\displaystyle l,m}$ the two numbers not represented in it ${\displaystyle \{1,2,3,4,5\}}$. Then you do the math

${\displaystyle (i\,j\,k)=(i\,k\,j)\cdot (i\,k\,j)=(i\,k)\cdot (k\,j)\cdot (i\,k)\cdot (k\,j)}$
${\displaystyle =(i\,k)\cdot (l\,m)\cdot (k\,j)\cdot (l\,m)\cdot (l\,m)\cdot (i\,k)\cdot (l\,m)\cdot (k\,j)}$
${\displaystyle =[(i\,k)\cdot (l\,m),(k\,j)\cdot (l\,m)]}$,

that is, every three-cycle is a commutator and therefore off ${\displaystyle K^{1}(A_{5})}$. Since the cycle of three according to the above, the group${\displaystyle A_{5}}$ generate follows ${\displaystyle K^{1}(A_{5})=A_{5}}$ and thus ${\displaystyle K^{n}(A_{5})=A_{5}}$ for all ${\displaystyle n}$. thats why${\displaystyle A_{5}}$not resolvable. [6]

${\displaystyle A_{5}}$is the smallest non-resolvable group. It is well known that every p-group , that is to say group, is of order${\displaystyle p^{n}}$for a prime number ${\displaystyle p}$, resolvable. Furthermore, groups are of order${\displaystyle p^{n}q^{m}}$ with prime numbers ${\displaystyle p}$ and ${\displaystyle q}$resolvable according to Burnside's theorem . After all, groups are okay${\displaystyle pqr}$ with prime numbers ${\displaystyle p,q}$ and ${\displaystyle r}$dissolvable. [7] The smallest order that comes into question for a non-resolvable group is thus${\displaystyle 2^{2}\cdot 3\cdot 5=60}$. ${\displaystyle A_{5}}$ is therefore a non-resolvable group of the smallest possible order, one can even show that it is the only non-resolvable group of order 60 apart from isomorphism.

From the non-solubility of ${\displaystyle A_{5}}$ it easily turns out that all ${\displaystyle S_{n}}$ and all ${\displaystyle A_{n}}$ With ${\displaystyle n\geq 5}$ are not resolvable, because subgroups of resolvable groups are resolvable again and all these groups contain one to ${\displaystyle A_{5}}$ isomorphic subgroup.

## A 5 is easy

A group ${\displaystyle G}$is called simple when it is next to the trivial normal divisors ${\displaystyle G}$ and ${\displaystyle \{e\}}$contains no further normal dividers. Since commutator groups are normal divisors, solvable groups that are not cyclically prime always have normal divisors, but non-resolvable groups can also have normal divisors, as in the example${\displaystyle S_{5}}$ show the ${\displaystyle A_{5}}$has as normal divisor. Therefore, the following statement is an intensification of the non-dissolvability:

• ${\displaystyle A_{5}}$ is simple.

That follows easily from the fact that ${\displaystyle A_{5}}$is a non-resolvable group of the smallest possible order. That would be${\displaystyle N\subset A_{5}}$ a non-trivial normal divisor, so would have ${\displaystyle N}$ and ${\displaystyle A_{5}/N}$a really smaller order and would therefore be resolvable. From the well-known theorems about solvable groups it followed the solvability of${\displaystyle A_{5}}$which gives the desired contradiction.

The argument just given for the simplicity of the ${\displaystyle A_{5}}$is by no means trivial, because it uses Burnside's theorem, which is the minimality of 60 for the order of a non-resolvable group. However, one does not need Burnside's theorem in its full strength, the solvability of groups of order , which can be proven without representation theory${\displaystyle p^{a}q^{b}}$ With ${\displaystyle a,b\leq 2}$is sufficient. [8th]

In a simpler proof one shows first that all three-cycles are conjugated and then that each normal subgroup different from the one-element subgroup must contain at least one three-cycle. The normal divisor then contains all conjugates of this cycle of three, because normal divisors are by definition stable under conjugation, and therefore all cycles of three. But since this already${\displaystyle A_{5}}$ generate follows ${\displaystyle N=A_{5}}$, that is, there are no non-trivial normal divisors in ${\displaystyle A_{5}}$. [9] This proof applies to everyone${\displaystyle A_{n},\,n\geq 5}$.

Another simpler and on the ${\displaystyle A_{5}}$tailored proof using Sylow's theorems can be found in the textbook by B. Huppert given below. [10] It is also shown there:

• Is ${\displaystyle G}$ a simple group of order 60, so is ${\displaystyle G\cong A_{5}}$.

## Charactertafel

The character board of the${\displaystyle A_{5}}$looks like this: [11]

${\displaystyle A_{5}}$ ${\displaystyle 1}$ ${\displaystyle 15}$ ${\displaystyle 20}$ ${\displaystyle 12}$ ${\displaystyle 12}$
${\displaystyle 1}$ ${\displaystyle (1,2)\,(3,4)}$ ${\displaystyle (1,2,3)}$ ${\displaystyle (1,2,3,4,5)}$ ${\displaystyle (1,3,5,2,4)}$
${\displaystyle \chi _{1}}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$
${\displaystyle \chi _{2}}$ ${\displaystyle 4}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle -1}$ ${\displaystyle -1}$
${\displaystyle \chi _{3}}$ ${\displaystyle 5}$ ${\displaystyle 1}$ ${\displaystyle -1}$ ${\displaystyle 0}$ ${\displaystyle 0}$
${\displaystyle \chi _{4}}$ ${\displaystyle 3}$ ${\displaystyle -1}$ ${\displaystyle 0}$ ${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$ ${\displaystyle {\frac {1-{\sqrt {5}}}{2}}}$
${\displaystyle \chi _{5}}$ ${\displaystyle 3}$ ${\displaystyle -1}$ ${\displaystyle 0}$ ${\displaystyle {\frac {1-{\sqrt {5}}}{2}}}$ ${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$

## Occurrence

### Symmetry group

As mentioned above, the group occurs ${\displaystyle A_{5}}$as a rotation group of the icosahedron. In order to get an overview of the possible rotations that transform the icosahedron into itself, let us consider how they affect the edges. The 30 edges of the icosahedron are divided into 5 classes of parallel edges, each of these classes containing 6 parallel edges. Since rotations of the icosahedron must preserve parallelism of edges, they permute these 5 classes and one obtains a homomorphism from the icosahedron group into the${\displaystyle S_{5}}$. A closer look then shows that it is an injective homomorphism whose image is currently${\displaystyle A_{5}\subset S_{5}}$is. Therefore the icosahedral group is isomorphic to${\displaystyle A_{5}}$.[12]

The elements of the ${\displaystyle A_{5}}$ thus correspond to the following rotations:

The 30 edges define 15 axes of rotation through the centers of pairs of opposite edges, and about each axis is a rotation about ${\displaystyle 180^{\circ }}$possible. These are the 15 elements of order 2.

The 20 side surfaces define 10 axes of rotation through the centers of pairs of opposing side surfaces, and about each of these axes is a rotation about ${\displaystyle 120^{\circ }}$ or ${\displaystyle 240^{\circ }}$ possible, these are the 20 elements of the 3rd order.

The 12 corners determine 6 axes of rotation through pairs of opposite corners, for each axis there are 4 rotations around ${\displaystyle k\cdot 72^{\circ }}$, ${\displaystyle k=1,2,3,4}$ of order 5, that is a total of 24 rotations of order 5.

### Galoisgruppe

The polynomial

${\displaystyle p(X)=X^{5}+20X+5}$

has one for ${\displaystyle A_{5}}$isomorphic Galois group . [13] According to theorems of Galois theory , because of the non-solubility of the group established above, this means that the zeros of the polynomial cannot be represented by radicals of the coefficients. This is confirmed by Abel-Ruffini's theorem , according to which there are no general solution formulas for polynomials of degree 5 or higher, which consist of roots and arithmetic operations of the coefficients.

### PSL 2 (4) and PSL 2 (5)

The projective linear groups ${\displaystyle \mathrm {PSL} _{n}(q)}$ for a finite body ${\displaystyle K}$ With ${\displaystyle q}$ Elements are except for ${\displaystyle \mathrm {PSL} _{2}(2)\cong S_{3}}$ and ${\displaystyle \mathrm {PSL} _{2}(3)\cong A_{4}}$ easy and have ${\displaystyle (q^{n}-1)\cdot (q^{n}-q)\cdot \ldots \cdot (q^{n}-q^{n-2})\cdot q^{n-1}/\mathrm {ggT} (n,q-1)}$Elements. Therefore applies

${\displaystyle |\mathrm {PSL} _{2}(4)|=(4^{2}-1)\cdot 4^{1}/\mathrm {ggT} (2,3)=15\cdot 4/1=60}$
${\displaystyle |\mathrm {PSL} _{2}(5)|=(5^{2}-1)\cdot 5^{1}/\mathrm {ggT} (2,4)=24\cdot 5/2=60}$.

Since all simple groups of order 60 are isomorphic to the ${\displaystyle A_{5}}$ are, follows

${\displaystyle \mathrm {PSL} _{2}(4)\cong \mathrm {PSL} _{2}(5)\cong A_{5}}$.[14]

## Individual evidence

1. You need all letters of the alphabet and then letter pairs from AA to BH, just like with table title lines in Microsoft Excel .
2. MathWorld: Alternating Group Please note one difference: In the color graphic of the link table shown above, the neutral element is highlighted as a black square, which is not the case in the color graphic in MathWorld.
3. Permutations This website contains the code to generate permutations in a defined order, in 97 programming languages.
4. ^ B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, Chapter I, Example 19.9.
5. B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, Chapter II, Proposition 8.25.
6. ^ Kurt Meyberg: Algebra. Part 1. With 287 exercises. 2nd Edition. Hanser, Munich / Vienna 1980, ISBN 3-446-13079-9 , sentence 2.6.5.
7. B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, chapter I sentence 8.9, sentence 8.13 and chapter V sentence 7.3.
8. Derek J. S. Robinson: A Course in the Theory of Groups. Springer-Verlag, 1996, ISBN 0-387-94461-3, Abschnitt 5.4.1.
9. ^ Kurt Meyberg: Algebra. Part 1. With 287 exercises. 2nd Edition. Hanser, Munich / Vienna 1980, ISBN 3-446-13079-9 , sentence 2.4.16.
10. ^ B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, Chapter I, Sentence 8.14.
11. JL Alperin, RB Bell: Groups and Representations , Springer-Verlag (1995), ISBN 0-387-94525-3 , chap. 6, example 9.
12. K. Lamotke: Regular Solids and Isolated Singularities. Vieweg-Verlag, Braunschweig 1986, ISBN 3-528-08958-X, §5: The Rotation Groups of the Platonic Solids.
13. John Swallow: Exploratory Galois Theory. Cambridge University Press, Cambridge, UK / New York 2004, ISBN 0-521-83650-6, S. 176 (hinter Theorem 34.7).
14. ^ B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, Chapter II, Sentence 6.14.