# A 5 (group)

The group considered in the mathematical sub-area of group theory is the alternating group of the 5th degree. 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 . ${\ displaystyle A_ {5}}$

## Definitions

The cycle (234) as an illustration

We consider the set of all bijective mappings of the 5-element set 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 or without a link symbol. This is the symmetric group with elements. ${\ displaystyle \ {1,2,3,4,5 \}}$${\ displaystyle \ cdot}$ ${\ displaystyle S_ {5}}$${\ displaystyle 5! = 120}$

Such mappings are called permutations and use the so-called cycle notation with different elements . The figure 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 maps 2 to 3, 3 to 4 and 4 to 2 and leaves elements 1 and 5 fixed. A cycle of length 2 therefore only interchanges and and leaves all other elements fixed, such mappings are called transpositions . Different cycles can describe the same permutation, for example , uniqueness is obtained by agreeing to put the smallest number occurring in the cycle at the beginning. ${\ displaystyle (i_ {1} \ ldots i_ {k})}$${\ displaystyle i_ {1}, \ ldots, i_ {k} \ in \ {1,2,3,4,5 \}}$${\ displaystyle (i_ {1} \ ldots i_ {k})}$${\ displaystyle (2 \, 3 \, 4)}$${\ displaystyle (i \, j)}$${\ displaystyle i}$${\ displaystyle j}$${\ displaystyle (2 \, 3 \, 4) = (3 \, 4 \, 2)}$

Every 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 is applied first, then . (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. ${\ displaystyle (2 \, 3)}$${\ displaystyle (1 \, 2)}$

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 . ${\ displaystyle A_ {5}}$

Of course, analogous formulations of terms for are possible instead of , which then leads to the alternating group A n . In this article we cover the case . ${\ displaystyle \ {1, \ ldots, n \}}$${\ displaystyle \ {1,2,3,4,5 \}}$${\ displaystyle n = 5}$

## Elementary properties

### Number of elements

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

### Cycle of three

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

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

A cycle of three is evidently a mapping that maps each of the elements from onto another element of this set of three and leaves the other two elements from fixed; there are exactly two such images, namely and . Since there are such sets of three in total, we arrive at a total of 20 cycles of three. The other way around ${\ displaystyle (i \, j \, k)}$${\ displaystyle \ {i, j, k \}}$${\ displaystyle \ {1,2,3,4,5 \}}$${\ displaystyle (i \, j \, k)}$${\ displaystyle (i \, k \, j)}$${\ displaystyle \ textstyle {\ binom {5} {3}} = 10}$

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

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

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

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

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

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 15th ${\ displaystyle (i \, j) \ cdot (k \, l)}$ two transpositions foreign to the element
3 20th ${\ displaystyle (i \, j \, k)}$ 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

In the case of 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 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 icosahedron rotating group and alternatively denotes it with the letter . It is a subgroup of the full icosahedral group . ${\ displaystyle A_ {5}}$${\ displaystyle I}$ ${\ displaystyle I_ {h}}$

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 or symbols in the table. However, it is possible to represent the elements using colored squares and, accordingly, also the link table, as is done in the online encyclopedia for mathematics MathWorld, for example.

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 . A permutation generator can be used to generate all 120 permutations of 5 objects in an orderly order. In this way the elements of the symmetrical group are obtained . To get to the alternating group , you just have to delete all odd permutations. Now the 59 x 59 group multiplications have to be carried out with these elements. ${\ displaystyle S_ {5}}$${\ displaystyle A_ {5}}$

Since the sequence 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. Consequently, the first three elements form an alternating subgroup of type and elements 1 to 12 form an alternating subgroup of type . These two subgroups of the alternating group can be clearly seen as diagonal blocks from the graphic on the link table. Furthermore, one immediately notices a grouping in blocks of 12 elements each. ${\ displaystyle A_ {3}}$${\ displaystyle A_ {4}}$${\ displaystyle A_ {5}}$

## presentation

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

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

Are defined. This means that every group that is created by two elements that also satisfy the above relations is isomorphic to . ${\ displaystyle x, y}$${\ displaystyle A_ {5}}$

The self is generated by and , and these elements satisfy the given relations. ${\ displaystyle A_ {5}}$${\ displaystyle x = (1 \, 2 \, 3 \, 4 \, 5)}$${\ displaystyle y = (1 \, 2) \ cdot (3 \, 4)}$

## Transitive operation on 6 elements

The group 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 overall that it operates transitively on a set with alternate elements. This operation is even faithful . The following reversal applies: ${\ displaystyle A_ {5}}$ ${\ displaystyle A_ {5}}$

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

## A 5 cannot be resolved

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

The group cannot be dissolved. If there is a cycle of three, then the two numbers not represented in it are out . Then you do the math ${\ displaystyle A_ {5}}$${\ displaystyle (i \, j \, k)}$${\ displaystyle l, m}$${\ displaystyle \ {1,2,3,4,5 \}}$

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

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

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

From the non-solubility of it follows easily that all and all with cannot be resolved, because subgroups of resolvable groups can be resolved again and all these groups contain a subgroup that is too isomorphic. ${\ displaystyle A_ {5}}$${\ displaystyle S_ {n}}$${\ displaystyle A_ {n}}$${\ displaystyle n \ geq 5}$${\ displaystyle A_ {5}}$

## A 5 is easy

A group is simple if it besides the trivial normal subgroups and contains no other normal subgroup. 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 the example shows, which has normal divisors. Therefore the following statement is a tightening of the non-dissolvability: ${\ displaystyle G}$ ${\ displaystyle G}$${\ displaystyle \ {e \}}$${\ displaystyle S_ {5}}$${\ displaystyle A_ {5}}$

• ${\ displaystyle A_ {5}}$ is simple.

This follows easily from the fact that a non-resolvable group is of the smallest possible order. If there were a non-trivial normal divisor, then and would have a really smaller order and could therefore be resolved. From the well-known theorems about solvable groups it followed the solvability of what results in the desired contradiction. ${\ displaystyle A_ {5}}$${\ displaystyle N \ subset A_ {5}}$${\ displaystyle N}$${\ displaystyle A_ {5} / N}$${\ displaystyle A_ {5}}$

The argument just given for the simplicity of the 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 full strength, the solvability of groups of the order with , which can be proven without representation theory, is sufficient. ${\ displaystyle A_ {5}}$${\ displaystyle p ^ {a} q ^ {b}}$${\ displaystyle a, b \ leq 2}$

In a simpler proof one shows first that all three-cycles are conjugate and then that every 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 these already generate, it follows , that is, there are no non-trivial normal divisors in . This proof applies to everyone . ${\ displaystyle A_ {5}}$${\ displaystyle N = A_ {5}}$${\ displaystyle A_ {5}}$${\ displaystyle A_ {n}, \, n \ geq 5}$

Another simpler and more tailored proof using the Sylow theorems can be found in the textbook by B. Huppert given below. In addition, there is shown: ${\ displaystyle A_ {5}}$

• If a simple group of order 60 is then .${\ displaystyle G}$${\ displaystyle G \ cong A_ {5}}$

## Character board

The character table of looks like this: ${\ displaystyle A_ {5}}$

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

## Occurrence

### Symmetry group

As mentioned above, the group occurs as the 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 gets a homomorphism from the icosahedron group into the . A closer look then shows that it is an injective homomorphism whose image is straight . Therefore the icosahedral group is isomorphic to . ${\ displaystyle A_ {5}}$${\ displaystyle S_ {5}}$${\ displaystyle A_ {5} \ subset S_ {5}}$${\ displaystyle A_ {5}}$

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

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

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

The corners 12 define rotation axes 6 by pairs of opposite corners, to each axis, there are 4 rotations , the order of 5, the total of the 24 rotations of the order. 5 ${\ displaystyle k \ cdot 72 ^ {\ circ}}$${\ displaystyle k = 1,2,3,4}$

### Galois group

The polynomial

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

has one to the isomorphic Galois group . 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. ${\ displaystyle A_ {5}}$

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

The projective linear groups for a finite body with elements are simple and have elements with the exception of and . Therefore applies ${\ displaystyle \ mathrm {PSL} _ {n} (q)}$${\ displaystyle K}$${\ displaystyle q}$${\ displaystyle \ mathrm {PSL} _ {2} (2) \ cong S_ {3}}$${\ displaystyle \ mathrm {PSL} _ {2} (3) \ cong A_ {4}}$${\ displaystyle (q ^ {n} -1) \ cdot (q ^ {n} -q) \ cdot \ ldots \ cdot (q ^ {n} -q ^ {n-2}) \ cdot q ^ {n -1} / \ mathrm {gcd} (n, q-1)}$

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

Since all simple groups of order 60 are isomorphic to , as mentioned above , it follows ${\ displaystyle A_ {5}}$

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

## 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 a 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 JS Robinson: A Course in the Theory of Groups. Springer-Verlag, 1996, ISBN 0-387-94461-3 , section 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 , p. 176 (behind Theorem 34.7).
14. ^ B. Huppert: Endliche Gruppen I. Springer-Verlag, 1967, chapter II, sentence 6.14.