Alternating group
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In mathematics, an alternating group is the group of even permutations of a finite set. The alternating group on a set of n elements is called the alternating group of degree n, or the alternating group on n letters and denoted by An or Alt(n).
Basic properties
[edit]For n > 1, the group An is the commutator subgroup of the symmetric group Sn with index 2 and has therefore n!/2 elements. It is the kernel of the signature group homomorphism sgn : Sn → {1, −1} explained under symmetric group.
The group An is abelian if and only if n ≤ 3 and simple if and only if n = 3 or n ≥ 5. A5 is the smallest non-abelian simple group, having order 60, and thus the smallest non-solvable group.
The group A4 has the Klein four-group V as a proper normal subgroup, namely the identity and the double transpositions { (), (12)(34), (13)(24), (14)(23) }, that is the kernel of the surjection of A4 onto A3 ≅ Z3. We have the exact sequence V → A4 → A3 = Z3. In Galois theory, this map, or rather the corresponding map S4 → S3, corresponds to associating the Lagrange resolvent cubic to a quartic, which allows the quartic polynomial to be solved by radicals, as established by Lodovico Ferrari.
Conjugacy classes
[edit]As in the symmetric group, any two elements of An that are conjugate by an element of An must have the same cycle shape. The converse is not necessarily true, however. If the cycle shape consists only of cycles of odd length with no two cycles the same length, where cycles of length one are included in the cycle type, then there are exactly two conjugacy classes for this cycle shape (Scott 1987, §11.1, p299).
Examples:
- The two permutations (123) and (132) are not conjugates in A3, although they have the same cycle shape, and are therefore conjugate in S3.
- The permutation (123)(45678) is not conjugate to its inverse (132)(48765) in A8, although the two permutations have the same cycle shape, so they are conjugate in S8.
Relation with symmetric group
[edit]- See Symmetric group.
As finite symmetric groups are the groups of all permutations of a set with finite elements, and the alternating groups are groups of even permutations, alternating groups are subgroups of finite symmetric groups.
Generators and relations
[edit]For n ≥ 3, An is generated by 3-cycles, since 3-cycles can be obtained by combining pairs of transpositions. This generating set is often used to prove that An is simple for n ≥ 5.
Automorphism group
[edit]| n | Aut(An) | Out(An) |
|---|---|---|
| n ≥ 4, n ≠ 6 | Sn | Z2 |
| n = 1, 2 | Z1 | Z1 |
| n = 3 | Z2 | Z2 |
| n = 6 | S6 ⋊ Z2 | V = Z2 × Z2 |
For n > 3, except for n = 6, the automorphism group of An is the symmetric group Sn, with inner automorphism group An and outer automorphism group Z2; the outer automorphism comes from conjugation by an odd permutation.
For n = 1 and 2, the automorphism group is trivial. For n = 3 the automorphism group is Z2, with trivial inner automorphism group and outer automorphism group Z2.
The outer automorphism group of A6 is the Klein four-group V = Z2 × Z2, and is related to the outer automorphism of S6. The extra outer automorphism in A6 swaps the 3-cycles (like (123)) with elements of shape 32 (like (123)(456)).
Exceptional isomorphisms
[edit]There are some exceptional isomorphisms between some of the small alternating groups and small groups of Lie type, particularly projective special linear groups. These are:
- A4 is isomorphic to PSL2(3)[1] and the symmetry group of chiral tetrahedral symmetry.
- A5 is isomorphic to PSL2(4), PSL2(5), and the symmetry group of chiral icosahedral symmetry. (See[1] for an indirect isomorphism of PSL2(F5) → A5 using a classification of simple groups of order 60, and here for a direct proof).
- A6 is isomorphic to PSL2(9) and PSp4(2)'.
- A8 is isomorphic to PSL4(2).
More obviously, A3 is isomorphic to the cyclic group Z3, and A0, A1, and A2 are isomorphic to the trivial group (which is also SL1(q) = PSL1(q) for any q).
Examples S4 and A4
[edit] The odd permutations are colored: Transpositions in green and 4-cycles in orange |
 Elements: The even permutations (the identity, eight 3-cycles and three double-transpositions (double transpositions in boldface)) Subgroups: .svg) .svg) .svg) .svg) .svg) |
 A3 = Z3 (order 3) |
 A4 (order 12) |
 A4 × Z2 (order 24) |
 S3 = Dih3 (order 6) |
 S4 (order 24) |
 A4 in S4 on the left |
Example A5 as a subgroup of 3-space rotations
[edit].gif)
icosahedron – radius 2π/5 – second half of split 5-cycles
A5 is the group of isometries of a dodecahedron in 3-space, so there is a representation A5 → SO3(R).
In this picture the vertices of the polyhedra represent the elements of the group, with the center of the sphere representing the identity element. Each vertex represents a rotation about the axis pointing from the center to that vertex, by an angle equal to the distance from the origin, in radians. Vertices in the same polyhedron are in the same conjugacy class. Since the conjugacy class equation for A5 is 1 + 12 + 12 + 15 + 20 = 60, we obtain four distinct (nontrivial) polyhedra.
The vertices of each polyhedron are in bijective correspondence with the elements of its conjugacy class, with the exception of the conjugacy class of (2,2)-cycles, which is represented by an icosidodecahedron on the outer surface, with its antipodal vertices identified with each other. The reason for this redundancy is that the corresponding rotations are by π radians, and so can be represented by a vector of length π in either of two directions. Thus the class of (2,2)-cycles contains 15 elements, while the icosidodecahedron has 30 vertices.
The two conjugacy classes of twelve 5-cycles in A5 are represented by two icosahedra, of radii 2π/5 and 4π/5, respectively. The nontrivial outer automorphism in Out(A5) ≃ Z2 interchanges these two classes and the corresponding icosahedra.
Example: the 15 puzzle
[edit]
It can be proved that the 15 puzzle, a famous example of the sliding puzzle, can be represented by the alternating group A15,[2] because the combinations of the 15 puzzle can be generated by 3-cycles. In fact, any 2k − 1 sliding puzzle with square tiles of equal size can be represented by A2k−1.
Subgroups
[edit]A4 is the smallest group demonstrating that the converse of Lagrange's theorem is not true in general: given a finite group G and a divisor d of |G|, there does not necessarily exist a subgroup of G with order d: the group G = A4, of order 12, has no subgroup of order 6. A subgroup of three elements (generated by a cyclic rotation of three objects) with any distinct nontrivial element generates the whole group.
For all n > 4, An has no nontrivial (that is, proper) normal subgroups. Thus, An is a simple group for all n > 4. A5 is the smallest non-solvable group.
Group homology
[edit]The group homology of the alternating groups exhibits stabilization, as in stable homotopy theory: for sufficiently large n, it is constant. However, there are some low-dimensional exceptional homology. Note that the homology of the symmetric group exhibits similar stabilization, but without the low-dimensional exceptions (additional homology elements).
H1: Abelianization
[edit]The first homology group coincides with abelianization, and (since An is perfect, except for the cited exceptions) is thus:
- H1(An, Z) = Z1 for n = 0, 1, 2;
- H1(A3, Z) = Aab
3 = A3 = Z3; - H1(A4, Z) = Aab
4 = Z3; - H1(An, Z) = Z1 for n ≥ 5.
This is easily seen directly, as follows. An is generated by 3-cycles – so the only non-trivial abelianization maps are An → Z3, since order-3 elements must map to order-3 elements – and for n ≥ 5 all 3-cycles are conjugate, so they must map to the same element in the abelianization, since conjugation is trivial in abelian groups. Thus a 3-cycle like (123) must map to the same element as its inverse (321), but thus must map to the identity, as it must then have order dividing 2 and 3, so the abelianization is trivial.
For n < 3, An is trivial, and thus has trivial abelianization. For A3 and A4 one can compute the abelianization directly, noting that the 3-cycles form two conjugacy classes (rather than all being conjugate) and there are non-trivial maps A3 ↠ Z3 (in fact an isomorphism) and A4 ↠ Z3.
H2: Schur multipliers
[edit]The Schur multipliers of the alternating groups An (in the case where n is at least 5) are the cyclic groups of order 2, except in the case where n is either 6 or 7, in which case there is also a triple cover. In these cases, then, the Schur multiplier is (the cyclic group) of order 6.[3] These were first computed in (Schur 1911).
- H2(An, Z) = Z1 for n = 1, 2, 3;
- H2(An, Z) = Z2 for n = 4, 5;
- H2(An, Z) = Z6 for n = 6, 7;
- H2(An, Z) = Z2 for n ≥ 8.
Notes
[edit]- ^ a b Robinson (1996), p. 78
- ^ Beeler, Robert. "The Fifteen Puzzle: A Motivating Example for the Alternating Group" (PDF). faculty.etsu.edu/. East Tennessee State University. Archived from the original (PDF) on 2021-01-07. Retrieved 2020-12-26.
- ^ Wilson, Robert (October 31, 2006), "Chapter 2.7: Covering groups", The finite simple groups, 2006 versions, archived from the original on May 22, 2011
References
[edit]- Robinson, Derek John Scott (1996), A course in the theory of groups, Graduate texts in mathematics, vol. 80 (2 ed.), Springer, ISBN 978-0-387-94461-6
- Schur, Issai (1911), "Über die Darstellung der symmetrischen und der alternierenden Gruppe durch gebrochene lineare Substitutionen", Journal für die reine und angewandte Mathematik, 1911 (139): 155–250, doi:10.1515/crll.1911.139.155, S2CID 122809608
- Scott, W.R. (1987), Group Theory, New York: Dover Publications, ISBN 978-0-486-65377-8
External links
[edit]Alternating group
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Definition and Fundamentals
Definition
In group theory, the alternating group $ A_n $ is the subgroup of the symmetric group $ S_n $, which consists of all permutations of $ n $ elements, comprising precisely those permutations that are even.[3] An even permutation is defined as one that can be expressed as a product of an even number of transpositions (2-cycles).[4] Formally, $ A_n $ is the kernel of the sign homomorphism $ \sgn: S_n \to { \pm 1 } $, where the sign of a permutation $ \sigma $ is $ \sgn(\sigma) = 1 $ if $ \sigma $ is even and $ \sgn(\sigma) = -1 $ if $ \sigma $ is odd; thus,
[5] This kernel property establishes $ A_n $ as a normal subgroup of $ S_n $ of index 2.[6]
The standard notation $ A_n $ (or sometimes $ \Alt(n) $) is used for $ n \geq 3 $, as these are the cases of primary interest in permutation group theory; for completeness, $ A_1 $ and $ A_2 $ are both the trivial group consisting solely of the identity permutation.[7]
The concept and terminology of the alternating group originated in the work of Camille Jordan, who introduced the term in his 1873 paper on transitive groups.[8]
Basic Properties
The alternating group $ A_n $ on $ n $ letters, for $ n \geq 2 $, consists of all even permutations in the symmetric group $ S_n $ and has order $ |A_n| = n!/2 $. This cardinality follows from the surjectivity of the sign homomorphism $ \operatorname{sgn}: S_n \to { \pm 1 } $, whose kernel is precisely $ A_n $.[9][10] A permutation is even if and only if it can be expressed as a product of an even number of transpositions, and this parity is independent of the particular decomposition into transpositions. The sign function provides the rigorous parity argument: for any permutation $ \sigma $, $ \operatorname{sgn}(\sigma) = (-1)^k $ where $ k $ is the number of transpositions in any decomposition of $ \sigma $, ensuring the even permutations form a well-defined subgroup of index 2 in $ S_n $.[10] For $ n \geq 3 $, the alternating group $ A_n $ is generated by the set of all 3-cycles. To see this, note that any even permutation is a product of an even number of transpositions, and a product of two disjoint transpositions such as $ (ab)(cd) $ equals $ (acb)(acd) $, a product of two 3-cycles; more generally, any even permutation reduces to the identity via conjugation and multiplication by 3-cycles. In particular, $ A_n $ is generated by the 3-cycle $ (1,2,3) $ and its conjugates under $ S_n $./10:_Normal_Subgroups_and_Factor_Groups/10.02:_The_Simplicity_of_the_Alternating_Groups)[11] The center of $ A_n $ is trivial, $ Z(A_n) = { e } $, for all $ n \geq 4 $. This holds because any non-identity even permutation fails to commute with some 3-cycle in $ A_n $, as conjugation by elements of $ S_n $ splits cycles in a way that disrupts centrality within the even permutations.[12] The derived subgroup (commutator subgroup) of $ A_n $ equals $ A_n $ itself for $ n \geq 5 $, making $ A_n $ a perfect group in these cases. For $ n = 4 $, the derived subgroup is the Klein four-group consisting of the identity and the three double transpositions.[12]Relation to Symmetric Group
Subgroup Structure
The alternating group is a normal subgroup of the symmetric group for , as it forms the kernel of the sign homomorphism , which maps even permutations to and odd permutations to .[3] This kernel property directly implies normality, since the kernel of any group homomorphism is a normal subgroup of the domain.[13] The index of in is 2, denoted , reflecting that partitions into exactly two cosets of .[14] Consequently, the order of is , half the order of , by Lagrange's theorem.[15] The coset decomposition is , where is any fixed odd permutation (such as the transposition ), and the second coset consists precisely of all odd permutations in .[14] By the first isomorphism theorem, the quotient group is isomorphic to , the cyclic group of order 2, as the sign homomorphism is surjective onto .[15] This quotient captures the parity distinction between even and odd permutations. Furthermore, acts on the set of its two cosets by left multiplication, yielding a transitive action that induces the sign homomorphism and underscores the structural role of as the unique subgroup of index 2 in .[14]Conjugacy Classes
The conjugacy classes of the alternating group $ A_n $ consist of the even permutations in $ S_n $, which are precisely those whose cycle decompositions contain an even number of even-length cycles.[16] In $ S_n $, conjugacy classes are partitioned solely by cycle type, but in the subgroup $ A_n $, even permutations of certain cycle types may form a single class, while others split into two equal-sized classes under conjugation by even permutations.[17] A conjugacy class of an even permutation in $ S_n $ splits into two classes in $ A_n $ if and only if its cycle type consists of disjoint cycles of distinct odd lengths (including at most one fixed point, as multiple fixed points would repeat the length-1 cycle).[18] In such cases, no odd permutation commutes with a representative of the class, so the centralizer $ C_{S_n}(\sigma) $ lies entirely within $ A_n $, causing the $ S_n $-class to divide evenly under the index-2 subgroup. For cycle types with at least one even-length cycle or repeated odd lengths, the centralizer in $ S_n $ contains odd permutations, and the $ A_n $-class remains unsplit with size equal to that of the corresponding $ S_n $-class.[18] For example, double transpositions of cycle type $ (2,2) $, such as $ (1,2)(3,4) $, involve even-length cycles and thus do not split; in $ A_4 $, the three such elements form a single conjugacy class of size 3.[17] In contrast, a 5-cycle in $ A_5 $ has cycle type $ (5) $ (a single odd length with no fixed points), so its $ S_5 $-class of 24 elements splits into two $ A_5 $-classes of size 12 each.[19] The size of any conjugacy class $ \mathrm{cl}(\sigma) $ in $ A_n $ is given by the class equation $ |\mathrm{cl}(\sigma)| = |A_n| / |C_{A_n}(\sigma)| $, where $ C_{A_n}(\sigma) = C_{S_n}(\sigma) \cap A_n $ is the centralizer of $ \sigma $ in $ A_n $.[17] For 3-cycles, which have cycle type $ (3,1^{n-3}) $, the presence of repeated length-1 cycles (for $ n \geq 5 $) prevents splitting, and all such elements are conjugate in $ A_n $. The total number of 3-cycles in $ A_n $ (for $ n \geq 3 $) is $ \binom{n}{3} \times 2 $, obtained by choosing 3 elements and forming either orientation of the cycle.[19] For a representative 3-cycle $ \sigma $ with $ n \geq 5 $, the centralizer in $ S_n $ has order $ 3(n-3)! $, half of which is even, so $ |C_{A_n}(\sigma)| = \frac{3(n-3)!}{2} $ and the class size is $ \frac{n!/2}{3(n-3)!/2} = \frac{n(n-1)(n-2)}{3} = \binom{n}{3} \times 2 $.[17]Presentation and Automorphisms
Generators and Relations
The alternating group $ A_n $ for $ n \geq 3 $ is generated by its 3-cycles; every even permutation can be expressed as a product of these elements, since any product of two disjoint transpositions is a product of two 3-cycles, and any even permutation is a product of an even number of transpositions.[12] Although the full set of 3-cycles is generally required, $ A_n $ admits a minimal generating set of size two for all $ n > 2 $ except $ n = 6, 7 $, where three generators are needed.[20] For $ n = 4 $, two 3-cycles suffice: $ A_4 = \langle (1, 2, 3), (1, 2, 4) \rangle $. This group has the presentation
where $ x = (1, 2, 3) $ and $ y = (1, 2, 4) $; the relations ensure the group order is 12, matching $ |A_4| $.[21]
For $ n \geq 5 $, $ A_n $ is generated by the 3-cycles, and a presentation can be derived from the Coxeter presentation of the symmetric group $ S_n $ using the Reidemeister-Schreier rewriting process to account for the index-2 subgroup of even permutations.[20] Explicit 2-generator presentations exist with $ O(\log n) $ relations of total length $ O(\log^2 n) $; for example, when $ n $ is odd, suitable generators are the $ n $-cycle $ y = (1, 2, \dots , n) $ and $ xyx $ where $ x = (1, 2) $.[20]
Automorphism Group
The inner automorphism group of the alternating group $ A_n $ for $ n \geq 4 $ is isomorphic to $ A_n $ itself, as the center $ Z(A_n) $ is trivial in these cases, yielding $ \operatorname{Inn}(A_n) \cong A_n / Z(A_n) \cong A_n $.[22] For $ n \geq 4 $ and $ n \neq 6 $, the full automorphism group $ \operatorname{Aut}(A_n) $ is isomorphic to the symmetric group $ S_n $. This isomorphism arises from the action of $ S_n $ on $ A_n $ by conjugation, which induces all automorphisms of $ A_n $; specifically, the natural homomorphism $ S_n \to \operatorname{Aut}(A_n) $ has trivial kernel, hence is an isomorphism. Conjugation by even permutations yields the inner automorphisms $ \operatorname{Inn}(A_n) \cong A_n $ of index 2, while conjugation by odd permutations provides the nontrivial outer automorphisms, with $ \operatorname{Out}(A_n) \cong C_2 $. Thus, $ |\operatorname{Aut}(A_n)| = n! $. For example, conjugation by an odd permutation in $ S_n $ provides the nontrivial outer automorphism of $ A_n $.[22] The case $ n = 6 $ is exceptional, where $ \operatorname{Aut}(A_6) $ has order $ 2 \cdot |S_6| = 1440 $, so $ \operatorname{Out}(A_6) \cong C_2 \times C_2 $. This enlargement stems from an additional outer automorphism of $ A_6 $, induced by the nontrivial outer automorphism of $ S_6 $, which itself arises from the enlarged automorphism group of the line graph of the complete graph $ K_6 $ (isomorphic to $ S_6 \times C_2 $, unlike for other $ n $).[23]Isomorphisms and Examples
Exceptional Isomorphisms
The exceptional isomorphisms of alternating groups with projective special linear groups for small represent non-standard identifications that arise in the classification of finite simple groups and have deep ties to geometry and number theory. These isomorphisms were first established in the 1870s by Camille Jordan and Felix Klein, who linked the permutation actions of to linear fractional transformations over finite fields, revealing structural parallels through shared orders and transitive actions on projective spaces.[24] For , the alternating group of order 12 is isomorphic to , the group of projective linear transformations over the field with 3 elements. This identification follows from the transitive action of on the 4 points of the projective line over , mirroring the even permutations on 4 letters, with the isomorphism constructed via coset decompositions of stabilizers.[24] The most prominent case is , where of order is isomorphic to , with denoting the special linear group over and its center. Jordan and Klein derived this by showing that the icosahedral rotation group, isomorphic to , acts equivalently to on the 6 points at infinity in the projective plane, via the Klein correspondence that equates quadratic forms to lines in projective geometry. The explicit isomorphism can be realized through the natural action of on the cosets of a Borel subgroup, yielding a primitive permutation representation of degree 6 identical to that of .[24] For , of order is isomorphic to , established through an outer automorphism of the double cover , where the isomorphism descends to the quotients by mapping even permutations to projective transformations over . This connection, also uncovered by Klein in his studies of modular equations, relies on the transitive action of on 10 points (the projective line over ), aligning with 's permutation degree via geometric realizations in the complex plane.[24]A₄ and S₄
The alternating group is the subgroup of the symmetric group consisting of all even permutations of four elements, and thus has order .[25] As a concrete realization within , contains all 3-cycles and all products of two disjoint transpositions.[25] Geometrically, is isomorphic to the group of rotational symmetries of a regular tetrahedron, which also has order 12, corresponding to the 12 possible orientations obtained by rotating the tetrahedron around its vertices, edges, or faces.[26] A defining feature of 's structure as a subgroup of is its unique normal Sylow 2-subgroup, known as the Klein four-group , which consists of the identity and the three double transpositions in .[25] This subgroup is normal in because conjugation by any element of permutes the double transpositions among themselves while preserving the even permutation structure.[25] The Sylow structure of further highlights its organization: it has a single Sylow 2-subgroup of order 4 (namely ), and four Sylow 3-subgroups of order 3, each generated by a 3-cycle such as , , , and .[25] The quotient is isomorphic to , the cyclic group of order 3 (equivalently, ), reflecting the action of on the cosets of .[25] This normal subgroup also serves as the derived subgroup , the subgroup generated by all commutators in , which implies that is not simple and has abelianization .[25] One explicit presentation of arises from its generation by the 3-cycles and , with relations that enforce the normality of :
where can be taken as a double transposition like and as a 3-cycle like , ensuring the structure collapses appropriately to yield the Klein four-group as the commutator subgroup.[27] This presentation underscores 's position as a semidirect product within .[25]
A₅ and 3D Rotations
The alternating group $ A_5 $ is isomorphic to the group of proper rotations (orientation-preserving symmetries) of the regular icosahedron and its dual, the regular dodecahedron, which embeds naturally into the special orthogonal group $ SO(3) $. This rotational icosahedral group has order 60, consistent with $ |A_5| = 5!/2 = 60 $.[28][29] The group is generated by rotations of specific angles about axes passing through pairs of opposite vertices, faces, or edges of the icosahedron. Specifically, there are rotations by $ 72^\circ $, $ 144^\circ $, $ 216^\circ $, and $ 288^\circ $ (order 5) about axes through opposite vertices (12 axes, contributing 24 non-identity rotations); rotations by $ 120^\circ $ and $ 240^\circ $ (order 3) about axes through centers of opposite faces (10 axes, contributing 20 non-identity rotations); and rotations by $ 180^\circ $ (order 2) about axes through midpoints of opposite edges (15 axes, contributing 15 rotations), plus the identity. These elements satisfy the relations of $ A_5 $, providing a geometric realization of its structure.[28][30] The binary icosahedral group, a central extension of $ A_5 $ by $ \mathbb{Z}/2\mathbb{Z} $, serves as its universal double cover and is isomorphic to the special linear group $ SL(2,5) $ of order 120. This double cover arises in the context of spin representations and the universal covering group of $ SO(3) $, where $ SL(2,5) \to A_5 $ is the quotient by the center $ { \pm I } $.[31][32] As the unique non-abelian simple group of order less than 100, $ A_5 $ plays a foundational role in the classification of finite simple groups, with its icosahedral realization highlighting connections between discrete geometry and abstract algebra. The action of $ A_5 $ on the 12 vertices of the icosahedron yields a transitive permutation representation of degree 12, faithful and primitive, embedding $ A_5 $ into $ S_{12} $.[33][34]The 15 Puzzle
The 15 puzzle is a classic sliding puzzle played on a 4×4 grid, featuring 15 square tiles numbered from 1 to 15 and one empty space, known as the blank. The goal is to rearrange the tiles into ascending order, row by row from left to right and top to bottom, by repeatedly sliding an adjacent tile into the blank space. Each move effectively transposes the blank with a neighboring tile, altering the positions of the pieces within the grid.[35] The solvability of a 15 puzzle configuration depends on the parity of the permutation. Treating the blank as tile 16, any configuration corresponds to a permutation σ in the symmetric group S_{16}. The puzzle is solvable if and only if σ is an even permutation, i.e., the number of inversions in σ is even. An inversion is a pair of positions (i, j) with i < j but σ(i) > σ(j) when reading the grid row by row. The sign of the permutation is given by
and solvability requires sgn(σ) = 1. This condition ensures the configuration lies within the reachable component of the puzzle's configuration space, which has size 16!/2 and corresponds to the even permutations in S_{16}. The arrangements of the 15 tiles with the blank in its target position form a set isomorphic to A_{15}.[35][36]
This parity invariant highlights the role of the alternating group in restricting reachable states: odd permutations, such as swapping tiles 14 and 15 while leaving others fixed (with blank in place), cannot be achieved. For the generalized n × n sliding puzzle with n² - 1 tiles, solvability requires an even permutation of all n² positions (including blank). When n is odd, this is equivalent to an even permutation of the tiles alone (i.e., the number of inversions among the tiles is even), as the blank's position parity aligns automatically. When n is even, as in the 15 puzzle, the condition accounts for the blank's position, but exactly half of all possible configurations are solvable.[37][35]
Subgroups and Homology
Normal Subgroups and Simplicity
The alternating group $ A_3 $ is isomorphic to the cyclic group $ \mathbb{Z}/3\mathbb{Z} $, which is simple as it has prime order and thus only trivial normal subgroups.
In contrast, $ A_4 $ is not simple, as it contains the Klein four-group $ V = { e, (12)(34), (13)(24), (14)(23) } $ as a nontrivial proper normal subgroup of order 4.
For $ n \geq 5 $, the alternating group $ A_n $ is simple, meaning its only normal subgroups are the trivial subgroup $ {e} $ and $ A_n $ itself.
This theorem establishes $ A_n $ (with $ n \geq 5 $) as non-abelian simple groups, forming one of the infinite families in the classification of finite simple groups alongside cyclic groups of prime order, groups of Lie type, and sporadic groups.
To sketch the proof, first note that all 3-cycles in $ A_n $ are conjugate for $ n \geq 3 $, and $ A_n $ is generated by its 3-cycles for $ n \geq 3 $.
\] Suppose $ N \trianglelefteq A_n $ is a nontrivial [normal subgroup](/page/Normal_subgroup). Then $ N $ must contain some even [permutation](/page/Permutation); analysis of possible cycle types shows that $ N $ contains a 3-cycle, as other even permutations (like double transpositions for $ n \geq 5 $) lead to contradictions via centralizer or index arguments.\[https://www.math.purdue.edu/~jlipman/503/A_n.pdf
By conjugacy, $ N $ then contains all 3-cycles, and thus $ N = A_n $ since the 3-cycles generate $ A_n $.
\] More precisely, if $ N \trianglelefteq A_n $ is normal, then either $ N $ contains all 3-cycles (hence $ N = A_n $) or $ N = \{ e \} $.\[https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf
The base case $ n=5 $ follows by direct computation: any proper normal $ N $ would have order dividing 20 but cannot contain 3-cycles, 5-cycles, or double transpositions without being the full group.
For $ n > 5 $, induction applies using the simplicity of $ A_{n-1} $ and examining $ N \cap A_{n-1} $, showing nontrivial elements in $ N $ force it to be the full $ A_n $.
$$]