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The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus

Act on the four labelled vertices. A proper rotation gives an even permutation, identifying the rotational tetrahedral symmetry group with $A_4$, an alternating group. Geometrically there are one identity, three half-turns about opposite-edge midpoint axes, and two nontrivial rotations about each of four vertex-face axes. ThereforeThe tetrahedral rotation conjugacy classes areTo justify the splitting, the centralizer of a 3-cycle within $A_4$ is its three-element cyclic subgroup, so its conjugacy class has size $12/3=4$. Its inverse is not conjugate to it in $A_4$: the permutations that reverse that cycle while preserving its fixed vertex are odd. Conjugating $(123)$ by the even permutations produces the displayed $C_{+}$, and taking inverses gives $C_{-}$. The three half-turns are conjugate, since rotations move their axes into one another. Their centralizer has order four and their class therefore has size three.

A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

For a finite group action, Burnside's lemma givesPair each vertex with its opposite face. This correspondence is equivariant under every tetrahedral symmetry. Each pair has six possible combined color states: three face colors and two vertex colors. Thus we are coloring four positions with six symbols; the two kinds of coloring must be counted together.

The group of rotations is the alternating group $A_4$. The identity has four cycles and fixes $6^4$ colorings. Its three half-turns have cycle type $(2,2)$ and each fix $6^2$. Its eight third-turns have cycle type $(3,1)$ and each also fix $6^2$. Consequently the rotation count isAllowing mirror equivalence gives the full tetrahedral symmetry group $S_4$. Now all permutations of the four paired positions are allowed, so a coloring is determined by a multiset of four symbols chosen from six. stars and bars givesFor an independent Burnside check, the five cycle types of $S_4$ give numerator $6^4+6\cdot 6^3+3\cdot 6^2+8\cdot 6^2+6\cdot 6=3024$, whose quotient by 24 is 126. This is paired face and vertex colorings of a tetrahedron.

Over ${\mathbb{F}}_5$, putThese determinant-one matrices satisfy $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. Conjugation by $r$ cycles $i,ij,j$. Thus $⟨i,j⟩$ is a quaternion group of order eight, normalized by an element of order three. Their generated subgroup has order $24$ in $S{L}_2(5)$, hence order $12$ after quotienting by $\{\pm I\}$. Its index-five coset action, together with simplicity and perfectness, identifies $PS{L}_2(5)$ with the alternating group $A_5$.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.

The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

For a separable polynomial $f\in F[T]$ in characteristic two, the Vandermonde determinant of its roots is fixed by every Galois automorphism: root permutations change it only by a sign, and $-1=1$ in $F$. It therefore lies in $F$. Its square, multiplied by the square of the appropriate leading-coefficient power for a nonmonic polynomial, is the polynomial discriminant. Thus being a square does not detect containment of the Galois group in the alternating group. For example $T^2+T+1$ over ${\mathbb{F}}_2$ has discriminant one and Galois group $C_2$, while $A_2$ is trivial. The discriminant criterion for an alternating Galois group requires characteristic different from two.

In the alternating group $A_6$, of order $360$, a Sylow subgroup of order five is generated by a five-cycle. There are $6\cdot 4!=144$ five-cycles, four in each such subgroup, giving $36$ subgroups. A Sylow subgroup of order nine is generated by disjoint three-cycles, hence is $C_3\times C_3$. The unordered partitions of six letters into two triples give all such subgroups by Sylow conjugacy, and their single three-cycles recover the partition uniquely. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions.

The permutation group of a three-element set has six elements: the identity, three transpositions, and two three-cycles. A three-cycle and any transposition generate it, with relations $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$. Its alternating group is the normal cyclic subgroup of order three. It is the unique nonabelian isomorphism type of a group of order six.

A regular tetrahedron has a proper rotational symmetry group $T$ of order 12, isomorphic to the alternating group $A_4$ acting on its four vertices. Its full spatial symmetry group $T_d$ has order 24 and is isomorphic to the symmetric group $S_4$; the additional operations reverse orientation. Three-fold vertex/face-axis rotations and half-turns about opposite-edge midpoints generate the proper group. The distinction matters when a field symmetry uses only proper rotations, but a scalar density also admits reflections.