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For an even permutation $g$, its symmetric group class splits into two equal alternating group classes exactly when its symmetric-group centralizer contains no odd permutation. If an odd commuting element exists, an odd conjugator can be multiplied by it to become even; otherwise the index-two subgroup has two conjugation orbits. In cycle notation the splitting types are precisely distinct odd cycle lengths, counting fixed points as length-one cycles: rotating odd cycles is even, whereas an even cycle or the exchange of two equal odd cycles supplies an odd centralizer element.

Cayley theorem embeds a finite group of order $n$ into $S_n$. Extend $\sigma \in S_n$ to $n+2$ letters and multiply it by the disjoint transposition $(n+1n+2)$ exactly when $\sigma$ is odd. Additivity of parity makes this an injective group homomorphism into the alternating group $A_{n+2}$. Thus every finite group is isomorphic to a subgroup of some alternating group.

An ordinary Specht module of non-self-conjugate shape stays irreducible on the alternating group, and conjugate shapes give the same restriction. A self-conjugate shape splits into two irreducibles of equal degree, interchanged by an odd permutation. The index-two inner-product identity is $⟨{\chi }^{\lambda }\downarrow ,{\chi }^{\mu }\downarrow ⟩={\delta }_{\lambda \mu }+{\delta }_{{\lambda }^{\prime }\mu }$, and Frobenius reciprocity proves exhaustion. The statement applies for $n\ge 2$; $A_1$ is trivial.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

In the conjugacy class order $1,(123),(12)(34),(12345),(13452)$, the five complex irreducible characters of the alternating group $A_5$ areThe class sizes are $1,20,15,12,12$. Restricting the degree-four and degree-five irreducible characters of $S_5$ gives rows of squared character inner product one. The degree-six restriction has squared norm two and is orthogonal to those rows and the trivial row, so it splits into the two remaining irreducibles. The sum of squared degrees forces their degrees to be three. Orthogonality forces their last two values to have sum one and product minus one, producing the displayed roots. Since a normal subgroup is a union of classes and its order divides sixty, these class sizes also prove the Simplicity of the alternating group A5.

For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs:This is the first isomorphism theorem applied to the sign of a permutation. Once commutativity is forced, all three transpositions become indistinguishable, and both three-cycles become the identity element. The surviving information is exactly whether the permutation is even or odd. Compare the very different center of a group $Z(S_3)=\{e\}$.

On the six-dimensional exterior square of ${\mathbb{F}}_2^4$, the Pfaffian $Q=x_{12}{x}_{34}+x_{13}{x}_{24}+x_{14}{x}_{23}$ is a nonsingular plus-type quadratic form. The action of $G{L}_4(2)$ preserves it and is faithful: fixing every decomposable exterior line fixes every two-space and then every one-space. The even-weight binary subset module on eight letters, modulo the all-one vector, gives another six-dimensional plus-type space with $Q=\mathrm{wt}(x)/2\mathrm{mod}2$. Coordinate permutations embed $S_8$ faithfully into its orthogonal group. That orthogonal group has order $40320$, while $G{L}_4(2)$ has order $20160$, identifying its index-two image with the alternating group $A_8$.

Divisibility of a proposed subgroup order is necessary by Lagrange's theorem but is not sufficient in arbitrary finite groups. The alternating group $A_4$ has order $12$ and no subgroup of order $6$. Such a subgroup would have index two, hence be a normal subgroup. The quotient would have order two, so every element of order three would map to its identity. But $A_4$ has eight distinct three-cycles, which together with the identity cannot fit in a subgroup of order six. This obstruction contrasts with cyclic groups, which have a subgroup of every divisor order.

A nontrivial finite group whose only normal subgroups are the trivial group and the whole group. A cyclic group of prime number order gives an abelian example; $A_5$ gives a nonabelian alternating group example. Finite versions of classical groups can produce further examples after taking suitable subgroups and quotients; small parameters require separate checks.

A group homomorphism is a map $\phi :G\to H$ preserving the group operation: $\phi (gh)=\phi (g)\phi (h)$. It also preserves the identity element and inverse elements. In an additive group, the same requirement reads $\phi (g+h)=\phi (g)+\phi (h)$. A group homomorphism preserves a way of combining inputs, even when several inputs have the same output.

Reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$ is a group homomorphism: reducing a sum agrees with adding the reduced residues. For example, $2+5=1$ modulo six, and both sides reduce to one modulo three; the outputs $2$ and $2$ also add to one modulo three. This map is surjective but not injective, since zero and three both map to zero.

The sign of a permutation gives another example, from the symmetric group $S_3$ to the multiplicative group $\{1,-1\}$. A product of two odd permutations is even, matching $(-1)(-1)=1$. Its group kernel is $A_3$, the alternating group of even permutations. This shows how a group homomorphism can retain a useful feature while forgetting other details.

Let $G=S_3$, the symmetric group on three letters, and take the normal subgroup $N=A_3=\{e,(123),(132)\}$, the alternating group of even permutations. Its two left cosets areProducts of permutations apply the rightmost factor first. Left multiplication by an even permutation preserves both cosets, while left multiplication by an odd permutation exchanges them. For example,Consequently all six group elements give only two distinct permutations of $\{E,O\}$. Both stabilizer subgroups equal $A_3$, so their intersection, the group kernel, is $A_3$. This is a transitive group action because $(12)$ reaches the other coset, but it is not a faithful group action because $(123)$ fixes both.

The orbit-stabilizer theorem still works perfectly: $6=2\cdot 3$. It counts reachable labels, without asserting that the induced group homomorphism is injective. The quotient group $S_3/A_3\cong C_2$ captures exactly the two visible operations, identity and exchange. This is the same group homomorphism as the sign of a permutation, with $+1$ represented by fixing both labels and $-1$ by exchanging them. Contrast the three-label example in faithful coset action of the symmetric group on three letters.

A subgroup $N\le G$ is a normal subgroup when $gN{g}^{-1}=N$ for every $g\in G$. In other words, conjugation by any group element preserves the subgroup as a set, although it may exchange its individual elements. Equivalently, $gN=Ng$ for every $g$: the left and right cosets of $N$ agree. Write $N⊴G$.

Every subgroup of an abelian group is a normal subgroup, because conjugation fixes each element. Normality can also hold in a nonabelian group: in $S_3$, the three even permutations form the alternating group $A_3=\{e,(123),(132)\}$, a normal subgroup as the group kernel of the sign of a permutation. Its two cosets separate the even and odd permutations.

In contrast, $H=\{e,(12)\}$ is a subgroup of $S_3$ but is not a normal subgroup. With function composition of permutations acting rightmost first, choose $g=(123)$. Thenwhich are different cosets. Also $g(12)g^{-1}=(23)\notin H$, directly contradicting invariance under conjugation. Thus a subgroup alone does not suffice for quotient group multiplication.

The practical value of normality is coset multiplication is well-defined exactly for normal subgroups. Each coset can act as one element of a quotient group, with results independent of representatives. Conversely every normal subgroup is the group kernel of its quotient map of groups, so normality also identifies which distinctions a group homomorphism can consistently forget.

Pair a tetrahedron vertex with its opposite face. This is an equivariant correspondence under all tetrahedral symmetries, so three face colors and two vertex colors become six combined symbols at each of four positions. Burnside's lemma on the alternating group $A_4$ gives $(6^4+3\cdot 6^2+8\cdot 6^2)/12=141$. Including reflections gives $S_4$ and reduces the count to multisets of four symbols from six, namely $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$.

Compose permutations from right to left. The indicated cycle lengths immediately give $X^3=P^2=Q^2=e$. Under conjugation, $X^{-1}(ab)X=(X^{-1}a{X}^{-1}b)$. Since $X^{-1}$ sends $1,2,3$ to $3,1,2$ and fixes $4$,Direct composition also gives $PQ=QP=(14)(23)$. Thus $V=\{e,P,Q,PQ\}$ is a Klein four-group. Conjugation by $X$ cycles its three nonidentity elements, so $X$ normalizes $V$.

The cosets $V,VX,V{X}^2$ have four elements each and are distinct. If two coincided, a nontrivial power of $X$ would belong to $V$, impossible because it has order three whereas every nonidentity element of $V$ has order two. Hence the subgroup generated by $X,P,Q$ contains at least twelve elements. These generators are even permutations, so the subgroup lies in the alternating group $A_4$, whose order is $4!/2=12$; multiplying by one fixed transposition pairs even and odd permutations to justify this count. ThereforeThe calculation also describes $A_4$ as the semidirect product $V⋊⟨X⟩$, with the order-three factor permuting the three involutions.

First prove the primitive three-cycle criterion directly. The useful generation fact is that adding a three-cycle whose support meets a set $S$, with $|S|\ge 3$, to the alternating group $A_S$ generates the alternating group on the enlarged set. If the support is $\{a,b,x\}$ with $a,b\in S$ and $x\notin S$, conjugation by $A_S$ and taking inverses produces every three-cycle with $x$ and two points of $S$. For $|S|\ge 4$ this uses two-transitivity of $A_S$; for $|S|=3$, its cyclic action on pairs together with inversion gives both pair orientations. These cycles, together with the cycles supported inside $S$, are all the three-cycles on $S\cup \{x\}$ and generate $A_{S\cup \{x\}}$. Every even permutation is a product of three-cycles: pair its transpositions, and use $(ab)(bc)=(abc)$ and $(ab)(cd)=(acb)(acd)$, with rightmost factors acting first.

If instead the new support is $\{a,x,y\}$ with just $a\in S$, choose $b\in S\setminus \{a\}$ and conjugate $(axy)$ by an element of $A_S$ carrying $a$ to $b$. This gives $(bxy)$, andThe two-point-overlap case first adjoins $x$; the original cycle then adjoins $y$. This proves connected triple supports generate an alternating group by successively adjoining intersecting supports.

Now take the supports of every conjugate in $G$ of a given three-cycle, and make these triples the edges of a hypergraph on the permutation domain. Its connected components are permuted by $G$, so they form a block system. Since at least one component contains three points, primitivity forces the single component to be the entire domain. Start with one triple and adjoin intersecting triples until all points have been reached. The generation argument shows that the subgroup generated by the conjugate cycles is $A_n$. Hence

To classify the proper maximal subgroups containing a three-cycle, distinguish the three possible types of action. A primitive maximal subgroup contains $A_n$, and thus equals $A_n$, since $[S_n:A_n]=2$. An intransitive maximal subgroup lies in the full stabilizer $S_k\times S_{n-k}$ of one of its orbits, and maximality makes it that full stabilizer. When $k=n-k$, this stabilizer lies properly inside the subgroup that can interchange the two parts, so it is not maximal. Take the smaller part to obtain $1\le k<n/2$. Such an unequal-part stabilizer is maximal: its transpositions connect all points within each part, and any element outside it carries at least one within-part transposition to a cross-part transposition. The resulting connected graph of transpositions generates $S_n$. To see that last assertion, transpositions along a path generate the transposition of its endpoints by successive conjugations. The stabilizer contains a three-cycle exactly when $n-k\ge 3$.

A transitive imprimitive maximal subgroup preserves a nontrivial partition into $b$ blocks of size $a$, with $n=ab$. It equals the full permutation wreath product $S_a\wr S_b$. This full block stabilizer is maximal: any permutation not preserving the partition sends some same-block pair to a different-block pair, giving a cross-block transposition by conjugation. The block stabilizer is transitive on cross-block pairs, so all cross-block transpositions follow; together with its within-block transpositions they generate $S_n$. A three-cycle in this group must act trivially on the blocks, since any nontrivial block permutation moves at least two entire blocks, and a block three-cycle moves $3a$ points. Thus a single three-cycle exists exactly when $a\ge 3$. The full classification, up to conjugation, is thereforeThis is the maximal symmetric-group subgroups containing a three-cycle classification. For $n<3$ there are no such subgroups.

For the first family of counterexamples to extending the criterion to longer cycles, let $G=PS{L}_2(p)$ act on the projective line ${\mathbb{F}}_p\cup \{\infty \}$. The determinant-one matricesact respectively as $x\mapsto x+1$ and $x\mapsto -1/x$. The first fixes infinity and is one $p$-cycle on the finite points. Its translations make the infinity stabilizer transitive on the other points; the second moves infinity, so the full action is two-transitive, hence primitive. The action is faithful after quotienting by scalar matrices. Counting a nonzero first column and the $p$ second columns with determinant one gives $|S{L}_2(p)|=p(p^2-1)$, and the scalar group center has order two. Thussince the ratio on the right to the left is $(p-2)!>1$. Consequently $PS{L}_2(p)$ is the required example for every prime $p>3$.

For the final example take $p=59$, and let the direct product of groups $A_5\times A_5$ act on the sixty elements of $A_5$ byThe action is transitive and the stabilizer of the identity is the diagonal subgroup $\Delta =\{(z,z):z\in A_5\}$. To prove that the diagonal is a maximal subgroup, no classification theorem is needed. If $\Delta <K\le A_5\times A_5$, choose $(x,y)\in K\setminus \Delta$. Multiplying by $(y^{-1},y^{-1})$ gives $(x{y}^{-1},1)\in K$ with $x{y}^{-1}\ne 1$. The subgroup $L=\{z:(z,1)\in K\}$ is normal in $A_5$, by diagonal conjugation, and is nontrivial. Simplicity of $A_5$ gives $L=A_5$, and then $K=A_5\times A_5$. Thus the coset action is primitive. It is also faithful: if $xt{y}^{-1}=t$ for every $t$, then $x=y$ by taking $t=1$, and $x$ is central in the centerless simple group $A_5$, so $x=y=1$. Finally,The prime $59$ is odd, and its primality follows by testing the primes $2,3,5,7$, the only ones below its square root. This is a faithful diagonal primitive permutation group whose order is not divisible by the degree minus one.

The rotation $g$ has order six. The possible cycle types in the alternating group $A_4$ are the identity, a three-cycle, and a product of two disjoint transpositions; their orders are one, three and two. Since a group isomorphism preserves the order of a group element, $D_{12}$ is not isomorphic to $A_4$.

We prove Simplicity of the alternating group A5 by finding all its conjugacy classes. The alternating group has group order $|A_5|=5!/2=60$. Its even permutations have cycle types $1$, $(abc)$, $(ab)(cd)$ and $(abcde)$. There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles, $5\cdot 3=15$ products of disjoint transpositions, and $4!=24$ five-cycles.

For a three-cycle, a commuting permutation must rotate its three-element support and may interchange the two fixed points. Its centralizer in $S_5$ has order six, and only the three rotations are even. Thus its conjugacy class in $A_5$ has size $60/3=20$ and contains every three-cycle. The centralizer of $(12)(34)$ in $S_5$ has order eight: one may interchange the two pairs and independently interchange the elements within each pair. It contains the odd element $(12)$, so exactly half of these commuting elements are even. The conjugacy class in $A_5$ consequently has size $60/4=15$. A permutation commuting with a five-cycle is determined by the image of one point and must be a power of that cycle. All five powers are even, so each such conjugacy class has size $60/5=12$. The 24 five-cycles form exactly two classes.

A normal subgroup contains the identity and is a union of whole conjugacy classes. Its possible orders therefore lie inBy Lagrange's theorem, its order must divide 60. Only 1 and 60 in this list do so. Hence $A_5$ has no proper nontrivial normal subgroup and is a simple group.

Adjoin $\infty$ to $\Omega =\{1,\dots ,n\}$ and take $G=A_n$ fixing it. Choose three distinct points $a,b,c\in \Omega$ and put $t=(\infty a)(bc)$. This is an even involution interchanging $\infty$ and $a$. Conjugation by it preserves $H=G_a$, because it permutes the remaining $n-1$ points and preserves parity. Thus $tHt=H$.

If $g\in G\setminus H$, the even permutation $k=tgt$ moves $\infty$ into $\Omega$. The alternating group $A_n$ is transitive for $n\ge 3$: a suitable 3-cycle carries any prescribed point to another. Choose $u\in G$ with $u(a)=k(\infty )$. Then $v=t{u}^{-1}k$ is even and fixes $\infty$, so $v\in G$ and $k=utv\in GtG$. All conditions of the double-coset criterion for a one-point extension hold, including when $n=3$ and $H$ is trivial.

The constructed extension lies in $A_{n+1}$ and has order $(n+1)|A_n|=(n+1)!/2$. Therefore