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A connected finite hypergraph of three-element supports, with a three-cycle on each support, generates the full alternating group. To merge a new triple meeting an old alternating support in two letters, conjugate over unordered pairs of old letters. For one-letter overlap compare $(abc)$ and $(a^{\prime }bc)$: their quotient is $(ab{a}^{\prime })$, reducing to two two-letter-overlap steps.

Up to conjugation, the proper maximal subgroups of $S_n$ containing a three-cycle are $A_n$; the unequal-part subset stabilizers $S_k\times S_{n-k}$ for $1\le k<n/2$ with $n-k\ge 3$; and the full block stabilizers $S_a\wr S_b$ for $n=ab$, $a\ge 3$, $b\ge 2$. The primitive three-cycle criterion treats the primitive case. A nontrivial block system gives the full permutation wreath product in the imprimitive case. These block stabilizers are maximal because any extra permutation produces a transposition connecting two different blocks; conjugating by the block stabilizer then gives every cross-block transposition, hence the symmetric group. Unequal-part subset stabilizers are maximal by the same connected-transposition argument. Equal-part subset stabilizers are not maximal: they lie in the larger stabilizer that can interchange the two parts.

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.

Use the following standard facts: $A_5$ has order $60$ and is a nonabelian simple group; the coset action on the cosets of an index-$j$ subgroup is a transitive homomorphism into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its kernel is a normal subgroup, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. There are no subgroups of indices 2, 3 or 4.

An index-5 subgroup has order $12$ by Lagrange's theorem. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the orbit-stabilizer theorem. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a subgroup of the permutations preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 permutations are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.

Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even permutations of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. ConsequentlyThese five subgroups are distinct: for $i\ne j$, a three-cycle on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.

A composition series is a subnormal chain whose nontrivial successive quotients are simple. The Jordan–Hölder theorem asserts that any two composition series of a finite group have the same length and the same multiset of simple factors, up to isomorphism. Existence follows by successively choosing maximal proper normal subgroups until the identity is reached.

For uniqueness, induct on the group order. Let $A,B$ be the penultimate terms of two series. If $A=B$, apply induction to that subgroup and add the common top factor. If $A\ne B$, their normality and maximality give $AB=G$. Put $C=A\cap B$. The isomorphism theorems giveBoth are simple. Take any composition series of $C$. Appending $A$ to it gives a series for $A$, so induction identifies the factors of the original series inside $A$ with the factors of $C$ together with $G/B$. Similarly the factors inside $B$ are those of $C$ together with $G/A$. Adding the respective top factors produces the same multiset in both cases. This proves uniqueness, including multiplicities and length.

A normal subgroup $K$ of order two is central because its automorphism group is trivial. By Jordan–Hölder, a composition series through $K$ leaves the simple quotient $G/K\cong A_5$. The image of $Z(G)$ in this quotient is central and hence trivial, since $A_5$ is nonabelian simple. ThusThe symmetric group $S_5$ has trivial group center: conjugation of each transposition by a central permutation must preserve its two-element support, forcing the permutation to fix every point. Since its Jordan–Hölder factors are $A_5$ and $C_2$, a normal subgroup of order two would equal that trivial group center, a contradiction.

Identify $C_2=\{1,-1\}$ and put $D=1\times A_5$ in $C_2\times S_5$. Its three maximal normal subgroups areThey are the three kernels of nonzero maps from the abelianization $C_2^2$ to $C_2$. There is no quotient $A_5$: a central $C_2$ must map trivially, while $S_5$ has no quotient $A_5$. The latter follows from its normal subgroups $1,A_5,S_5$, which can themselves be obtained by intersecting with the simple subgroup $A_5$ and using its trivial centralizer. The complete list of composition series isWithin $M_1$ the two direct factors are the only maximal normal subgroups; within each of $M_2,M_3\cong S_5$ the only such subgroup is its copy of $A_5$. Hence there are exactly four series.

For $S_5\times S_5$, put $D=A_5\times A_5$, $X=A_5\times 1$, $Y=1\times A_5$ andAgain these are exactly the maximal normal subgroups: they are the three cyclic quotient kernels, and there is no simple quotient $A_5$ of the whole product. All series aretogether withTo verify exhaustiveness, $M_L$ has maximal normal subgroups $D$ and $1\times S_5$, and $M_R$ has $D$ and $S_5\times 1$. A quotient $A_5$ of a direct product can receive a nontrivial image from only one simple factor, since the images commute and $A_5$ has trivial group center. The remaining maximal-normal possibility is $M_D$. Its derived subgroup is $D$, because both alternating factors are perfect. A hypothetical quotient $A_5$ would project one alternating factor isomorphically and kill the other, but conjugation by an odd diagonal element induces an outer automorphism on that surviving factor, impossible inside $A_5$. Odd conjugation is outer because otherwise an odd permutation times an even one would centralize $A_5$; a permutation commuting with every three-cycle fixes every three-element support and is the identity. Thus $D$ is the only maximal normal subgroup of $M_D$. Finally the only maximal normal subgroups of $D$ are $X,Y$: normal subgroups of a product of two centerless simple groups are products of the factors, as follows by commutating with each factor. This proves exactly eight series and the classification of composition chains in products with S5.

For $S{L}_2(5)$, commuting with both elementary upper and lower transvections forces a central matrix to be scalar. A scalar $\lambda I$ has determinant ${\lambda }^2$, soIf $g^2=I$, characteristic five makes $g$ diagonalizable with eigenvalues in $\{1,-1\}$. Determinant one rules out one of each, so the unique nonidentity involution is $-I$. An index-two subgroup would have order $60$ and, by the Cauchy theorem for groups, contain that involution and hence the group center. Its image in $PS{L}_2(5)$ would have index two. The simplicity and identification $PS{L}_2(5)\cong A_5$ are established by the independent matrix argument in Solution 4; a simple group has no proper index-two subgroup. Therefore $S{L}_2(5)$ has no subgroup of index two.

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

We prove the primitive three-cycle criterion through supports. First, suppose a subgroup contains $A_U$, the alternating group on a set $U$ of at least three letters, and a three-cycle with support $E$ meeting $U$.

If $E$ adds one letter, it has two letters in $U$. The group $A_U$ is transitive on unordered pairs of $U$: for three letters its cyclic action permutes the three pairs, and for at least four letters one can adjust the parity of a permutation sending one pair to another while preserving its target pair. Conjugating the new cycle, and using inverses, supplies every three-cycle on two old letters and the new letter. Together with $A_U$, these generate $A_{U\cup E}$.

If $E$ adds two letters, write its cycle as $\tau =(abc)$ with $a\in U$. Choose $a^{\prime }\in U\setminus \{a\}$ and conjugate within $A_U$ to obtain ${\tau }^{\prime }=(a^{\prime }bc)$. ThenThe preceding case first adds $b$; applying it again to $\tau$ adds $c$. This proves connected triple supports generate an alternating group, by successively merging the intersecting triples of a connected support hypergraph. We used that three-cycles generate alternating groups: pairs of transpositions with a common letter are three-cycles, while a pair of disjoint transpositions is a product of two three-cycles.

Now take all conjugates in $G$ of the given three-cycle. Their support hypergraph is $G$-invariant, and its connected components form a block system. Transitivity makes their union the whole set. There is an edge, so a component has more than one vertex. A primitive group action forces a single component. The preceding merging argument proves

For the first family of counterexamples, letIts elements are Möbius transformations $x\mapsto (ax+b)/(cx+d)$, modulo nonzero scalar matrices. Any ordered triple of distinct projective points is the image of $(\infty ,0,1)$ under exactly one such transformation: representatives of the first two target lines form a basis, and scaling the two columns sends their sum to the third target line. Thus the action is sharply three-transitive, hence two-transitive and primitive. Its degree is $p+1$ and order $p(p^2-1)$. It is divisible by $p$, butso it cannot contain $A_{p+1}$.

For the second example takeThe same projective-line construction over ${\mathbb{F}}_8$ is sharply three-transitive. Its order is divisible by 7 and is less than $|A_9|=181440$, so it does not contain $A_9$. For a concrete field one can use ${\mathbb{F}}_8={\mathbb{F}}_2[u]/(u^3+u+1)$.

A cyclic group has one generator of a group: $G=⟨a⟩=\{a^j:j\in \mathbb{Z}\}$. An abelian group has $xy=yx$ for every pair of elements. Powers of one element commute because $a^r{a}^s=a^{r+s}=a^{s+r}=a^s{a}^r$, so every cyclic group is Abelian. The Klein four-group $C_2\times C_2$ is Abelian but is not a cyclic group: every nonidentity element has order two, whereas a generator of a group for a four-element cyclic group would have order four.

Fix a generator of a group $x$ of $C_n$. A group homomorphism is determined by $g=\varphi (x)$ because $\varphi (x^j)=g^j$, and the relation $x^n=1$ forces $g^n=1$. Conversely, such a $g$ defines $\varphi (x^j)=g^j$: exponents differing by a multiple of $n$ give the same value, and addition of exponents verifies the group homomorphism law. Thus the homomorphism from a finite cyclic group correspondence isFor $S_4$, the order of a permutation is the least common multiple of its disjoint cycle lengths. The condition ${\sigma }^4=1$ allows identity, transpositions, two disjoint transpositions, and four-cycles. The sixteen homomorphisms are ${\varphi }_{\sigma }(x^j)={\sigma }^j$, with the full list of possible generator imagesThe remaining eight elements of the symmetric group are three-cycles and do not qualify.

The product of two transpositions has three possible forms. Equal transpositions produce the identity, transpositions meeting in one letter produce a three-cycle, and disjoint transpositions produce two disjoint two-cycles. There are two three-cycles on each chosen triple and three pairings of each chosen four-set. Therefore the number of distinct permutations representable this way is

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

A group homomorphism from an additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.

For ${\mathbb{Z}}_{18}\to {\mathbb{Z}}_{15}$, the relation is $18a\equiv 0(\mathrm{mod}15)$, equivalently $3a\equiv 0(\mathrm{mod}15)$. The possibilities are $a=0,5,10$, and each defines a homomorphism between cyclic groups. Thus there are three such maps, an instance of the number of homomorphisms between finite cyclic groups.

For ${\mathbb{Z}}_9\to S_4$, the image of $1$ must be a permutation whose order of a group element divides $9$. A permutation on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the three-cycles. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles: choose their three symbols and one of the two cyclic orientations. Hence there are nine maps.

Finally, each transposition in $S_5$ has order $2$, so its image in ${\mathbb{Z}}_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every permutation is a product of transpositions, so every element must map to the identity. There is one map. In the order asked, the counts are

By the preceding relabelling formula, every conjugate of $(123)$ is a three-cycle, and any three-cycle on four symbols can be obtained by a suitable relabelling. ThereforeThere are eight elements, agreeing with the count $\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 2$.

Likewise, conjugation preserves the pattern of two disjoint transpositions, and every partition of four symbols into two pairs is realized by relabelling. ThusThese have the respective cycle types $(3,1)$ and $(2,2)$ in the symmetric group $S_4$.

A product of two transpositions is the identity when they coincide, a three-cycle when their supports meet in one letter, and a product of two disjoint permutation cycles when their supports are disjoint.