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For any perfect group $H$ and $m\ge 3$, the permutation wreath product $H\mathrm{wr}{A}_m$ is generated by elements of order three. The top $A_m$ is generated by 3-cycles. With $\sigma =(123)$, the product of the conjugate $h_1\sigma h_1^{-1}$ and ${\sigma }^{-1}$ is $h_1{h}_2^{-1}$; using ${\sigma }^{-1}$ gives $h_1{h}_3^{-1}$. The commutator of $h_1{h}_2^{-1}$ and $k_1{k}_3^{-1}$ is $[h,k{]}_1$, since distinct coordinate groups commute. Perfectness gives the entire first coordinate group, and transitivity gives all coordinates. No finite-generation assumption on $H$ is needed.

Commutation with a fixed-point-free involution preserves its partner relation. A commuting permutation can permute the $m$ pairs and independently swap each pair. This gives the permutation wreath product with order $2^{m}m!$.

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.

Let $B=H_1\times \cdots \times H_m$, with each $H_i$ a copy of $H$. The permutation group $F\le S_m$ acts on $B$ byDefine the permutation wreath product to be the semidirect productExplicitly,Writing $h_i$ for the element with entry $h$ in coordinate $i$ and identity elsewhere, and writing $\sigma$ for $(1,\sigma )$, conjugation satisfiesThus the top permutation group permutes the coordinate subgroups in exactly its given action. Distinct coordinate subgroups commute, including when $H$ itself is nonabelian.

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.

Conjugation in the symmetric group sends $(12)$ to $(\sigma (1)\sigma (2))$. Hence $\sigma$ commutes with $g$ exactly when it preserves the unordered pair $\{1,2\}$. It can independently swap that pair and arbitrarily permute the remaining points. Therefore the centraliser of a transposition isLet $m=n/2$. The orbits of a group action generated by $h$ are its $m$ two-point blocks. Any centraliser element must permute those blocks, and can choose independently whether to swap the two points in each image block. Conversely every such choice commutes with $h$, since it preserves the partner relation. This centraliser of a fixed-point-free involution is the permutation wreath product $C_2\wr S_m$, with