Rank-two conjecture for groups even on maximal proper subsets

Claim. Let $m\geq n\geq2$, and let

$$\Gamma_{m\oplus n} = \left\{ (\sigma,\tau)\in S_m\times S_n: \operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau) \right\}. \tag{1}$$

Then the complete minimum-generator classification is

$$\operatorname{rank}(\Gamma_{m\oplus n}) = \begin{cases} 1,&(m,n)=(2,2),\\ 3,&(m,n)\in\{(3,3),(4,3),(4,4)\},\\ 2,&\text{otherwise}. \end{cases} \tag{2}$$

In particular, this proves Conjecture 1 of Vítor H. Fernandes, Groups of permutations that are even on maximal proper subsets, and related monoids, arXiv:2605.12342v1. The source identifies its group with (1), supplies a three-generator upper bound, and proves the two-generator assertion only for certain coprime cycle lengths. Here all cycle lengths, including equal block sizes, are covered uniformly.

Step 1: Convenient symmetric-group generators. For every $d\geq2$, define

$$t_d=(1\ 2), \qquad u_d= \begin{cases} (1\ 2\ \cdots\ d),&d\text{ odd},\\ (2\ 3\ \cdots\ d),&d\text{ even}. \end{cases} \tag{3}$$

For $d=2$, the second expression means $u_2=1$. In every case $t_d$ is odd, $u_d$ is even, and

$$\langle t_d,u_d\rangle=S_d. \tag{4}$$

Indeed, when $d$ is odd, conjugating $t_d$ by powers of the full cycle gives the transpositions along a connected cycle. When $d$ is even, the same conjugation gives the transpositions joining the fixed point $1$ to every other point. Transpositions along either connected graph generate the full symmetric group.

We will use the standard subdirect-product form of Goursat's lemma. If

$$H\leq G_1\times G_2 \tag{5}$$

projects onto both factors, then there are normal subgroups $N_i\trianglelefteq G_i$ and an isomorphism

$$G_1/N_1\cong G_2/N_2 \tag{6}$$

such that $H$ consists exactly of pairs whose images in this common quotient agree. Explicitly,

$$N_1=\{x\in G_1:(x,1)\in H\}, \qquad N_2=\{y\in G_2:(1,y)\in H\}. \tag{7}$$

The common quotient is trivial exactly when $H=G_1\times G_2$.

For $d\geq5$, the normal subgroups of $S_d$ are

$$1,\quad A_d,\quad S_d. \tag{8}$$

Thus its nontrivial quotients are $S_d$ and $C_2$. The smaller cases are

$$\begin{array}{c|c} G&\text{nontrivial quotient groups}\\ \hline S_2&C_2\\ S_3&S_3,\ C_2\\ S_4&S_4,\ S_3,\ C_2, \end{array} \tag{9}$$

where the additional $S_3$ quotient comes from

$$S_4/V_4\cong S_3. \tag{10}$$

The exceptional outer automorphism of $S_6$ does not alter (8) or its quotient list.

Step 2: Unequal block sizes. Suppose $m>n\geq2$ and $(m,n)\neq(4,3)$. Consider the two elements

$$g=(t_m,t_n), \qquad h=(u_m,u_n) \tag{11}$$

of $\Gamma_{m\oplus n}$. By (4), the subgroup

$$H=\langle g,h\rangle \tag{12}$$

projects onto both $S_m$ and $S_n$.

Since $H$ is contained in the parity fiber product (1), it cannot be the entire direct product. Therefore its Goursat quotient is nontrivial. By (8)–(9), the only common nontrivial quotient of $S_m$ and $S_n$ is $C_2$: distinct symmetric groups have different orders, and the sole additional coincidence for unequal sizes is the excluded pair

$$S_4/V_4\cong S_3. \tag{13}$$

For every $d\geq2$, the unique quotient homomorphism from $S_d$ onto $C_2$ is the sign homomorphism. Consequently Goursat's description gives

$$H = \{(\sigma,\tau): \operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau)\} = \Gamma_{m\oplus n}. \tag{14}$$

This includes all cases $n=2$, since $u_2=1$ and (4) still holds.

Step 3: Equal block sizes. Suppose $m=n=d\geq5$, and put

$$c=(3\ 4\ 5)\in S_d. \tag{15}$$

The permutations $t_d$ and $c$ have disjoint supports, so

$$(t_dc)^3=t_d, \qquad \operatorname{ord}(t_d)=2, \qquad \operatorname{ord}(t_dc)=6. \tag{16}$$

Consider instead

$$g=(t_d,t_dc), \qquad h=(u_d,u_d). \tag{17}$$

Both coordinates of $g$ are odd, both coordinates of $h$ are even, and hence both generators belong to $\Gamma_{d\oplus d}$. The first projection generates $S_d$ by (4). The second projection also generates $S_d$, because (16) recovers $t_d$ from $t_dc$, after which (4) applies again.

Let $H=\langle g,h\rangle$. By (8), its nontrivial Goursat quotient is either $C_2$ or $S_d$. If that quotient were $S_d$, then $H$ would be the graph of an automorphism

$$\varphi:S_d\longrightarrow S_d. \tag{18}$$

Since $g\in H$, this would imply

$$\varphi(t_d)=t_dc. \tag{19}$$

But automorphisms preserve element orders, whereas (16) gives orders $2$ and $6$. This is impossible, even for $d=6$, where an outer automorphism exists. Hence the common quotient is $C_2$, and the same sign argument as in (14) proves

$$\langle(t_d,t_dc),(u_d,u_d)\rangle = \Gamma_{d\oplus d} \qquad(d\geq5). \tag{20}$$

Whenever $m>2$, the group $\Gamma_{m\oplus n}$ projects onto the noncyclic group $S_m$. Thus it is itself noncyclic, so the two generators in (14) and (20) are minimal.

Step 4: Determine all exceptional ranks. First,

$$\Gamma_{2\oplus2} = \{(1,1),(t_2,t_2)\} \cong C_2, \tag{21}$$

so its rank is one.

Next let

$$E=\Gamma_{3\oplus3}. \tag{22}$$

Its even-even subgroup is

$$V=A_3\times A_3\cong C_3^2, \tag{23}$$

and any diagonal odd-odd transposition acts on $V$ by inversion in both coordinates. Therefore

$$E\cong C_3^2\rtimes C_2, \tag{24}$$

with the nonidentity element of $C_2$ acting as multiplication by $-1$.

No two elements generate $E$. If both belong to $V$, their subgroup misses the odd coset. If exactly one belongs to $V$, conjugation by the other only changes that single vector to its negative. If both belong to the odd coset, their product is a single vector of $V$, and conjugation again only changes its sign. In the latter two cases the generated subgroup meets $V$ in a subspace of dimension at most one, whereas $V$ has dimension two. Thus

$$\operatorname{rank}(E)\geq3. \tag{25}$$

Because $V_4\subset A_4$, the quotient map in (10) preserves sign. Applying it in the $S_4$ coordinate, or in both $S_4$ coordinates, gives surjections

$$\Gamma_{4\oplus3}\twoheadrightarrow E, \qquad \Gamma_{4\oplus4}\twoheadrightarrow E. \tag{26}$$

Consequently both groups also require at least three generators.

For completeness, the universal three-generator upper bound follows directly from

$$(t_m,t_n), \qquad (u_m,1), \qquad (1,u_n). \tag{27}$$

Indeed, the normal closure of $u_d$ in $S_d=\langle t_d,u_d\rangle$ is exactly $A_d$: it is contained in $A_d$, while its quotient is generated by the image of the involution $t_d$ and therefore has order at most two. Thus the subgroup generated by (27) contains

$$A_m\times1, \qquad 1\times A_n, \qquad (t_m,t_n), \tag{28}$$

which together generate the entire parity fiber product. Therefore all three groups in (25)–(26) have rank exactly three. Combining this with (14), (20), and (21) proves the complete classification (2).