Conjecture on the largest noncyclic abelian group with a 2-element spanning set

Put

$$B_s=\{(x,y)\in\mathbb Z^2:|x|+|y|\leq s\}, \qquad L=2s+1.$$

Let $G$ be a noncyclic finite abelian group with an $s$-spanning pair $a,b$. The surjection

$$\pi:\mathbb Z^2\longrightarrow G,\qquad (x,y)\longmapsto xa+yb,$$

has kernel $\Lambda$. Since $G$ is noncyclic and two-generated,

$$G\cong\mathbb Z_c\times\mathbb Z_{ck}$$

for some $c\geq2$. Choose a prime $p\mid c$. Smith normal form implies

$$\Lambda\subseteq p\mathbb Z^2.$$

Therefore reduction modulo $p$ induces a surjection

$$G\twoheadrightarrow(\mathbb Z/p\mathbb Z)^2,$$

each fiber having size $|G|/p^2$. Since $\pi(B_s)=G$, every residue class contains at least $|G|/p^2$ points of $B_s$. Hence

$$|G|\leq p^2\mu_p(s), \qquad \mu_p(s)= \min_{\rho\in(\mathbb Z/p\mathbb Z)^2} |B_s\cap(\rho+p\mathbb Z^2)|. \tag{1}$$

For odd $p$, write

$$L=pm+r,\qquad 0\leq r<p.$$

The invertible change of coordinates modulo $p$

$$u=x+y,\qquad v=x-y$$

identifies $B_s$ with pairs

$$(u,v)\in[-s,s]^2,\qquad u\equiv v\pmod2.$$

For a residue $\alpha\pmod p$, let $E_\alpha,O_\alpha$ count respectively the even and odd integers in $[-s,s]$ congruent to $\alpha$. Set

$$n_\alpha=E_\alpha+O_\alpha,\qquad \delta_\alpha=E_\alpha-O_\alpha.$$

Then the occupancy of the residue pair $(\alpha,\beta)$ equals

$$E_\alpha E_\beta+O_\alpha O_\beta =\frac{n_\alpha n_\beta+\delta_\alpha\delta_\beta}{2}. \tag{2}$$

Exactly $r$ residue classes have size $m+1$, and $p-r$ have size $m$. Furthermore,

$$\delta_\alpha=0\ \text{if }n_\alpha\text{ is even}, \qquad \delta_\alpha\in\{-1,1\}\ \text{if }n_\alpha\text{ is odd}.$$

Ordering residue classes by first appearance in the interval, the $p-r$ minimum-size classes occur consecutively, and their $\delta$-signs alternate whenever $m$ is odd. It follows from (2) that

$$\mu_p(s)= \begin{cases} m^2/2,&m\text{ even},\\[2mm] (m^2-1)/2,&m\text{ odd},\ r\leq p-2,\\[2mm] (m^2+1)/2,&m\text{ odd},\ r=p-1. \end{cases} \tag{3}$$

Indeed, the middle case has two minimum-size classes with opposite signs. In the last case there is exactly one minimum-size class, and its self-pair gives $(m^2+1)/2$; every pair involving a size-$(m+1)$ class has occupancy at least $m(m+1)/2$.

If $p\mid L$, then $r=0$ and $m$ is odd, so

$$|G| \leq\frac{p^2(m^2-1)}2 =\frac{L^2-p^2}{2} = 2s^2+2s-\frac{p^2-1}{2}. \tag{4}$$

If $p\nmid L$, then $r\geq1$. The first two cases of (3) give

$$p^2\mu_p(s)\leq\frac{(L-1)^2}{2}=2s^2.$$

In the third case, $r=p-1$, the same conclusion reduces to

$$p^2(m^2+1)\leq(pm+p-2)^2.$$

The difference between right and left is

$$2pm(p-2)-4p+4,$$

which is positive for $p\geq5,m\geq1$, and for $p=3$ whenever the odd integer $m\geq3$. The sole exceptional case $p=3,m=1$ gives $s=2$, outside the hypothesis. Thus

$$p\nmid L,\ s\geq3 \quad\Longrightarrow\quad |G|\leq2s^2. \tag{5}$$

For $p=2$, exactly $s^2$ points of $B_s$ satisfy

$$x+y\not\equiv s\pmod2.$$

If $s$ is even, these split equally between the residue classes $(0,1)$ and $(1,0)$, giving

$$\mu_2(s)\leq s^2/2.$$

If $s$ is odd, they occupy $(0,0)$ and $(1,1)$, giving

$$\mu_2(s)\leq(s^2-1)/2.$$

Equation (1) again yields $|G|\leq2s^2$.

If $L=2s+1$ is prime, a prime $p\mid c$ cannot equal $L$: formula (3) would give $\mu_L(s)=0$, contradicting spanning. Therefore $p\nmid L$, and

$$|G|\leq2s^2.$$

If $L$ is composite and $p_0$ is its least prime divisor, either $p\nmid L$, yielding (5), or $p\mid L$, in which case $p\geq p_0$ and (4) gives

$$|G| \leq 2s^2+2s-\frac{p_0^2-1}{2}.$$

Since $p_0^2\leq L$, this bound exceeds $2s^2$ and therefore covers both cases.

Finally, both upper bounds are attained by the constructions already established in Theorem 1.7. The group

$$\mathbb Z_s\times\mathbb Z_{2s}$$

has the $s$-spanning pair

$$\{(0,1),(1,1)\}$$

and order $2s^2$. If $L=p_0t$, the group

$$\mathbb Z_{p_0} \times \mathbb Z_{p_0(t^2-1)/2}$$

has the $s$-spanning pair

$$\left\{ \left(1,\frac{t-1}{2}\right), \left(1,\frac{t+1}{2}\right) \right\}$$

and order

$$\frac{L^2-p_0^2}{2} = 2s^2+2s-\frac{p_0^2-1}{2}.$$

These matching upper and lower bounds prove both cases of the conjecture for every $s\geq3$.