Divisible subdivisions of graphs in subdivisions of complete graphs

Abstract

Let $\mathbb{Z}_q$ denote the cyclic group of order $q$. A $\mathbb{Z}_q$-edge-weighted $K_f$ is the complete graph $K_f$ equipped with a weight function $\omega : E(K_f) \to \mathbb{Z}_q$. A subdivision of a graph $H$ in a $\mathbb{Z}_q$-edge-weighted $K_f$ is called a $q$-divisible subdivision of $H$ if every subdivision path has weight congruent to zero modulo $q$. Let $q\ge 2$ be an integer and let $H$ be a graph with $n$ vertices and $m$ edges. Define $s_q(H)$ to be the smallest number $f$ such that every $\mathbb{Z}_q$-edge-weighted $K_{f}$ contains a $q$-divisible subdivision of $H$. Das, Dragani\'c, and Steiner raised the following question (Problem 4.1 in [Tight bounds for divisible subdivisions, J. Combin. Theory, Ser. B 165 (2024) 1-19]): Given $q\in\mathbb{N}$ and a subcubic graph $H$ with $n$ vertices and $m$ edges, is it true $s_q(H)= m(q - 1) + n$? They also established the upper bound $s_q(H)\le 7mq+8n+14q$ for such a graph $H$. In this paper, we improve this bound by showing that $s_q(H)\le (2q - 1)m + 2n - 1 + 4q$, and establishing a sharper bound $s_p(H)\le \frac{3p - 1}{2}m - \frac{p - 1}{2}n + \frac{p + 1}{2}$ for prime $p$ and connected $H$. We resolve this problem in the case $q=2$ by proving that $s_2(H) = m + n$ for any 5-degenerate graph $H$, and in the case $q\ge 2$ and $T$ being a tree, by showing that $s_q(T) = nq - q + 1$. Let $s_q(H,t)$ be the minimum number $f$ such that every $\mathbb{Z}_q$-edge-weighted $K_f$ contains a $q$-divisible $t$-subdivision of $H$, where a $t$-subdivision of $H$ is a subdivision of $H$ such that each edge of $H$ is subdivided exactly $t$ times. We also prove that $s_2(H,1)= m + n$, where $H$ is a tree or a cycle on $n$ vertices with $m$ edges.