Diameter of a finite group with respect to a generating set (${\mathrm{diam}}_S(G)$)

For a finite group $G$ with a finite symmetric generating set $S$, its diameter is the maximum word metric distance between two elements, equivalently the diameter of its Cayley graph. PutThenTranslation invariance reduces the maximum pairwise distance to the maximum distance from the identity, which proves the equality.

If $1\in S$, one may replace $B_S(n)$ by $S^n$. Otherwise exact powers may never equal $G$: for $G=C_2=\{1,a\}$ and $S=\{a\}$, even powers are $\{1\}$ and odd powers are $\{a\}$, while the actual Cayley graph has diameter one. This is why the ball convention matters. The trivial diameter bound for a finite group gives ${\mathrm{diam}}_S(G)\le |G|-1$.

A connected Cayley graph on a finite group $G$ has $|G|$ vertices, and a shortest path repeats no vertex. Therefore

For every $\epsilon >0$ there is $\lambda (\epsilon )$ such that every non-abelian finite simple group $G$ and every finite symmetric generating set $S$ containing the identity satisfyThe proof applies the one-scale virtual nilpotence theorem below the diameter and then uses simplicity to eliminate both the bounded-index subgroup and the finite normal subgroup.