Fiber-counting lemma for a quotient map
= Fiber-counting lemma for a quotient map
Let $\pi:G\to G/N$ be a <quotient group> map, let $A$ be finite and symmetric, and suppose $P\subseteq\pi(A^m)$ has $|P|\geq\delta|\pi(A)|$. Then
$$
|\pi^{-1}(P)\cap A^{m+2}|\geq\delta|A|.
$$
Choose one lift in $A^m$ of each member of $P$ and multiply those lifts by $A^2\cap N$. Distinct fibers are disjoint, while $|\pi(A)|\,|A^2\cap N|\geq|A|$.