Suppose we know $G$ and $N$ are groups with $G$ non-abelian, $N\triangleleft G$ and $N\ne G$.
Further suppose that for any known $g,h\in G$ we can compute $gh$, compute $g^{-1}$ and decide if $g=1$ (that is $G$ is a black box group).
Given $a,b\in G\setminus\{1\}$ suppose we know that exactly one of these elements is in $N$ but we do not know which. Can we compute an element $g$ of $G$ that is likely to be a non-trivial element of a proper normal subgroup of $G$, that is with probability strictly greater than $1/2$?
In the case $[a,b]=aba^{-1}b^{-1}$ is non-trivial we have $[a,b]\in N$ is non-trivial as required with probability $1$, but what can we do in the case $a$ and $b$ commute?
For those who want background, this is a simplified case for the last exercise of chapter 2 in 'Permutation Group Algorithms' by Seress. I can't quite do the exercise and think this will help.