# Generating pairs of a finite group with $(a,b)$ conjugate to $(a,a^kb)$ - can we control $k$?

Let $$G$$ be a finite group admitting a pair of generators $$a,b$$. I'm happy to assume that $$G$$ is finite simple.

Sometimes the pair $$(a,b)$$ is (simultaneously) conjugate to $$(a,a^kb)$$ for some $$k \ge 1$$ (ie, there is a $$g\in G$$ with $$gag^{-1} = a, gbg^{-1} = a^kb$$). Of course this is always true for $$k = |a|$$. Let $$k_{a,b} := \min\{k\ge 1\;|\; (a,b) \text{ is (simultaneously) conjugate to }(a,a^kb)\}$$

I'm looking for statements of the form: Under some conditions on $$G$$, as $$(a,b)$$ ranges over all generating pairs of $$G$$, $$k_{a,b}$$ can't be too small (relative to $$|a|$$) too often. In particular I'd be interested in an upper bound for the integer $$\text{lcm}\left(\left\{\frac{|a|}{k_{a,b}}\right\}_{a,b}\right)$$ as $$(a,b)$$ ranges over all generating pairs of $$G$$.

I'm interested in anything in this direction.

I'm also happy to make an additional restriction: Let $$(a,b)$$ be a generating pair of a finite simple group $$G$$. Suppose $$a$$ has prime order. Could $$(a,b)$$ be conjugate to $$(a,ab)$$ by an element of the form $$[a^{-1},b^{-1}]^r := (a^{-1}b^{-1}ab)^r$$? ($$r$$ is an integer).

I believe I can show that for $$PSL(2,q)$$, we always have $$k_{a,b} = |a|$$, but the proof is quite involved and uses some structure which is special to $$PSL(2,q)$$. Can we say anything nontrivial for more general finite simple groups?