For finite groups $L \leq K$ we define $d(L,K)$ to be the least $n \in \mathbb{N}$ for which there exist $a\_1, \dots, a\_n \in K$ such that $\langle L, a\_1 \dots, a\_n \rangle = K$. 


Is there some $m \in \mathbb{N}$ for which the following claim holds:


Let $G$ be a finite group and let $H \leq G$ be a subgroup with $d(H,G) = 1$. Suppose that $G\_0$ is a subgroup of index $2$ in $G$ such that $H \leq G\_0 \leq G$. Then $d(H,G\_0) \leq m$.



Generating finite groups using subgroups
I don't think so. 


Let $K$ be any group, and let $e\neq t\in C\_2$ act on $K\times K$ by exchanging coordinates. Let $G=(K\times K)\rtimes \langle t\rangle$, and let $H$ be generated by the elements $(k,1)$ with $k\in K$. Then $\langle H,t\rangle = G$, so $d(H,G) = 1$.


Now let $G\_0 = K\times K$. Then $d(H,G\_0)=d(1,K)$, and that can be arbitrarily large by a suitable choice of $K$.


