Generators of a group and normal subgroups

finite-groupsgr.group-theorynormal-subgroups

Can we say anything about a minimal generating set of a finite group based on its normal subgroups? For example, can we bound their order, or say whether they come from the same conjugacy class?

An easy example is that when $$G$$ is cyclic, if $$x$$ generates $$G$$ then $$x$$ does not generate any (normal) subgroup of $$G$$.

If it isn't possible in general to relate a minimal set of generators of a finite group to its normal subgroups, can we do this when $$G$$ is generated by only 2 or 3 elements?

The arXiv paper The Minimum Generating Set Problem by Lucchini and Thakkar (to be published in Journal of Algebra) describes an algorithm for finding a smallest-sized generating set of (a finite group) $$G$$ starting from a chief series $$G = N_u > N_{u-1} > \dotsb N_1 > N_0= \{1\}$$ of $$G$$.
The idea is to successively find smallest-sized generating sets of the quotients $$G/N_{u-1}, G/N_{u-2},\dotsc,G/N_1,G/N_0 \cong G$$. The top factor is simple and either cyclic with one generator, or nonabelian simple with two generators, which can quickly found by a random search.
Defining $$d(G)$$ to be the smallest size of a generating set of $$G$$, there is a useful result that, for a minimal normal subgroup $$N$$ of $$G$$, we always have $$d(G/N) \le d(G) \le d(G/N)+1$$ and, furthermore, if $$d(G/N) = d$$ with $$G/N = \langle g_1N,\ldots,g_dN\rangle$$, then either $$d(G) = d(G/N)$$ and there exist $$n_1,\dotsc,n_d \in N$$ with $$G = \langle g_1n_1,\dotsc,g_dn_d \rangle$$; or $$d(G) = d(G/N)+1$$ and there exist $$n_1,\dotsc,n_d,n_{d+1} \in N$$ with $$G = \langle g_1n_1,\dotsc,g_dn_d,n_{d+1} \rangle$$.
The bulk of the paper is describing methods to decide which of the two cases we are in when moving from $$G/N_i$$ to $$G/N_{i-1}$$ without recourse to too many exhaustive searches through all $$d$$- or ($$d+1$$)-tuples of elements of $$N$$. Implementations run very fast in practice.