# Generators of a group and normal 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?

• One cannot talk about the generators of a group. For instance, one can take the entire group as a generating set. 2 days ago
• Every finite simple group is generated by two elements. There is an extensive research on this. Maybe if you have only a few normal subgroups one might be able to restrict the number of generators? 2 days ago
• Please do not post simultaneously here and in math.stackexchange. Especially if you do not alert readers of the double post. 2 days ago
• math.stackexchange copy has been deleted on that site. 2 days ago
• I should add that for a $p$-group if you have $k$ normal subgroups, then in particular, you have at most $k$ maximal normal subgroups, Thus, if $\Phi(G)$ is the Frattini subgroup, then $G/Phi(G)$ has at most $k$ maximal subspaces, so its dimension, which is also the number of generators, is at most about $\log_p k$ (I am too tired now to figure out the $\pm 1$). 2 days ago

This arXiv paper 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 $$G = N_u > N_{u-1} > \cdots N_1 > N_0= \{1\}$$ of $$G$$.
The idea is to successively find smallest-sized generating sets of the quotients $$G/N_{u-1}, G/N_{u-2},\ldots,G/N_1,G/N_0 \cong G$$. 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 $$n_1,\ldots,n_d \in N$$ with $$G = \langle g_1n_1,\ldots,g_dn_d \rangle$$; or $$d(G) = d(G/N)+1$$ and there exist $$n_1,\ldots,n_d,n_{d+1} \in N$$ with $$G = \langle g_1n_1,\ldots,g_dn_d,n_{d+1} \rangle$$.
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.
• Does it mean that $d(G)$ is bounded by something linear in the number of normal groups because $u$ is bounded by the number of normal subgroups? Is that best? yesterday
• @YiftachBarnea Yes it clearly implies that $d(G)$ is bounded by the length of a chief series, and that bound is attained in an elementary abelian group, for example. There could be a better bound in terms of the number of normal subgroups. yesterday