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?

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    $\begingroup$ One cannot talk about the generators of a group. For instance, one can take the entire group as a generating set. $\endgroup$ 2 days ago
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    $\begingroup$ 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? $\endgroup$ 2 days ago
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    $\begingroup$ Please do not post simultaneously here and in math.stackexchange. Especially if you do not alert readers of the double post. $\endgroup$ 2 days ago
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    $\begingroup$ math.stackexchange copy has been deleted on that site. $\endgroup$ 2 days ago
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    $\begingroup$ 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 $\pm 1$). $\endgroup$ 2 days ago

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ yesterday
  • $\begingroup$ @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. $\endgroup$
    – Derek Holt

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