Recently I learned that the cardinality of a minimal set of generators of a finite $p$-group $G$ is well defined namely it is equal to the dimension of $H^1(G,\mathbb{F}_p)$. Moreover, if $S:=\{g_1,\ldots,g_s\}$ is a minimal generating set of $G$ then the cardinality of a minimal set of relations with respect to $S$ is also a well defined integer, namely it is equal to the dimension of $H^2(G,\mathbb{F}_p)$.

For a general group, the cardinality of a minimal set of generators is not well defined. Take for example the symmetric group of degree $n$. For instance you may get a cardinality equal to $2$ or $n-1$.

Q1: So for what class of finite groups do we expect the cardinality of a minimal generating set to be well defined?

P.S. By "a minimal set of generators" I mean "irredundant", that is the set that generates the group, but no proper subset does.

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    $\begingroup$ I suppose that by "minimal" you mean "irredundant", that is the set generates the group, but no proper subset does. In which case I would not expect so many finite groups to have all minimal generating sets of the same cardinality. For example, all nonabelian finite simple groups have one minimal generating set of cardinality 2. They can all also be generated by a conjugacy class of involutions. Refine this to a minimal generating set, and the resulting minimal generating set can't just have two elements, as any two involutions generate a dihedral group. $\endgroup$ – Geoff Robinson Jun 4 '11 at 14:34
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    $\begingroup$ Cyclic groups (other than p-groups) don't have the property, but dihedral groups of order 2p do. I wouldn't expect any nice characterization. $\endgroup$ – Jack Schmidt Jun 4 '11 at 15:40
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    $\begingroup$ A more natural consideration than minimal generating set is minimal Nielsen equivalence class of generators. en.wikipedia.org/wiki/Nielsen_transformation $\endgroup$ – Ian Agol Jun 4 '11 at 16:31

This is an answer for the abelian case only.

For a finite abelian group (let us exclude the trivial one) there are two standard ways to decompose it as a direct sum of cyclic groups. One into cyclic groups of order $n_1,\dots,n_r$ (not $1$) such that $n_i \mid n_{i+1}$ and the other one into cyclic groups of order $m_1, \dots, m_s$ such that each $m_i$ is a prime power (not $1$). [Under each of the assumptions 'divisibility' and 'prime power' the repective decomposition is unique, in the latter case of course up to the ordering.]

The parameter $r$ is sometimes called the rank and the parameter $s$ the total rank of the group (although terminology here is not completely uniform).

Now, it is known that the rank is the minimal cardinality of a generating set, in the sense that there does not exists a set of a smaller cardinality that generates the group. And, that the total rank is the maximal cardinality of a minimal generating set, that is there exists a generating set of cardinailty $s$ such that no subset of this set generates the group.

Thus, the cardinality of all minimal/irredundant generating set is uniquely determined if and only if the rank equals the total rank. This is the case if and only if the group is an (abelian) $p$-groups.

Note that for a non-$p$-group on can first consider the first decomposition into cycylic groups, and then decompose each cyclic component as the sum of cyclic groups of prime power order (more or less Chines Remainder Theorem); so only if all the orders in the first decomposition are also prime powers (and thus necessarily powers of the same prime) does one not get a different value for rank and total rank.

  • $\begingroup$ I should have thought about it! Thanks! $\endgroup$ – Hugo Chapdelaine Jun 4 '11 at 18:36
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    $\begingroup$ For a nilpotent group $G$ a set $S$ is generating $G$ iff it does modulo the commutator subgroup. Hence, it is reduced to the abelian case. $\endgroup$ – Yiftach Barnea Jun 4 '11 at 19:19
  • $\begingroup$ I think the argument extends to show that $G$ modulo the commutator subgroup must be a $p$-group. $\endgroup$ – Frieder Ladisch Jun 7 '11 at 23:00
  • $\begingroup$ More on “Now, it is known …” part: mathoverflow.net/questions/82603/… $\endgroup$ – Tsuyoshi Ito Feb 19 '12 at 22:55

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