The Minimum Generating Set Problem

Let $G$ be a finite group. In order to determine the smallest cardinality $d(G)$ of a generating set of $G$ and a generating set with this cardinality, one should repeat many times the test whether a subset of $G$ of small cardinality generates $G$. We prove that if a chief series of $G$ is known, then the numbers of these generating tests can be drastically reduced. At most $|G|^{13/5}$ subsets must be tested. This implies that the minimum generating set problem for a finite group $G$ can be solved in polynomial time.


Introduction
Let G be a finite group.A generating set of G with minimum size is called a minimum generating set.The size of a minimum generating set of a group G is denoted by d(G).In this paper, we consider the problem of computing d(G), known as the minimum generating set problem, and computing a minimum generating set of a given group G.We call "a generating test" for G, the check whether a given subset of G generates G.The computational cost of a single generating test for a given group G may depends of the size of G and of the subset that we are testing, and also of the way in which G is assigned (G could be a represented by permutation group representation, a multiplication table, a linear group representation, or an abstract group assigned with a presentation), but in any case an algorithm aimed to determine a minimum generating set becomes more efficient the smaller number of generating tests it requires.This motivates the following question.
Question 1.How many generating tests are needed to determine a minimum generating set of a finite group?Question 1 has been studied before, mainly for groups represented by their multiplication table or by their permutation representation.With these representations of groups, it is easy to see that the minimum generating set problem is in the complexity class NP, as there is a polynomial time algorithm to check if a given subset generates the given group [22].Arvind and Torán prove that the minimum generating set problem is in DSPACE(log 2 n) [2].Recently, the problem has been studied for general and special classes of groups by Das and Thakkar [9].However, it is still open if the minimum generating set problem admits a polynomial time algorithm or equivalently if Question 1 could be solved using at most polynomial many generating tests even for groups given by their multiplication tables.Information on the complexity classes NP and DSPACE(log 2 n) can be found for example in [1].
Notice that, for every finite group G, d(G) ≤ log p |G| being p the smallest prime factor of |G|, so at least one set {g 1 , . . ., g d } with d ≤ log p (|G|) and 1 = g i ∈ G for 1 ≤ i ≤ d is a minimum generating set.Hence a crude bound for the number of generating tests that are needed is t≤log p (|G|) (|G| − 1) t .
A better strategy is suggested by the following nice remark due to Gaschütz [11].
Suppose indeed that a chief series of G is available.The factor G/N u−1 is a simple group, in particular d(G/N u−1 ) ≤ 2, so at most |G/N u−1 | 2 generating tests are needed to find a subset Y of minimum size with respect to the property that G = Y N u−1 .This can be considered a first step of an algorithm.For 0 < i < u suppose that {g 1 , . . ., g d } is a minimum generating set for G modulo N i .By the main result in [15], In other words, if we know a minimum generating set for generating tests are needed to determine a minimum generating set for G modulo N i−1 .This approach reduces the number of generating tests, but this number remains in general quite large, and it does not give any evidence that a polynomial bound in terms of the order of G is possible.Our main contribution is to show that the number of generating tests needed to obtained a minimum generating set of G/N i−1 from a minimum generating set of G/N i can be drastically reduced (see in particular Corollaries 10 and 14).As a consequence, we obtain we following unexpected result.
Theorem 2. The number of generating tests needed to determine a minimum generating set of a finite group G is at most |G| 13 5 .
Our procedure requires the knowledge of a chief series of the group, that in any case can be computed in polynomial time (see for example [22, 6.2.7]), so we may conclude that the problem of finding a minimum generating set for a finite group G has a solution which is polynomial on |G|.
Our proof of Theorem 2 uses the theory of crowns of finite groups (see Section 4) and some results about the generation of finite groups with a unique minimal normal subgroup that strongly depend on the classification of the finite non-abelian simple groups.

Chief series and chief factor
Let G be a nontrivial finite group.Recall that a chief series of a finite group G is a normal series of finite length with the property that for i ∈ {0, . . ., u − 1}, N i+1 /N i is a minimal normal subgroup of G/N i .The integer u is called the length of the series and the factors N i+1 /N i , where 0 ≤ i ≤ u − 1, are called the chief factors of the series.A nontrivial finite group G always possesses a chief series.Moreover, two chief series of G have the same length, and any two chief series of G are the same up to permutation and isomorphism.Thus, adopting the notation above, we may define the chief length of G to be u, and the chief factors of G to be the groups N i+1 /N i .
In the following we will denote by Frat(G) the Frattini subgroup of a finite group G. Moreover if H/K is a chief factor of G, we say that Since the Frattini subgroup of a finite group G is nilpotent, and the only subgroup supplementing Frat(G) is G itself, the following lemma is immediate.Lemma 3. Let G be a nontrivial finite group and let A = H/K be a chief factor of G.

Monolithic group and crown-based power
Let L be a monolithic primitive group and let A be its unique minimal normal subgroup.For each positive integer k, let L k be the k-fold direct product of L. The crown-based power of L of size k is the subgroup L k of L k defined by We also define L 0 := 1. Assume that A = soc L is non-abelian and let It follows from the main theorem in [17], that The elements of G are of the form l(y 1 , . . ., y δ ) with l ∈ L and y 1 , . . ., y δ ∈ A. We may identify N with the subgroup {(1, . . ., 1, y) | y ∈ A} of A δ = soc(G).Moreover, for 1 ≤ i ≤ d and 1 ≤ j ≤ δ, there exist l i ∈ L and y ij in A such that g 1 =l 1 (y 1,1 , . . ., y 1,δ ), . . . . . . . . .
Moreover let Ω be the set of (l 1 , . . ., l d ) in L d such that l 1 , . . ., l d = L and l i ≡ l i mod A for 1 ≤ i ≤ d.The condition G = g 1 , . . ., g d N implies that ω i ∈ Ω for 1 ≤ i ≤ δ − 1 and if i = j then ω i and ω j belong to different Γ-orbits for the action of Γ in Ω.
Let ∆ be the subset of Ω consisting of the d-tuples (l 1 , . . ., l d ) with l i = l i y iδ for each i > t.Assume A ∼ = S n , with n ∈ N and S a finite non-abelian simple group.By Lemma 5, Notice that the action of Γ on Ω is semiregular.Since (see the proof of [6, Lemma 1] and the upper bound for | Out(S)| given in [14]), the number of Γ-orbits with a representative in ∆ is at least Since |A| t−8/5 = |N| t−8/5 ≥ δ, we can conclude that ∆ contains an element with the required property.

Crowns
The notion of crown was introduced by Gaschütz in [12] in the case of finite solvable groups and generalized in [13] to arbitrary finite groups.A detailed exposition of the theory is also given in [3, 1.3].
If a group G acts on a group A via automorphisms (that is, if there exists a homomorphism G → Aut(A)), then we say that A is a G-group.If G does not stabilise any nontrivial proper subgroup of A, then A is called an irreducible G-group.Two G-groups A and B are said to be Gisomorphic, or A ∼ = G B, if there exists a group isomorphism φ : A → B such that φ(g(a)) = g(φ(a)) for all a ∈ A, g ∈ G. Following [13], we say that two G-groups A and B are G-equivalent and we put A ≡ G B, if there are isomorphisms φ : A → B and Φ : A ⋊ G → B ⋊ G such that the following diagram commutes: Note that two G-isomorphic G-groups are G-equivalent.In the particular case where A and B are abelian the converse is true: if A and B are abelian and G-equivalent, then A and B are also G-isomorphic.It is proved (see for example [13,Proposition 1.4]) that two chief factors A and B of G are G-equivalent if and only if either they are G-isomorphic, or there exists a maximal subgroup M of G such that G/ Core G (M) has two minimal normal subgroups X and Y that are G-isomorphic to A and B respectively.For example, the minimal normal subgroups of a crown-based power L k are all L k -equivalent.
For an irreducible G-group A denote by L A the monolithic primitive group associated to A. That is If A is a non-Frattini chief factor of G, then L A is a homomorphic image of G.More precisely, there exists a normal subgroup N of G such that G/N ∼ = L A and soc(G/N) ≡ G A. Consider now all the normal subgroups N of G with the property that G/N ∼ = L A and soc(G/N) ≡ G A: the intersection R G (A) of all these subgroups has the property that G/R G (A) is isomorphic to the crown-based power In our proof we will use the following consequence of [8, Proposition 1.1].

Lemma 7.
Let N be a minimal normal subgroup of a finite group G.
The next result is an immediate consequence of [10, Lemma 10].

Abelian minimal normal subgroups
Proposition 9. [18,Theorem 4] Let G be a group and let N = e 1 , . . ., e l be an abelian minimal normal subgroup of G.If G = g 1 , . . ., g d N then one of the following occurs: (1) d(G) ≤ d and either G = g 1 , . . ., g d or there exist Remark 11.If N is an abelian minimal normal subgroup of a finite group G, then N ∼ = C l p is an elementary abelian p-group and a generating set e 1 , . . ., e l for N can be obtained in l steps.So we may improve the statement of the previous corollary stating that a minimum generating set of G can be obtained from a minimum generating set of G/N in at most l + dl + 1 = (d + 1)l + 1 steps.

Non-abelian minimal normal subgroups
Lemma 12. Let N be a non-abelian minimal normal subgroup of a finite group G. Suppose that First we test the |N| t u-tuples of kind (g 1 n 1 , . . ., g t n t , g t+1 , . . ., g u ), with n 1 , . . ., n t ∈ N. If one of them is a generating set, we are done.If not, it follows from Lemma 12 that u < t.In this case we have tested that g 1 n 1 , . . ., g u n u = G for every (n 1 , . . ., n u ) ∈ N u , so if follows from Lemma 1 that d(G) > u.Moreover, by the main result in [15], Hence d(G) = u + 1 ≤ t and, by Lemma 1, one of the (u + 1)-tuples of kind (g 1 n 1 , . . ., g d n u , n u+1 ) is a minimum generating set In applying the previous corollary to design an algorithm to compute a minimum generating set of a finite group G, an obstacle is given by the fact that, even if a chief series of G is available, it is not easy to recognize whether two factors of the series are G-equivalent.Hence, with the computation applications in mind, it is better to consider the weaker equivalence relation in which two factors are equivalent if they have the same order.Denoting with η G (A) the number of factors in a chief series of G with order |A|, we may state the weaker formulation of the previous corollary.
We consider the equivalence classes in B in which two factors are equivalent if and only if they have the same order.Let B 1 , . . ., B r be the equivalence classes in B and for every class choose a representative Y i for this class.Then we have = β(G) 13 5 .
If follows that the number of required generating tests is at most .

Permutation groups
The best bound for the cardinality of a generating set of a permutation group is due to A. In the following we will denote by λ(G) the maximum of the orders of the non-abelian chief factors of G (and we will set λ(G) = 1 if G is solvable).Combining Theorem 15 and Lemmas 16 and 17 we obtain: Theorem 18.Let G ≤ Sym(n).Then we may obtain a minimum generating set of G with at most n 2 λ(G) 13/5 generating test.Proof.Let H/K be a non-abelian chief factor of G. Then H/K ∼ = S u , with u ∈ N and S a finite non-abelian simple group.Let X/K be a Sylow 2-subgroup of H/K.We have X/K ∼ = P u with P a Sylow 2subgroup of S. Since P cannot be cyclic, 2u ≤ d(X/K) ≤ d(X) ≤ n/2, hence u ≤ n/4 and |H/K| ≤ c n/4 , so the conclusion follows from the previous Theorem.

Corollary 10 .
Let N be a minimal abelian normal subgroup of a finite group G and let g 1 N, . . ., g d N be a minimum generating set of G/N.We can find a minimal generating set for G testing at most d(|N|−1)+1 ≤ d|N| elements of |G| d .Proof.By the previous proposition, if d(G) = d, then {g 1 , . . ., g d } or {g 1 , . . ., g i n, . . .g d }, with 1 = n ∈ N and 1 ≤ i ≤ d, is a minimal generating set of G.If none of these sets generates G, then d(G) = d+1 and, for every 1 = n ∈ N, {g 1 , . . ., g d , n} is a generating set.

6 Corollary 13 . 8 5
and δ = δ G (N). Moreover let L be the monolithic primitive group associated to N. We use the bar notation ḡ to denote the elements of the factor group Ḡ = G/R.By Lemma 8, N = NR/R is a minimal normal subgroup of G/R.Moreover it follows from Lemma 7 that G = g 1 n 1 , . . ., g t n t , g t+1 , . . ., g d if and only if Ḡ = ḡ1 n1 , . . ., ḡt nt , ḡt+1 , . . .ḡd .Since Ḡ ∼ = L δ , with L ∼ = G/C G (N), the conclusion follows from Proposition Let N be a minimal non-abelian normal subgroup of a finite group G and let g 1 N, . . ., g d N be a minimum generating set of G/N.We may produce a minimum generating set for G testing at most |N| ⌈ 8 5 +log |N| δ G (N )⌉ elements of G d and at most |N| ⌈ +log |N| δ G (N )⌉ elements of G d+1 .Proof.Let u = max{d, 2}.If d < u, set g d+1 = • • • = g u = 1.Let t = min{u, ⌈ 8 5 + log |N | δ⌉}.
McIver and P. Neumann: the so call "McIver-Neumann Half-n Bound" says that if G is a subgroup of Sym(n) and G = Sym(3), then d(G) ≤ ⌊n/2⌋.This result is stated without a proof in [20, Lemma 5.2] and a sketch of the proof is given in [4, Section 4].The following result is crucial for our purposes.Theorem 15. [21, Theorem 10.0.5].Let G be a permutation group of degree n with s orbits.Then a chief series of G has length at most n − s.Lemma 16.Let G ≤ Sym(n) and N a minimal abelian normal subgroup of G.If we know a minimum generating set for G/N, we can find a minimum generating set for G in at most (⌊n/2⌋ + 1) 2 steps.Proof.We may assume n > 3. Then max{d(G), d(N)} ≤ n/2, so the conclusion follows from Remark 11.Lemma 17.Let G ≤ Sym(n) and N a minimum non-abelian normal subgroup of G.If we know a minimum generating set for G/N, we can find a minimum generating set for G with at most (n − 1)|N| 13 5 generating tests.Proof.By Corollary 13, we need at most |N| log |N| δ G (N )+ 13 5 = δ G (N)|N| 13 5 generating tests.Moreover from Theorem 15, it follows δ G (N) ≤ n − 1.

Corollary 19 .
Let G ≤ Sym(n).If every non-abelian composition factor of G has order at most c, then we may obtain a minimum generating set of G with at most n 2 c 13 20 n generating tests.
[16,and γ ∈ C. The following holds (see[5, Section  2]).Before to state the next result, we need to recall another important observation.Let H be a finite group, X a subset of H, M a normal subgroup of H and assume that h 1 , ..., h k ∈ H and X generate H modulo M, that is H = h 1 , ..., h k , X, M .It follows from[16,] that the number Φ H,M (X, k) of elements (u 1 , ..., u k ) ∈ M k with the property that H = h 1 u 1 , ..., h k u k , X is independent of the choice of h 1 , ..., h k .Let ∆ be the set of the t-tuples (a 1 , ..., a t ) ∈ A t with the property that L = l 1 a 1 , ..., l t a t , l t+1 , ..., l d .Then Set X = {l t+1 , ..., l d }.It follows from the proof of [19,Lemma  2], that there exist y 1 , . . ., y t ∈ L such that L = y 1 , . . ., y t , X .Hence |∆| = φ L,A (X, t).To conclude it suffices to notice that, by [7, Theorem 2.2], φ L,A (X, t) ≥ 53|A| t /90.Proposition 6.Let L be a monolithic primitive group, and assume that A = soc(L) is non-abelian.Consider the crown-based product G = L δ and let N ∼ = A be a minimal normal subgroup of G. Suppose that G = g 1 , . . ., g d N, with d ≥ 2. Let t = ⌈ 8 5 +log |N | δ⌉.If t ≤ d, then there exist n 1 , . . ., n t ∈ N such that G = g 1 n 1 , . . ., g t n t , g t+1 , . . ., g d .
Corollary 14.Let N be a minimal non-abelian normal subgroup of a finite group G.If we know a minimum generating set of G/N, then the number of generating tests needed to compute a minimum generating set for G is at most2|N| ⌈ 8 5 +log |N| η G (N )⌉ .7.Finding a minimal generating set in polynomial timeProof of Theorem 2. Let A and B be, respectively, the set of abelian and non-abelian factors in a given chief series of G. Moreover let α(G) = With iterated application of Corollaries 10 and 14, we deduced that the number of generating tests required to obtain a minimum generating set for G is at most X∈A |A| and β(G) = Y ∈B |B|.