Polynomial ideals and classes of finite groups

Serge Bouc

The object of this note is to discuss properties of some polynomials (on a countable set of indeterminates) attached to any finite group, which generalize the Eulerian functions of a group defined by P. Hall ([8]). In particular, I will define some classes of finite groups associated to prime ideals of the polynomial ring, and I will show that each finite group has a unique largest quotient in such a class of groups.

This work is a generalization of the notion of b-group introduced in [2], by a systematic use of the polynomial formalism of section 7.2.5 of [2]. For the reader’s convenience however, this paper is self-contained, and the proofs of the results already stated in [2] are included.

1. The polynomial ring

(1.1) Notation: I will consider the (countable) set S of isomorphism classes of finite (non-trivial) simple groups, and the polynomial ring over S

R = ℤ[(X_{S})_{S∈S}]

In other words, the ring R is the algebra over ℤ of the Grothendieck monoid of the category of finite groups.

(1.2) Notation: If p is a prime number, and S is the isomorphism class of the cyclic group C_{p} of order p, I denote by X_{p} the variable X_{S} of R.

It is convenient to turn R into a ℝ-graded ring by setting

deg(X_{S}) = log |S|

If G is a finite group, then I define the monomial P (G) ∈ R by

P (G) = ∏_{S∈S} X_{S}^{ν_{S}(G)}

where for each finite simple group S, the integer ν_{S}(G) is the multiplicity of S as a composition factor of G. In particular P (1) = 1 and P (S) = X_{S} if S ∈ S. Note that with this definition, the degree of P (G) is equal to log |G|, for any finite group G.

I denote by P̃(G) the polynomial obtained from P (G) by Möbius inversion on the poset of subgroups of G, i.e.

P̃(G) = ∑_{H≤G} μ(H, G)P (H)

where the notation H ≤ G means that H is a subgroup of G, and μ(H, G) is the Möbius function of the poset of subgroups of G. The monomial of highest degree in the expression of P̃(G) corresponds to the subgroup H = G, and the coefficient μ(G, G) is equal to 1. Thus

(1.3)  deg(P̃(G)) = log |G|

Finally, if N is a normal subgroup of G, I define

Q_{G,N} = ∑_{{H≤G \atop HN=G}} μ(H, G)P (H ∩ N)

This is a generalization of the previous formula, since Q_{G,G} = P̃(G). On the other hand Q_{G,1} = 1 for any G. The terms of highest degree in Q_{G,N} correspond to subgroups H such that H ∩ N = N and HN = G. Hence the only possible subgroup is H = G, and again the coefficient of P (H ∩ N) = P (N) is equal to 1. Thus

(1.4)  deg(Q_{G,N}) = log |N|

The polynomials P (G), P̃(G), and Q_{G,N} are invariant under group isomorphism: if φ : G → G′ is an isomorphism, then obviously P (G) = P (G′), P̃(G) = P̃(G′), and Q_{G,N} = Q_{G′,φ(N)}.

(1.5) Remark: The Eulerian function of the group G, defined by Hall ([8]), is the function ϕ(G, s) defined for a complex number s by

ϕ(G, s) = ∑_{H≤G} μ(H, G)|H|^{s}

Hence it is the evaluation of the polynomial P̃(G) when X_{S} is replaced by |S|^{s}, for all S ∈ S. The name “Eulerian” comes from the fact that when s is a positive integer, the value ϕ(G, s) is the number of sequences of s elements of G which generate G.

The polynomials P̃(G) and Q_{G,N} are also closely related to the probabilistic zeta function and its relative version, studied by K. Brown in [4]. The value ζ(G, s) of this zeta function at a complex number s is given by

1/ζ(G, s) = \frac{ϕ(G, s)}{|G|^{s}} = ∑_{H≤G} \frac{μ(G, H)}{|G : H|^{s}}

(Brown’s notation for 1/ζ(G, s) is P (G, s), but it is a bit confusing here, and I prefer to use another symbol).

Hence the value ζ(G, s)|G|^{−s} can also be recovered from 1/P̃(G) by replacing each variable X_{S} by |S|^{s}.

(1.6) Remark: Let D denote the ring of Dirichlet polynomials over ℤ, i.e. the ring of finite linear combinations with coefficients in ℤ of functions s ↦ n^{−s} from ℂ to ℂ, for n ∈ ℕ − {0}. It is clear that the map sending X_{S} ∈ R for S ∈ S to the map s ↦ |S|^{−s} induces a surjective ring homomorphism from R to D.

On the other hand, the decomposition of an integer n ∈ ℕ−{0} as a product of prime factors yields an isomorphism from D to the algebra R′ over ℤ of the multiplicative monoid ℕ − {0}, sending the function s ↦ n^{−s} to ∏_{p∈P} Y_{p}^{v_{p}(n)}, where P is the set of prime numbers and v_{p}(n) is the p-valuation of n, for p ∈ P, and Y_{p} is the image of the prime p in R′.

Hence the ring D is isomorphic to the quotient of R by the ideal generated by the elements X_{S} − ∏_{p∈P} X_{p}^{v_{p}(|S|)}, for all non-abelian simple groups S.

The fundamental properties of the polynomial P (G), P̃(G), and Q_{G,N} are given in the following lemma:

(1.7) Lemma: [[2] Lemme 19] The G be a finite group, let N be a normal subgroup of G, and let R be a subgroup of G containing N. Then

(1.8)  P (G) = P (G/N)P (N)

(1.9)  ∑_{{H≤G \atop HN=R}} μ(H, G)P (H) = μ(R, G)P (R/N)Q_{G,N}

(1.10)  ∑_{{H≤G \atop HN=R}} μ(H, G)P (H ∩ N) = μ(R, G)Q_{G,N}

(1.11)  P̃(G) = P̃(G/N)Q_{G,N}

Proof: The first equality is a trivial consequence of the definition, since the multiplicity of the simple group S as a composition factor of G is the sum of its multiplicities as a composition factor of N and G/N.

For the second one, by Crapo complementation formula (see [6] Theorem 5 or [5] pp. 420-421), for H ≤ G

μ(H, G) = ∑_{H≤K≤G ; {KN=G \atop K∩HN=H}} μ(H, K)μ(K, G) = ∑_{H≤K≤G ; {KN=G \atop K∩HN=H}} μ(HN, G)μ(K, G)

Now if N ≤ R ≤ G, the conditions KN = G, K ∩ HN = H and HN = R are equivalent to KN = G and H = K ∩ R. This gives

∑_{{H≤G \atop HN=R}} μ(H, G)P (H) = ∑_{{K≤G \atop KN=G}} μ(R, G)μ(K, G)P (K ∩ R)

= μ(R, G) ∑_{{K≤G \atop KN=G}} μ(K, G)P (K ∩ R)

Moreover (K ∩ R)N = R, thus R/N ≃ (K ∩ R)/(K ∩ N), and

P (K ∩ R) = P (R/N)P (K ∩ N)

It follows that for N ≤ R ≤ G

∑_{{H≤G \atop HN=R}} μ(H, G)P (H) = μ(R, G)P (R/N) ∑_{{K≤G \atop KN=G}} μ(K, G)P (K ∩ N)

and equality 1.9 follows. Equality 1.10 is a consequence, since if HN = R, then P (H) = P (R/N)P (H ∩ N).

Since moreover μ(R, G) = μ(R/N, G/N), the summation of 1.9 for subgroups R with N ≤ R ≤ G gives

P̃(G) = ∑_{N≤R≤G} μ(R, G)P (R/N)Q_{G,N} = P̃(G/N)Q_{G,N}

as was to be shown.

(1.12) Corollary: If M and N are normal subgroups of G such that G/M ≃ G/N, then Q_{G,M} = Q_{G,N}.

Proof: Indeed by 1.11 Q_{G,N} = P̃(G)/P̃(G/N).

2. The polynomials P̃(G)

In view of 1.11, it is natural to ask when the polynomial P̃ is irreducible:

(2.1) Proposition: Let G be a finite group. Then P̃ is irreducible if and only if G is simple. Proof: Let S be a simple group. Then

P̃(S) = X_{S} + R

where R is a polynomial in the variables X_{T}, for simple groups T not isomorphic to S. Hence P̃(S) is irreducible.

Conversely, if G is a finite group and P̃(G) is irreducible, let N be a normal subgroup of G. Then by 1.11 one of the polynomial P̃(G/N) or Q_{G,N} has degree 0. By 1.3 and 1.4, it follows that N = G or N = 1. Hence G is simple.

(2.2) Notation: If N is a normal subgroup of G, I denote by K_{G}(N) the set of complements of N in G, i.e.

K_{G}(N) = {L ≤ G | LN = G, L ∩ N = 1}

If M and N are normal subgroups of G, I denote by K_{G}(M, N) the set of subgroups of G which are complements of M and N, i.e.

K_{G}(M, N) = K_{G}(M) ∩ K_{G}(N)

(2.3) Proposition: Let G be a finite group, and N be a minimal (non-trivial) abelian normal subgroup of G, isomorphic to (C_{p})^{n}, for p prime and n > 0. Then

Q_{G,N} = X_{p}^{n} − |K_{G}(N)|

Proof: Let H be a subgroup of G such that HN = G. Then H∩N is normalized by H, and by N since N is abelian. Hence either H ∩ N = N, and then H = G, or H ∩ N = 1, and H ∈ K_{G}(N). In this case H is a maximal subgroup of G, and μ(H, G) = −1. The proposition follows.

(2.4) Corollary:

1. Let G be a solvable finite group, and

1 = N_{−1} < N_{0} < N_{1} < … < N_{k} = G

be a chief series for G. Then for 0 ≤ i ≤ k, there exist a prime number p_{i} and a positive integer n_{i} such that N_{i}/N_{i−1} ≃ (C_{p_{i}})^{n_{i}}. Denote by m_{i} the number of complements of N_{i}/N_{i−1} in the group G/N_{i−1}. Then

(2.5)  P̃(G) = ∏_{i=0}^{k}(X_{p_{i}}^{n_{i}} − m_{i})

2. Conversely, if G is a finite group, and if there exists an integer k, if there exist prime numbers p_{i} and integers m_{i}, for 0 ≤ i ≤ k such that 2.5 holds, then G is solvable.

Proof: Assertion 1) follows from an obvious induction argument. For assertion 2), suppose that G is a finite group such that 2.5 holds. Then the monomial of highest degree in P̃(G), which is equal to P (G), is the product ∏_{i=0}^{k} X_{p_{i}}^{n_{i}}. This means that all the composition factors of G are cyclic, i.e. that G is solvable.

(2.6) Remark: Corollary 2.4 can be viewed as a generalization of the Eulerian product formula obtained by Gaschütz ([7]) for the zeta function of a solvable group (see also [1]). It should also be compared with the following question, cited by Brown (Question 1 of [4]):

Question: If G is a finite group such that ζ(G, s) has an Euler product expansion with factors of the form _{\frac{1}{1−c_{i}q_{i}^{−s}}}, is G solvable?

3. The two normal subgroups formula

(3.1) Proposition: [[2] Lemme 20] Let G be a finite group. If N and M are normal subgroups of G then

Q_{G,N} = ∑_{{L≤G \atop LM=LN=G}} μ(L, G)P (L ∩ M ∩ N)Q_{G/M,(L∩N)M/M}

Proof: By definition

Q_{G,N} = ∑_{{H≤G \atop HN=G}} μ(H, G)P (H ∩ N)

As noted above

μ(H, G) = ∑_{H≤L≤G ; {LM=G \atop L∩HM=H}} μ(H, L)μ(L, G)

This gives

Q_{G,N} = ∑_{{H≤G \atop HN=G}} ∑_{H≤L≤G ; {LM=G \atop L∩HM=H}} μ(H, L)μ(L, G)P (H ∩ N)

Now the conditions

HN = G  H ≤ L  LM = G  L ∩ HM = H

are equivalent to the conditions

LM = LN = G  H ≤ L  H(L ∩ N) = L  H ≥ L ∩ M

It follows that

Q_{G,N} = ∑_{{L≤G \atop LM=LN=G}} μ(L, G) ∑_{H≤L ; {H≥L∩M \atop H(L∩N)=L}} μ(H, L)P (H ∩ N)

The inner summation is equivalent to a sum over subgroups K = H/(L ∩ M) of \overline{L} = L/(L ∩ M) such that K.J = \overline{L}, where J denotes the normal subgroup (L ∩ N)(L ∩ M)/(L ∩ M) of \overline{L}.

Moreover since HN = G

P (H ∩ N) = P (H)P (N)/P (G) = P (K)P (N)P (L ∩ M)/P (G)
= \frac{P (K ∩ J)P (\overline{L})}{P (J)}P (N)P (L ∩ M)/P (G)
= \frac{P (K ∩ J)P (L)P (N)}{P (J)P (G)}
= \frac{P (K ∩ J)P (L ∩ N)}{P (L ∩ N/L ∩ N ∩ M)}
= P (K ∩ J)P (L ∩ M ∩ N)

Thus

Q_{G,N} = ∑_{{L≤G \atop LM=LN=G}} μ(L, G)P (L ∩ M ∩ N) ∑_{{K≤\overline{L} \atop KJ=\overline{L}}} μ(K, \overline{L})P (K ∩ J)

= ∑_{{L≤G \atop LM=LN=G}} μ(L, G)P (L ∩ M ∩ N)Q_{\overline{L},J}

The formula follows, since \overline{L} ≃ G/M, and since the image of J under this isomorphism is (L ∩ N)M/M.

(3.2) Corollary: If M ≤ N, then

Q_{G,N} = Q_{G,M}Q_{G/M,N/M}

Proof: This follows from the fact that if LM = G and if M ≤ N, then (L ∩ N)M = N. This corollary has also an obvious direct proof, using 1.11.

4. I-groups

One of the methods used in [2] was to replace each variable X_{S} by |S|, and to look whether the resulting number Q_{G,N} is zero. This can be generalized by considering an ideal I of R, and looking whether Q_{G,N} is in I or not.

(4.1) Convention: In the sequel, the expression ”the group H is a quotient of the group G” means that H is isomorphic to a factor group of G.

(4.2) Definition: Let I be an ideal of R. A finite group G is called an I-group if for any non-trivial normal subgroup N of G, the polynomial Q_{G,N} belongs to I.

(4.3) Proposition: Let G be a finite group, and I be an ideal of R. If M and N are normal subgroups of G, and if G/M is an I-group, then

Q_{G,N} ≡ ∑_{L≤G ; {LM=LN=G \atop L∩N≤L∩M}} μ(L, G)P (L ∩ N) (mod. I)

In particular, if Q_{G,N} ∉ I, then G/M is a quotient of G/N. Proof: The first assertion follows from proposition 3.1, and from the definition of an I-group. Now if Q_{G,N} ∉ I, then there exists a subgroup L of G such that LM = LN = G and L ∩ N ≤ L ∩ M. Now G/M ≃ L/(L ∩ M) is a quotient of G/N ≃ L/(L ∩ N).

(4.4) Proposition: Let G be a finite group, and I be a prime ideal of R. There exists a factor group β_{I}(G) of G, characterized uniquely up to isomorphism by the following properties:

1. The group β_{I}(G) is an I-group.

2. If K is a quotient of G, and if K is an I-group, then K is a quotient of β_{I}(G).

Moreover if N is a normal subgroup of G, then the following conditions are equivalent:

a) The group β_{I}(G) is a quotient of G/N.

b) β_{I}(G/N) ≃ β_{I}(G).

c) Q_{G,N} ∉ I.

Proof: Properties 1) and 2) clearly show that the group β_{I}(G) is unique up to isomorphism, if it exists.

Let M be a normal subgroup of G, maximal subject to Q_{G,M} ∉ I. Then by Corollary 3.2, if N is a normal subgroup of G strictly containing M

Q_{G,N} = Q_{G,M}Q_{G/M,N/M} ∈ I

Since Q_{G,M} ∉ I and since I is prime, it follows that Q_{G/M,N/M} ∈ I. This holds for any non-trivial normal subgroup N/M of G/M, hence G/M is an I-group.

By Proposition 4.3, since Q_{G,M} ∉ I, it follows that any I-group which is a quotient of G is a quotient of G/M. Thus β_{I}(G) = G/M has properties 1) and 2).

Note that by construction, and by 1.12, if M is a normal subgroup of G such that G/M ≃ β_{I}(G), then Q_{G,M} ∉ I.

Now if N is a normal subgroup of G, then the group β_{I}(G/N) is an I-group, which is quotient of G/N, hence of G. Thus β_{I}(G/N) is always a quotient of β_{I}(G).

If a) holds, then β_{I}(G) is an I-group, which is a quotient of G/N. Hence β_{I}(G) is a quotient of β_{I}(G/N), and b) holds.

If b) holds, and if M/N is a normal subgroup of G/N such that

(G/N)/(M/N) ≃ β_{I}(G/N)

then G/M ≃ β_{I}(G/N) ≃ β_{I}(G). Hence as noted above Q_{G,M} ∉ I, and since Q_{G,M} is a multiple of Q_{G,N}, it follows that Q_{G,N} ∉ I. Hence c) holds.

If c) holds, then by proposition 4.3 the group β_{I}(G) is a quotient of G/N, and a) holds.

There are generally several normal subgroups M of G such that G/M ≃ β_{I}(G). They are related as follows:

(4.5) Proposition: Let G be a finite group, and I be a prime ideal of R. Let M and N be normal subgroups of G such that

G/M ≃ G/N ≃ β_{I}(G)

Let ]M∩N, M[^{G} denote the poset of normal subgroups of G which contain strictly M ∩ N and are strictly contained in M, and let n_{G}(M, N) denote its reduced Euler-Poincaré characteristic (with the convention n_{G}(M, N) = 1 if M = N). Then:

1. There exists an automorphism θ of the group G/M ∩ N such that

θ(M/M ∩ N) = N/M ∩ N

Hence the posets ]M ∩ N, M[^{G} and ]M ∩ N, N[^{G} are isomorphic, and in particular n_{G}(M, N) = n_{G}(N, M).

2. The group M/M ∩ N ≃ N/M ∩ N is isomorphic to a direct product of simple groups.

3. Q_{G,N} ≡ Q_{G,M∩N}n_{G}(M, N)|K_{G/M∩N}(M/M ∩ N, N/M ∩ N)| (mod. I).

Proof: Since Q_{G,N} = Q_{G,M∩N}Q_{G/M∩N,N/M∩N} by Corollary 3.2, replacing G by G/M ∩ N shows that it suffices to consider the case M ∩ N = 1. In this case the groups M and N centralize each other.

By proposition 4.3, since moreover G/M ≃ G/N implies L ∩ M = L ∩ N whenever LM = LN = G and L ∩ N ≤ L ∩ M

Q_{G,N} ≡ ∑_{L≤G ; {LM=LN=G \atop L∩M=L∩N=1}} μ(L, G) (mod. I)

In other words

Q_{G,N} ≡ ∑_{L∈K_{G}(M,N)} μ(L, G) (mod. I)

Since Q_{G,N} ∉ I, this shows in particular that K_{G}(M, N) ≠ ∅.

Now fix L ∈ K_{G}(M, N), and define a map ϕ from M to N by ϕ(m) = n if mn ∈ L. This is well defined since L ∈ K_{G}(M, N). Moreover if m and m′ are in M, and if n = ϕ(m) and n′ = ϕ(m′), then

mnm′n′ = mm′nn′ ∈ L

since M and N commute. This shows that ϕ(mm′) = nn′, hence ϕ is a group homomorphism from M to N. By symmetry, the map ψ from N to M defined by ψ(n) = m if nm ∈ L is also a group homomorphism, which is clearly the inverse of ϕ, since M and N commute. Moreover the maps ϕ and ψ are clearly L equivariant.

Now define a map θ : G → G by

θ(ml) = ϕ(m)l ∀m ∈ M, ∀l ∈ L

Then for m, m′ in M and l, l′ in L

θ(mlm′l′) = θ(m.^{l}m′.ll′) = ϕ(m).^{l}ϕ(m′).ll′ = ϕ(m)lϕ(m′)l′ = θ(ml)θ(m′l′)

This shows that θ is a group homomorphism, which is clearly an automorphism of G.

By construction θ(M) = N, thus θ induces an isomorphism of posets from ]1, M[^{G} to ]1, N[^{G}, and n_{G}(M, N) = n_{G}(N, M). This shows assertion 1).

Now the maps

X ∈ [1, N]^{L} ↦ LX ∈ [L, G] Y ∈ [L, G] ↦ Y ∩ N ∈ [1, N]^{L}

are inverse isomorphisms of posets. Moreover [1, N]^{L} = [1, N]^{LM} = [1, N]^{G} since M and N commute. Thus μ(L, G) is equal to the reduced Euler-Poincaré characteristic of the poset ]1, N[^{G}= [1, N]^{G} − {1, N}, and this does not depend on the choice of L in K_{G}(M, N). Thus

Q_{G,N} ≡ χ̃(]1, N[^{G})|K_{G}(M, N)| (mod. I)

and this proves assertion 3).

Since Q_{G,N} ∉ I, it follows from Crapo complementation formula that every normal subgroup of G contained in N has a complement in N, invariant by G. This can only happen if N is a direct product of simple groups, and this completes the proof of the proposition.

(4.6) Example: The case of b-groups discussed in [2] corresponds to the ideal I generated by all X_{S} − |S|, for S ∈ S. In this case if G is a p-group, then β_{I}(G) is trivial if G is cyclic, and isomorphic to (C_{p})^{2} otherwise. This follows from Proposition 14 of [2], but I will give another more general proof here in Corollary 7.4.

If Φ(G) is the Frattini subgroup of G, and if the order of G/Φ(G) is at least p^{3}, then there are several normal subgroups N of G such that G/N ≃ (C_{p})^{2}, but all those subgroups contain Φ(G), and are conjugate by the automorphism group of G/Φ(G).

5. Example: the ideal of valuation

(5.1) Notation: I denote by V the ideal of R generated by all X_{S}, for S ∈ S.

In other words, the ideal V is the ideal of polynomial with constant term equal to zero. Clearly the quotient ring R/V is isomorphic to ℤ, thus V is a prime ideal.

By definition, if N is a normal subgroup of the finite group G, then

Q_{G,N} = ∑_{{L≤G \atop LN=G}} μ(L, G)P (L ∩ N)

In this sum, the monomial P (L ∩ N) is in V, unless L ∩ N = 1. It follows that

Q_{G,N} ≡ ∑_{L∈K_{G}(N)} μ(L, G) (mod. V)

If N is a minimal normal subgroup of G, and if N is abelian, then by Proposition 2.3

Q_{G,N} ≡ −|K_{G}(N)| (mod. V)

In particular Q_{G,N} ∈ V if and only if K_{G}(N) = ∅. This is equivalent to requiring N to be contained in each maximal subgroup L of G, i.e. N to be contained in Φ(G).

(5.2) Notation: If G is a finite group, I denote by M(G) the subgroup generated by all minimal normal subgroups of G.

(5.3) Proposition: Let G be a solvable finite group. Then G is a V-group if and only if M(G) ≤ Φ(G). Proof: This follows clearly from the previous discussion.

(5.4) Remark: If G is an arbitrary finite group, and if M(G) ≤ Φ(G), then G is a V-group, but the converse is false in general: for example, a simple group S is a V-group if and only if μ(1, S) = 0. This happens for instance if S is the simple group of order 168. Still M(S) = S is not contained in Φ(S) = 1.

This remark and Proposition 4.4 suggest the following variation:

(5.5) Notation: I denote by A the ideal of R generated by all X_{p}, for p prime.

Clearly R/A is isomorphic to the ring ℤ[(X_{S})_{S∈S^{0}}] of polynomials on the set S^{0} of isomorphism classes of non−abelian simple groups. Hence A is a prime ideal of R.

(5.6) Proposition: Let G be a finite group. Then G is an A-group if and only if M(G) ≤ Φ(G). Proof: If M(G) ≤ Φ(G), since Φ(G) is nilpotent, all minimal normal subgroups of G are abelian, and have no complement in G. If N is a minimal normal subgroup of G, isomorphic to (C_{p})^{n}, for a prime number p and a positive integer n, then by Proposition 2.3

Q_{G,N} = X_{p}^{n} − |K_{G}(N)| = X_{p}^{n} ∈ A

Hence G is an A-group.

Conversely, if G is an A-group and N is a minimal normal abelian subgroup of G, isomorphic to (C_{p})^{n}, then

Q_{G,N} = X_{p}^{n} − |K_{G}(N)| ≡ −|K_{G}(N)| (mod. A)

Thus Q_{G,N} ∈ A if and only if K_{G}(N) = ∅, or equivalently since N is abelian, if N ≤ Φ(G).

Now suppose N is a non-abelian minimal normal subgroup of G. Then N ≃ S^{n}, for some non-abelian simple group S and some positive integer n. But

Q_{G,N} = ∑_{{H≤G \atop HN=G}} μ(H, G)P (H ∩ N)

and the term of highest degree in this expression is obtained for H = G, and it is equal to P (N) = X_{S}^{n}. But the ideal A consists of linear combinations of monomials ∏_{S∈S} X_{S}^{α_{S}} for which the exponent α_{S} is positive at least for one abelian simple group S (i.e. the monomials in which some variable X_{p} for p prime appears). This shows that Q_{G,N} cannot belong to A if N is non-abelian.

Hence all the minimal normal subgroups of G are abelian, and it follows that M(G) ≤ Φ(G), as was to be shown.

(5.7) Corollary: Let G be a finite group. Then there exists a factor group H of G, characterized uniquely up to isomorphism by the following properties:

1. M(H) ≤ Φ(H).

2. If K = G/N is a quotient of G such that M(K) ≤ Φ(K), and if M \underline{◁} G is such that G/M ≃ H, then there exists a subgroup L of G with

LM = LN = G L ∩ M ≤ L ∩ N

and in particular K is a quotient of H.

Proof: Of course H = β_{A}(G).

(5.8) Example: When G is a p-group, for some prime p, it is possible to describe explicitly the quotient β_{V}(G) = β_{A}(G). Indeed, this quotient is obtained by considering a normal subgroup N of G, maximal subject to Q_{G,N} ∉ V. With the notation of Proposition 2.4, if

1 = N_{−1} < N_{0} < … < N_{k} = G

is a chief series for G, and if N = N_{l}, for −1 ≤ l ≤ k, then

Q_{G,N} = P̃(G)/P̃(G/N) = ∏_{i=0}^{l}(X_{p}^{n_{i}} − m_{i}) ≡ (−1)^{l+1} ∏_{i=0}^{l} m_{i} (mod. V)

Thus N is a maximal normal subgroup such that all the m_{i}’s are non-zero. In particular, every minimal normal subgroup of G contained in N must have a complement in G, hence in N. It follows that N is elementary abelian, and G-semi-simple. Hence N is central.

Moreover if L_{i}/N_{i−1} ∈ K_{G/N_{i−1}}(N_{i}/N_{i−1}), for 0 ≤ i ≤ l, then it is easy to see that L_{0} ∩ L_{1} ∩ … ∩ L_{l} is a complement of N_{l} = N in G. Thus G can be split as a direct product

G = N × L

Moreover L is a V-group, thus M(L) ≤ Φ(L). Since L is a p-group, all the minimal normal subgroups of L are central of order p. It follows that no central subgroup of order p can have a complement in L. Hence N is a maximal elementary abelian central subgroup of G having a complement in G, and β_{V}(G) is the quotient of G by its “largest elementary abelian direct summand”.

In this case, one can say exactly how many subgroups N such that G/N ≃ β_{V}(G) there are: indeed with the previous notations, the group M(L) is equal to the subgroup Ω_{1}(Z(L)) generated by the central elements of order p of L. But

Φ(G) = 1 × Φ(L)

and

(5.9)  Ω_{1}(Z(G)) = N × Ω_{1}(Z(L))

Taking the intersection of those two equations gives

Φ(G) ∩ Ω_{1}(Z(G)) = 1 × Ω_{1}(Z(L))

It follows from 5.9 that N must be a complement of Φ(G) ∩ Ω_{1}(Z(G)) in Ω_{1}(Z(G)).

Conversely, if N is such a complement, there exists a subgroup K of G, containing Φ(G), such that K/Φ(G) is a complement of NΦ(G)/Φ(G) in G/Φ(G), since G/Φ(G) is elementary abelian. In other words

KN = G K ∩ NΦ(G) = Φ(G)

It follows that K ∩ N ≤ N ∩ Φ(G) = 1, thus K is a complement of N. Clearly now N is G-semi-simple, and it follows that Q_{G,N} ∉ V.

Thus the subgroups N such that G/N ≃ β_{I}(G) are exactly the complements of Φ(G) ∩ Ω_{1}(Z(G)) in Ω_{1}(Z(G)).

6. Direct products

(6.1) Notation: Let G and H be finite groups. Denote by p_{1} and p_{2} the projections from G × H to G and H respectively. If L is a subgroup of G × H, set

p_{1}(L) = {g ∈ G | ∃h ∈ H, (g, h) ∈ L} k_{1}(L) = {g ∈ G | (g, 1) ∈ L}

p_{2}(L) = {h ∈ H | ∃g ∈ G, (g, h) ∈ L} k_{2}(L) = {h ∈ H | (1, h) ∈ L}

Then k_{i}(L) \underline{◁} p_{i}(L) for i = 1, 2, and the quotients q(L) = L/(k_{1}(L) × k_{2}(L)), p_{1}(L)/k_{1}(L) and p_{2}(L)/k_{2}(L) are canonically isomorphic.

(6.2) Proposition: Let G and H be finite groups. Then

P̃(G)P̃(H) = ∑_{L≤G×H ; {p_{1}(L)=G \atop p_{2}(L)=H}} P̃(L)

Proof: This is a consequence of the definition of the polynomials P̃ by Möbius inversion. Indeed for any finite groups G and H

(6.3)  P (G × H) = ∑_{L≤G×H} P̃(L)

Now setting

σ(G, H) = ∑_{L≤G×H ; {p_{1}(L)=G \atop p_{2}(L)=H}} P̃(L)

the right hand side of equation 6.3 can be written as

∑_{{A≤G \atop B≤H}} σ(A, B)

∑_{{C≤G \atop D≤H}} μ(C, G)μ(D, H)P (C × D) = ∑_{{C≤G \atop D≤H}} ∑_{{A≤C \atop B≤D}} μ(C, G)μ(D, H)σ(A, B)

= ∑_{{A≤G \atop B≤H}} ( ∑_{{A≤C≤G \atop B≤D≤H}} μ(C, G)μ(D, H))σ(A, B)

= σ(G, H)

The proposition follows, since the left hand side is equal to P̃(G)P̃(H), because P (C × D) = P (C)P (D) for any C ≤ G and D ≤ H.

(6.4) Corollary: If G and H have no non-trivial isomorphic factor group, then

P̃(G × H) = P̃(G)P̃(H)

Proof: Indeed in this case, the only subgroup L of G × H such that p_{1}(L) = G and p_{2}(L) = H is G × H itself, since q(L) is a quotient of both G and H, hence it is trivial.

(6.5) Proposition: [[2] Lemme 22] Let G and H be finite groups. Then

(6.6)  P̃(G × H) = P̃(G)P̃(H) ∑_{L≤G×H ; {p_{1}(L)=G \atop p_{2}(L)=H}} \frac{χ̃(]1, q(L)[^{q(L)})}{P̃(q(L))}

Proof: Denote by K the group G × H, and by M and N the normal subgroups G × 1 and 1 × H of K. Then by proposition 3.1 since M ∩ N = 1

Q_{K,N} = ∑_{{L≤K \atop LM=LN=K}} μ(L, K)Q_{K/M,(L∩N)M/M}

The condition LM = LN = K is equivalent to p_{1}(L) = G and p_{2}(L) = H. In this case moreover, the maps

X ∈]L, K[↦ k_{1}(X)/k_{1}(L) ∈]1, G/k_{1}(L)[^{G}

Y/k_{1}(L) ∈]1, G/k_{1}(L)[^{G}↦ {(g, h) ∈ G × H | ∃a ∈ G, (a, h) ∈ L, ga^{−1} ∈ Y }

are mutual inverse isomorphisms of posets. Since G/k_{1}(L) ≃ q(L), it follows that μ(L, G) = χ̃(]1, q(L)[^{q(L)}).

Moreover

Q_{K/M,(L∩N)M/M} = Q_{L/L∩M,(L∩N)(L∩M)/L∩M} = \frac{P̃(L/L ∩ M)}{P̃(L/(L ∩ N)(L ∩ M))}

But L/L ∩ M ≃ K/M ≃ H and

L/(L ∩ N)(L ∩ M) = L/k_{1}(L) × k_{2}(L) = q(L)

It follows that

Q_{K,N} = P̃(H) ∑_{L≤G×H ; {p_{1}(L)=G \atop p_{2}(L)=H}} \frac{χ̃(]1, q(L)[^{q(L)})}{P̃(q(L))}

The proposition follows, since Q_{K,N} = P̃(K)/P̃(K/N) = P̃(K)/P̃(G).

(6.7) Corollary: Let G and H be finite groups, and denote by G_{s}, H_{s} and (G × H)_{s} ≃ G_{s} × H_{s} the respective largest semi-simple quotients of G, H and G × H. Then

\frac{P̃(G × H)}{P̃((G × H)_{s})} = \frac{P̃(G)}{P̃(G_{s})} \frac{P̃(H)}{P̃(H_{s})}

Proof: The only non-zero terms in the formula 6.6 correspond to subgroups L of G × H such that q(L) is semi-simple. Moreover since p_{1}(L) = G and p_{2}(L) = H, the group q(L) is a quotient of G and H. It follows that if M and N are normal subgroups of G and H respectively, such that G/M ≃ G_{s} and H/N ≃ H_{s}, then L ≥ M × N. Then the group L′ = L/M × N is a subgroup of G_{s} × H_{s} ≃ (G × H)_{s}, and

q(L) = L/k_{1}(L) × k_{2}(L) ≃ (L/M × N)/(k_{1}(L) × k_{2}(L)/M × N)
≃ L′/(k_{1}(L)/M) × (k_{2}(L)/N) = q(L′)

The corollary follows, since the correspondence L ↦ L′ from

{K ≤ G × H | p_{1}(K) = G, p_{2}(K) = H, K ≥ M × N}

to the set

{K′ ≤ G_{s} × H_{s} | p_{1}(K′) = G_{s}, p_{2}(K′) = H_{s}}

is clearly one to one.

Corollary 6.7 gives a way to compute P̃(G × H) knowing P̃(G), P̃(H), and the groups G_{s} and H_{s}. The following notation will be convenient:

(6.8) Notation: If G and H are finite groups, I denote by s(G, H) the number of surjective group homomorphisms from G to H.

If n, m ∈ ℕ, I set

F (G, n) = ∏_{i=0}^{n−1} (P̃(G) − s(G^{i}, G))

B(G, m, n) = \frac{F (G, m + n)}{F (G, m)F (G, n)}

Here the letter F stands for “factorial”, and the letter B for “binomial”. Note that B(G, m, n) is an element of the field of fractions of R.

(6.9) Proposition:

1. If S is a finite simple group and n ∈ ℕ, then P̃(S^{n}) = F (S, n). Moreover s(S^{n}, S) is equal to p^{n} −1 if S ≃ C_{p}, and to n|Aut(S)| if S is non-abelian.

2. Let G and H be finite groups. If S ∈ S, denote by a_{S} = ν_{S}(G_{s}) and b_{S} = ν_{S}(H_{s}) the multiplicity of S as a factor of G_{s} and H_{s} respectively. Then

P̃(G × H) = P̃(G)P̃(H) ∏_{S∈S} B(S, a_{S}, b_{S})

Proof: The first formula follows from an easy induction argument, using 6.2 or 6.6. Using this, the second formula follows from 6.4 and 6.7.

(6.10) Lemma: Let I be a prime ideal of R. If G and H are I-groups, and if they have no non-trivial isomorphic factor group, then G × H is an I-group. Proof: Since G and H are quotients of G×H, it follows that they are quotients of β_{I}(G × H). Hence there is a group homomorphism

ϕ : β_{I}(G × H) → G × H

such that p_{1} ∘ ϕ and p_{2} ∘ ϕ are surjective. Let K denote the image of ϕ. Then p_{1}(K) = G and p_{2}(K) = H. Thus q(K) is a quotient of both G and H, hence it is trivial. It follows that K = G × H, hence β_{I}(G × H) ≃ G × H. Thus G × H is an I-group.

(6.11) Proposition: Let I be a prime ideal of R, and G and H be finite groups having no non-trivial isomorphic factor group.

1. If M \underline{◁} G and N \underline{◁} H, then

Q_{G×H,M×N} = Q_{G,M}Q_{H,N}

2. Moreover

β_{I}(G × H) ≃ β_{I}(G) × β_{I}(H)

3. The group G × H is an I-group if and only if G and H are I-groups.

Proof: Let M be a normal subgroup of G and N be a normal subgroup of H. Then by Corollary 6.4, since G/M and H/N have no non-trivial isomorphic factor group

P̃(G × H) = P̃(G)P̃(H)
P̃(G × H/M × N) = P̃(G/M × H/N)
= P̃(G/M)P̃(H/N)

Taking the quotient of those equations gives assertion 1)

Q_{G×H,M×N} = Q_{G,M}Q_{H,N}

Clearly β_{I}(G) and β_{I}(H) have no common non-trivial factor group. Hence by Lemma 6.10 β_{I}(G) × β_{I}(H) is an I-group, and it is a quotient of G × H. Hence it is a quotient of β_{I}(G × H).

Conversely if M \underline{◁} G is such that G/M ≃ β_{I}(G), and N \underline{◁} H is such that H/N ≃ β_{I}(H), then Q_{G,M} ∉ I and Q_{H,N} ∉ I, thus Q_{G×H,M×N} ∉ I by assertion 1) since I is prime. By Proposition 4.3 the group β_{I}(G × H) is a quotient of

(G × H)/(M × N) ≃ β_{I}(G) × β_{I}(H)

Assertion 2) follows.

Finally G×H is an I-group if and only if G×H = β_{I}(G×H). By assertion 2), this is equivalent to G = β_{I}(G) and H = β_{I}(H), which proves assertion 3).

7. Nilpotent I-groups

It is possible to describe all the nilpotent I-groups, when I is a prime ideal of R. Since any nilpotent group is isomorphic to the direct product of its Sylow p-subgroups, for prime numbers p, it follows from proposition 6.11 that a nilpotent group is an I-group if and only if all its Sylow subgroups are Igroups. Thus in order to describe the nilpotent I-groups, it suffices to describe the p-groups which are also I-groups.

(7.1) Proposition: Let I be a prime ideal of R, let p be a prime number, and let G be a finite p-group. Then:

1. If G is elementary abelian of order p^{n}, then G is an I-group if and only if n = 0 or n ≥ 1 and X_{p} − p^{n−1} ∈ I.

2. If G is not elementary abelian, then G is an I-group if and only if X_{p} ∈ I and one of the following holds:

(a) p ∈ I.

(b) Ω_{1}(Z(G)) ≤ Φ(G), or equivalently, the group G cannot be written G ≃ C_{p} × H, for some finite group H.

Proof: Suppose G is non-trivial, and let N be a minimal normal subgroup of G. Then N is central of order p, and

Q_{G,N} = X_{p} − |K_{G}(N)|

since any proper subgroup L of G such that LN = G is a complement of G, and L is a maximal subgroup of G in this case.

Suppose that G is elementary abelian of order p^{n}. If n = 0, then G is trivial and G is an I-group. If n ≥ 1, and if N is a subgroup of G of order p, then

Q_{G,N} = X_{p} − p^{n−1}

since there are p^{n−1} complements of N in G. This proves assertion 1).

If G is not elementary abelian, then there exists a central subgroup N of G of order p contained in Φ(G). If N is such a subgroup, then K_{G}(N) = ∅, and

(7.2)  Q_{G,N} = X_{p}

Now if M is a central subgroup of G of order p, and M ≰ Φ(G), then

(7.3)  Q_{G,N} = X_{p} − |K_{G}(M)|

Moreover in this case K_{G}(M) ≠ ∅, thus there exists a group H such that G ≃ M × H, and

|K_{G}(M)| = |Hom(H, M)|

is a power of p, and different from 1 since H is non trivial (if H = 1, then G has order p and G is elementary abelian).

Hence if G is an I-group, then X_{p} ∈ I by 7.2. Now if Ω_{1}(Z(G)) ≰ Φ(G), it follows from 7.3 that |K_{G}(M)| ∈ I. This is a power of p, different from 1, thus p ∈ I since I is prime.

Conversely, if X_{p} ∈ I and p ∈ I, then clearly Q_{G,N} ∈ I by 7.2 and 7.3 for any central subgroup N of order p of G. Then G is an I-group.

And if X_{p} ∈ I and Ω_{1}(Z(G)) ≤ Φ(G), then any central subgroup N of order p of G is contained in Φ(G), hence Q_{G,N} ∈ I by 7.2. Thus G is an I-group, and this completes the proof of the proposition.

Proposition 7.1 can be reformulated as follows:

(7.4) Corollary: Let I be a prime ideal of R, let p be a prime number, and let G be a finite p-group. Then:

• If X_{p} ∈ I and p ∈ I, then G is an I-group if and only if G ≄ C_{p}.

• If X_{p} ∈ I and p ∉ I, then G is an I-group if and only if Ω_{1}(Z(G))≤Φ(G).

• If X_{p} ∉ I, then let h, k ∈ ℕ ∪ {∞} defined by

h = Inf {l ∈ ℕ | X_{p} − p^{l} ∈ I}
k = Inf {l ∈ ℕ − {0} | 1 − p^{l} ∈ I}

Then G is an I-group if and only if G is elementary abelian of order p^{n} with n = 0, or n ≡ h + 1 (k) if h ≠ ∞ and k ≠ ∞, or n = h + 1 if h ≠ ∞ and k = ∞.

Proof: Indeed, if X_{p} ∈ I and p ∈ I, then by Proposition 7.1, the only case where G is not an I-group is the case G ≃ C_{p}.

If X_{p} ∈ I and if a non-trivial elementary abelian p-group of order p^{n} is an I-group, then p^{n−1} ∈ I, hence 1 ∈ I or p ∈ I. Thus if p ∉ I, then no non-trivial elementary abelian p-group can be an I-group, and G is an I-group if and only if Ω_{1}(Z(G)) ≤ Φ(G).

Finally if X_{p} ∉ I, then G is an I-group if and only if G is elementary abelian of order p^{n} with n = 0 or n ≥ 1 and X_{p} − p^{n−1} ∈ I. This cannot happen if h = ∞, and if h ∈ ℕ, this is equivalent to

p^{h} − p^{n−1} = p^{h}(1 − p^{n−h−1}) ∈ I

But p^{h} ∉ I, since otherwise X_{p} = (X_{p} − p^{h}) + p^{h} ∈ I, and this is equivalent to 1 − p^{n−h−1} ∈ I, or n ≡ h + 1 (k), which has to be viewed as an equality if k = ∞.

(7.5) Remark: The last case of Corollary 7.4 should be compared with Theorem 8.2 of [3], which deals with “b-groups in characteristic q”, i.e. with the case where I is generated by a prime number q ≠ p (or q = 0), and by X_{S} − |S|, for all simple groups S.

References

[1] S. Bouc. Homologie de certains ensembles ordonnés. C. R. Acad. Sc. Paris, t.299(2):9–12, 1984. Série I.

[2] S. Bouc. Foncteurs d’ensembles munis d’une double action. J. of Algebra, 183(0238):664–736, 1996.

[3] S. Bouc and J. Thévenaz. The group of endo-permutation modules. Invent. Math., 1998. To appear.

[4] K. Brown. The coset poset and the probabilistic zeta function of a finite group. Preprint, 1999.

[5] K. Brown and J. Thévenaz. A generalization of Sylow’s third theorem. Journal of Algebra, 115:414–430, 1988.

[6] H. Crapo. The Möbius function of lattices. J. Comb. Theory., 1:126–131, 1966.

[7] W. Gaschütz. Die Eulersche Funktion endlicher auflösbarer Gruppen. Illinois J. Math., 3:469–476, 1959.

[8] P. Hall. The Eulerian functions of a group. Quart. J. Math., 7:134–151, 1936.