Bounding the size of permutation groups and complex linear groups of odd order

Geoffrey R. Robinson, Institute of Mathematics, University of Aberdeen

August 2010

Introduction: Numerous results exist in the literature bounding the orders of finite odd order subgroups of permutation groups and complex linear groups (see, for example, (Gow, [2]), where such bounds were used to bound the number of characters in a p-block of a finite group). Our aim here is to point out that these bounds can be substantially improved if we place restrictions on the smallest prime divisor of the group order.

Given an odd prime p, we define constants α(p) and β(p) as follows:

i) α(3) = \sqrt{3}, α(5) = 5^{\frac{1}{4}} and α(p) = [p(2p + 1)]^{\frac{1}{2p}} for p ≥ 7.

ii) β(p) = [p(2p + 1)^{2}]^{\frac{1}{2p}} if p ≡ 2 (mod 3) and β(p) = [p(2p − 1)^{2}]^{\frac{1}{2p−2}} otherwise.

We will prove:

Theorem: a) Let G be a subgroup of odd order of the symmetric group S_{n}. Then |G| ≤ α(p)^{n−1}, where p is the smallest prime divisor of |G|.

b) Let G be a finite subgroup of odd order of GL(n, ℂ) and let p be the smallest prime divisor of |G|. Then:

i) [G : F (G)] ≤ α(p)^{n−1} and ii) G has an Abelian normal subgroup A with [G : A] ≤ β(p)^{n−1}.

Remark: Notice that β(3) = 75^{\frac{1}{4}} < 3, α(5) < 1.5 and β(5) = 605^{\frac{1}{10}} < 1.9. Also, α(p) = max\{p^{\frac{1}{p−1}} , [p(2p + 1)]^{\frac{1}{2p}} \} for all odd primes p. For when p = 3, we have \sqrt{3} > 21^{\frac{1}{6}} , when p = 5, we have 5^{\frac{1}{4}} > 55^{\frac{1}{10}} , and when p ≥ 7, we have 2^{p−1} > p^{2}, so that [p(2p + 1)]^{p−1} > p^{2p}. We will see later that α(p) < p^{\frac{3}{2(p−1)}} and β(p) < p^{\frac{5}{2(p−1)}} for all odd primes p.

We also remark that when n = p^{k} for some prime p and positive integer k, a Sylow p-subgroup of S_{n} has order p^{\frac{n−1}{p−1}} .

The theorem above has some consequences for the structure of general finite (solvable) groups of odd order.

For a finite group G, let F (G) denote the Fitting subgroup of G and Φ(G) denote the Frattini subgroup of G. Let F_{k}(G) denote the k-th Fitting subgroup of G, which is defined inductively via: F_{1}(G) = F (G), F_{k}(G)/F_{k−1}(G) = F (G/F_{k−1}(G)).

Corollary: Let G be a finite solvable group of odd order and let p be the smallest prime divisor of |G|. Then: i)

[G : F_{2}(G)] < [F (G) : Φ(G)]^{\frac{1}{p−1}}

if p ∈ {3, 5} and

[G : F_{2}(G)] < [F (G) : Φ(G)]^{( \frac{1}{p} + \frac{log 2}{2p log p} + \frac{1}{4p^{2} log p})}

otherwise.

ii) [G : Φ(G)] < ^{\frac{[F (G):Φ(G)]^{\frac{2p−1}{p−1}}}{2}} if p ∈ \{3, 5\} and

[G : Φ(G)] < ^{\frac{[F (G):Φ(G)]^{\frac{4p−1}{2p−2}}}{2}} otherwise.

1 Preparatory results

General principles: When attempting to bound the size of subgroups of S_{n}, it is common to seek a bound of the form |G| < c^{n−1} for some constant c if G belongs to a chosen collection C of finite groups (we remark in passing that there is no such constant c when C contains all finite groups, since n! grows faster that c^{n} for any constant c). However, such bounds occur frequently: if G is a nilpotent subgroup of S_{n}, then it is well known that |G| ≤ 2^{n−1}, and equality is attained whenever n = 2^{k}, taking G as a Sylow 2-subgroup of S_{n}. The constant c = 24^{\frac{1}{3}} is known to be minimal choice of c when C is the collection of all finite solvable groups.

Such bounds are amenable to inductive proofs: if C is under taking normal subgroups and homomorphic images, and we know that the bound is known to hold for groups in C of order less than |G|. Then we may reduce to the primitive case as follows: If G is intransitive, but has r orbits of respective sizes n_{1}, n_{2}, … n_{r}, then G embeds in a direct product G_{1} × … × G_{r}, where G_{i} is a homomorphic image of G and is also a transitive subgroup of S_{n_{i}}, so |G| ≤ ∏_{i=1}^{r} |G_{i}| ≤ ∏_{i=1}^{r} c^{n_{i}−1}. Hence we may assume that G is transitive. If G is transitive, but imprimitive, let H be a point stabilizer, and let K be a subgroup with H max K < G, say [G : K] = a and b = [K : H], so that n = ab. Let N be the intersection of the G-conjugates of K. Then N ⊲ G and N is not contained in H, so that K = HN. By induction, we may suppose [G : N] ≤ c^{a−1}. Now [N : N ∩ H] = b so N has an orbit of size b on the cosets of H. Thus all orbits of N on the cosets of H have size b and |N| ≤ (c^{b−1})^{a}. Then |G| ≤ c^{ab−1}. So it suffices to consider the case that G is primitive.

If, in addition, C is closed under taking arbitrary subgroups, we can reduce the proof that if G ∈ C is a subgroup of GL(n, ℂ) for some n, then G has an Abelian normal subgroup A of index at most d^{n−1}, where d is a constant greater than or equal to c above, to the case that is G is a primitive linear group. If the given representation of G is induced from a primitive representation of degree b of a proper subgroup H of index m, then [G : N] ≤ c^{m−1}, where N = ∩_{g∈G}H^{g}. Now by Clifford’s theorem, N embeds in a direct product N_{1} × N_{2} × … × N_{m}, where N_{i} is a subgroup of GL(b, ℂ) and is a homomorphic image of N. For at least one value of i, there is an irreducible overgroup H_{i} of N_{i} which is the image of H under a primitive representation τ_{i} of degree b. Since H_{i} acts primitively, all its Abelian normal subgroups are central. Hence [H_{i} : Z(H_{i})] ≤ d^{b−1}. Thus [N_{i} : Z(N_{i})] ≤ d^{b−1} and [N : Z(N)] ≤ ∏_{i=1}^{m}[N_{i} : Z(N_{i})] ≤ d^{m(b−1)}. Hence [G : Z(N)] ≤ d^{mb−m}c^{m−1} ≤ d^{mb−1}. It is also possible in the linear case to work with a bound of the form d^{rn−s} for constants r and s when d^{s} ≤ c, such as is done in (Dornhoff, [1]) for finite solvable complex linear groups, but we restrict our discussion to the case r = s = 1 for ease of exposition.

A similar argument works to reduce the proof that [G : F (G)] ≤ t^{n−1} for a constant t ≥ c when G ∈ C is a finite subgroup of GL(n, ℂ). We do not repeat the details, save to remark that it is not necessary to assume that C is closed under taking arbitrary subgroups. Closure under taking normal subgroups (and homomorphic images) will do: if we have bounded the [N_{i} : F (N_{i})] in this way, for the N_{i} as above, then we bound [G : F (N)], so also bound [G : F (G)]. In the Abelian case, we needed the overgroup H_{i} of N_{i} to conclude that the particular choice of Abelian normal subgroup Z(N_{i}) could be used. Had we used an arbitrary Abelian normal subgroup A_{i} of N_{i}, the resulting Abelian normal subgroup of N need not have been normal in G.

In this paper, we wish to vary this type of argument slightly. We wish to allow the constants c, d and t to be replaced by functions which may depend on the group G ∈ C. Examination of the arguments shows that this is permissible if the following conditions are satisfied (when C is closed under taking subgroups and homomorphic images).

i) max{c(X), t(X)} ≤ d(X) for all X ∈ C. ii) c(X) ≥ c(Y) whenever X ∈ C and Y is a section of X. iii) d(X) ≥ d(Y) whenever X ∈ C and Y is a section of X. iv) t(X) ≥ t(Y) whenever X ∈ C and Y is a section of X.

The arguments above show that when these conditions are satisfied, if we wish to prove that whenever G ∈ C is a subgroup of S_{n}, we have |G| ≤ c(G)^{n−1}, it suffices to consider the case that G is primitive. Similarly, if we have established the bound for subgroups of S_{n} in C, and we wish to prove that [G : F (G)] ≤ t(G)^{n−1} and that there is an Abelian normal subgroup A of G with [G : A] ≤ d(G)^{n−1}, when G ∈ C is a finite subgroup of GL(n, ℂ), then it suffices to consider the case that G is (irreducible and) primitive as a linear group.

We will use the functions α(p) in the role of both d(G) and t(G), and β(p) in the role of c(G), where p is the smallest prime divisor of |G|. The collection C we are considering is the collection of finite (solvable) groups of odd order. If Y is a section of X ∈ C, then the smallest prime divisor of |Y | is certainly greater than or equal to the smallest prime divisor of |X|. We need to show,then,that functions α and β decrease as primes increase, and that we have α(p) ≤ β(p) for all odd primes p.

We need some elementary calculus. We define functions f, g, h, k on (e, ∞), by f(x) = x^{\frac{1}{x−1}} , g(x) = [x(2x + 1)]^{\frac{1}{2x}} , h(x) = [x(2x − 1)^{2}]^{\frac{1}{2x−2}} and k(x) = [x(2x + 1)^{2}]^{\frac{1}{2x}} .

Lemma 1: The real valued functions f, g, h, k are strictly decreasing on their domains. Furthermore, h(x) ≥ k(x) ≥ max{f(x), g(x)} for all x ≥ 3.

Proof: The monotonicity is an exercise in elementary calculus. We next notice that h(x) ≥ h(x + 1) > k(x) for x > e, and the last last inequality is clear since x^{\frac{3}{2x}} ≥ x^{\frac{1}{x−1}} for x ≥ 3.

Lemma 2: We have k(x) > h(x + 2) for x ≥ 4.

Proof: We have k(x) > (4x^{3})^{\frac{1}{2x}} and

h(x + 2) = [4x^{3}]^{\frac{1}{2x+2}} [(1 + \frac{2}{x})(1 + \frac{3}{2x})^{2}]^{\frac{1}{2x+2}} .

Since (1 + \frac{a}{x})^{x} ≤ e^{a} for a, x > 0, we have

h(x + 2) ≤ [4x^{3}]^{\frac{1}{2x+2}} e^{\frac{5}{2(x^{2}+x)}} .

Now \frac{k(x)}{h(x+2)} ≥ (4x^{3}e^{−5})^{\frac{1}{2(x^{2}+x)}} . When x ≥ 4, we have 4x^{3} ≥ 256 > 243 = 3^{5} > e^{5}, so the result follows.

Corollary 3: Let p and q be odd primes with p < q.

i) α(p) > α(q) and β(p) > β(q). ii) If \frac{q+1}{2} is odd and greater than p, then we have β(p)^{q−1} ≥ q^{2}(\frac{q+1}{2}). iii) α(q) ≤ q^{\frac{1}{4}} for all primes q > 3.

Proof: The first claim of i) follows because of the monotonicity of f and g. For the second, if β(p) = h(p), then we have β(p) = h(p) > h(q) ≥ k(q). If β(p) = k(p), then q ≥ p + 2 and p ≡ 2 (mod 3), so that p ≥ 5. Then β(p) = k(p) > h(p + 2) ≥ h(q) > k(q).

For ii), note that h(\frac{q+1}{2})^{q−1} = q^{2}(\frac{q+1}{2}). Also, in this case \frac{q+1}{2} ≥ (p + 2). It suffices to prove that β(p) ≥ h(\frac{q+1}{2}). If β(p) = h(p), this is clear. If not, then p ≡ 2 (mod 3) so p ≥ 5 and β(p) = k(p) > h(p + 2) ≥ h(\frac{q+1}{2}). iii) We may suppose that q ≥ 7. Then

α(q) = [q(2q + 1)]^{\frac{1}{2q}} = q^{\frac{1}{2q}} (2q + 1)^{\frac{1}{2q}} < q^{\frac{1}{2(q−1)}} q^{\frac{1}{q−1}} = q^{\frac{3}{2(q−1)}} .

2 Some assumed results

We summarise some well-known facts about representations of solvable groups which we will freely use. We outline a proof in some cases. If G is a finite solvable group, then F (G/Φ(G)) = F (G)/Φ(G) and this is a completely reducible module for G/F (G). For each prime divisor of |F (G)|, O_{p}(G)/O_{p}(Φ(G)) is a completely reducible GF(p)G-module, on which O_{p}(G)C_{G}(O_{p}(G)) acts trivially.

If q is odd, then an Abelian {2, q}′-subgroup of Sp(2m, q) has order at most \frac{q^{m}+1}{2}. If q is odd, then an Abelian \{2, q\}′-subgroup of GL(m, q) has order at most \frac{q^{m}−1}{2}.

If G is a nilpotent {2, q}′-subgroup of GL(n, q) where q is an odd prime, then |G| ≤ \frac{q^{n}−1}{2}. We may suppose that G is irreducible. If G is imprimitive, and the given representation is induced from a subgroup H of prime index r, then H ⊲G as G is nilpotent. By induction, we may suppose that |G| ≤ r(\frac{q^{b}−1}{2})^{r} ≤ \frac{q^{rb}−1}{2}. If G is primitive, then G is cyclic, and |G| ≤ \frac{q^{n}−1}{2}, as claimed.

In particular, if G is a completely reducible nilpotent odd order subgroup of GL(n, q) for an odd prime q, then |G| ≤ \frac{q^{n}−1}{2}.

Returning to the completely reducible action of G/F (G) on F (G)/Φ(G) we have [F_{2}(G) : F (G)] ≤ \frac{[F (G):Φ(G)]−1}{2} when G is solvable of odd order. For each prime divisor q of |F (G)|, we have a homomorphism ϕ_{q} : G/F (G) → Aut(O_{q}(G)/O_{q}(Φ(G))), and the image of ϕ_{q} is a completely reducible subgroup of GL(m, q) where q^{m} = [O_{q}(G) : O_{q}(Φ(G))]. The image of F_{2}(G)/F (G) is a completely reducible nilpotent subgroup of odd order of GL(m, q), so this image has order at most \frac{q^{m}−1}{2}. Since G/F (G) embeds in the direct product ∏_{q∈π(F (G))} Imϕ_{q}, the stated bound follows.

We will make implicit use of the fact that if G is a finite group, then |G| and |G/Φ(G)| have the same prime divisors. For if some Sylow q-subgroup Q of G is contained in Φ(G), then Q ⊲ G and there is a complement N to Q by the Schur-Zassenhaus theorem, a contradiction.

When q is an odd prime, a completely reducible subgroup G of odd order of Sp(2m, q) lifts to a finite subgroup of GL(m, ℂ). To see this, we may suppose that G is irreducible. Then, after extending the ground field to a splitting field, the natural module decomposes as a sum of r Galois conjugate absolutely irreducible modules of equal dimension d, each of which affords a faithful representation of G. By the Fong-Swan theorem, G is isomorphic to an irreducible subgroup of GL(d, ℂ). Hence d must be odd, so that r is even and d ≤ m.

Let G be a finite primitive solvable subgroup of odd order of GL(n, ℂ) where n > 1. Suppose that p_{1} < p_{2} < … < p_{r} are the prime divisors of [F (G) : Z(G)], and set P_{i} = Ω_{1}(O_{p_{i}}(G)) for each i. Then each P_{i} has nilpotence class 2 and is extra-special of exponent p_{i}, Z(G) is cyclic, and F (G) = Z(G)P_{1} … P_{r}. Also P_{i}/Z(P_{i}) is a completely reducible module for G (and O_{p_{i}}(G)C_{G}(O_{p_{i}}(G)) acts trivially on it), since whenever Q_{i} ⊲ G with Q_{i} ≤ P_{i}, we have P_{i} = Q_{i}C_{P_{i}}(Q_{i}).

Let |P_{i}| = p_{i}^{2d_{i}+1} for each i. Then ∏_{i=1}^{r} p_{i}^{d_{i}} divides n and for each i, there is a homomorphism ϕ_{i} : G → Sp(2d_{i}, p_{i}). Let G_{i} denote the image of G and K_{i} denote the kernel of ϕ_{i}. Then ∩_{i=1}^{r}K_{i} = F (G).

Hence G/F (G) is isomorphic to a subgroup of the direct product G_{1} × … × G_{r}. Letting V_{i} denote the natural module for Sp(2d_{i}, p_{i}), we note also that G/Z(G) is isomorphic to a subgroup of (V_{1}G_{1}) × … × (V_{r}G_{r}). Each G_{i} is a completely reducible odd order subgroup of Sp(2d_{i}, p_{i}), so lifts to a finite subgroup of GL(d_{i}, ℂ). By Clifford’s theorem, F (G_{i}) is also completely reducible, so, in particular, F (G_{i}) is a p′_{i}-group. We have [G : F (G)] ≤ ∏_{i=1}^{r} |G_{i}| and [G : Z(G)] ≤ ∏_{i=1}^{r} p^{2d_{i}}|G_{i}|. We remark here that d_{i} > 1 if p_{i} is the smallest prime divisor of |G|, since otherwise i = 1 and G_{1} is a isomorphic to an odd order subgroup of SL(2, p_{1}). Hence F (G_{1}) is a p′_{1}-group, so that |F (G_{1})| divides \frac{(p_{1}−1)(p_{1}+1)}{4}, contrary to the assumption that p_{1} is the smallest prime divisor of |G|.

Also, if d_{i} = 2 for any i, then G_{i} lifts to an odd order finite subgroup of GL(2, ℂ), so G_{i} is Abelian (hence G_{i} is a p′_{i}-prime group, since O_{p_{i}}(G_{i}) = 1).

We also remark that the symplectic action of G_{i} allows us to construct the semi-direct product P_{i}G_{i} with G_{i} acting trivially on Z(P_{i}). The complex irreducible character of degree p_{i}^{d_{i}} of P_{i} is uniquely determined by the linear character of Z(P_{i}) it covers, and extends to P_{i}Sp(2d_{i}, p_{i}), so in particular to P_{i}G_{i} (the extension to Sp(2d_{i}, p_{i}) is uniquely specified on p-regular elements if we insist (as we may) that they act with determinant 1. This determines the extension uniquely unless d_{i} = 1 and p_{i} = 3. But in the exceptional case, the unique unimodular extension to the normal subgroup P_{i}Q_{8} is SL(2, p_{i})-stable, so extends to P_{i}SL(2, p_{i}) in three ways as P_{i}Q_{8} is a normal subgroup of prime index.

Let A_{i} be an Abelian normal subgroup of G_{i} of maximal order (this need

not be unique). Then we have |A_{i}| ≤ \frac{p_{i}^{d_{i}} +1}{2} and |F (G_{i})| ≤ \frac{p_{i}^{2d_{i}} −1}{2} for each i.

3 Proof of the Theorem

Proof: We prove a) and b) together by induction. If |G| = p, the result is clear. Suppose that |G| > p, and that a) and b) have been established for groups of order less than |G|. Notice that for every odd prime p, we have β(p) ≤ β(3) = 75^{\frac{1}{4}} < 3.

To prove a), it suffices to treat the case that G is a (transitive and) primitive permutation group, so suppose this is the case. Then G = V M, where V is elementary Abelian of order n = q^{s} and M is an irreducible odd order subgroup of GL(s, q). Then M lifts to a subgroup of subgroup of GL(s, ℂ). By the inductive hypothesis, M has an Abelian normal subgroup A of index at most β(p)^{s−1} < 3^{s−1}. Now A⊲M and M acts faithfully and irreducibly on V, so that A acts completely reducibly on V by Clifford’s theorem. Hence A is a q′-group, so that |A| ≤ \frac{q^{s}−1}{2}. Then |G| ≤ q^{s}3^{s−1} \frac{q^{s}−1}{2}. Then certainly |G| < n^{3}.

If s ≥ 3, then 1 + q + … + q^{s−1} ≥ sq^{\frac{s−1}{2}} ≥ 3s by the arithmetic-geometric mean inequality. Then

α(p)^{n−1} ≥ α(q)^{n−1} ≥ q^{\frac{n−1}{q−1}} ≥ q^{3s} = n^{3}.

Hence we may suppose that s < 3.

If s = 2, then M is isomorphic to a subgroup of GL(2, ℂ). Thus M is an Abelian q′-group, and |G| < q^{4}. Then certainly

|G| < q^{q+1} = q^{\frac{n−1}{q−1}} ≤ p^{\frac{n−1}{p−1}} ≤ α(p)^{n−1}.

Hence we may suppose that s = 1.

Now n = q for some prime q and |G| divides \frac{q(q−1)}{2}. Then p ≤ \frac{q−1}{2} and

α(p)^{q−1} ≥ g(p)^{q−1} ≥ g(\frac{q − 1}{2})^{q−1} = \frac{q(q − 1)}{2}.

b) Now suppose that G is a subgroup (of odd order) of GL(n, ℂ), and parts a) and b) hold for groups of smaller (odd) order. We will prove that G has an Abelian normal subgroup of index at most β(p)^{n−1} and that [G : F (G)] ≤ α(p)^{n−1}. Since α(p) ≤ β(p) for all p, we may suppose that G is irreducible and primitive, and that n > 1, so we do.

Let [F (G) : Z(G)] = p_{1}^{2d_{1}} … p_{r}^{2d_{r}} where p_{1} < … < p_{r} are primes. Then ∏_{i=1}^{r} p_{i}^{d_{i}} divides n. Define the groups G_{i} and A_{i} as in the discussion of Section 2. Each G_{i} is isomorphic to a subgroup of GL(d_{i}, ℂ), so by the inductive hypothesis, we have [G_{i} : F (G_{i})] ≤ α(p)^{d_{i}−1} and [G_{i} : A_{i}] ≤ β(p)^{d_{i}−1} for each i.

Then

[G : Z(G)] ≤ ∏_{i=1}^{r} p^{2d_{i}}|G_{i}| ≤ ∏_{i=1}^{r} p_{i}^{2d_{i}}(\frac{p^{d_{i}} + 1}{2})β(p)^{d_{i}−1}.

Also,

[G : F (G)] ≤ ∏_{i=1}^{r} |G_{i}| ≤ ∏_{i=1}^{r}(\frac{p^{2d_{i}} − 1}{2})α(p)^{d_{i}−1}.

It suffices to prove that for each i we have p^{2d_{i}}|G_{i}| ≤ β(p)^{p_{i}^{d_{i}} −1} and |G_{i}| ≤ α(p)^{p_{i}^{d_{i}} −1}. In other words, it suffices to establish the stated bounds for each of the groups P_{i}G_{i}, viewed as subgroups of GL(p^{d_{i}}, ℂ), since P_{i} = F (P_{i}G_{i}) and Z(P_{i}) is the largest Abelian normal subgroup of P_{i}G_{i}.

Hence we may suppose that r = 1, that F (G) is a p_{1}-group, that G = P_{1}G_{1}, and that n = p_{1}^{d_{1}}. If we can prove that

p_{1}^{2d_{1}}(\frac{p^{d_{1}} + 1}{2})β(p)^{d_{1}−1} ≤ β(p)^{p_{1}^{d_{1}} −1},

for each i, we will have established part i) of b).

This is certainly true if d_{1}[3 log_{β(p)}(p_{1})+1] ≤ p_{1}^{d_{1}}. Now β(p) ≥ α(p) ≥ p1^{\frac{1}{p_{1}−1}} , so that log_{β(p)}(p_{1}) ≤ p_{1} − 1. The required inequality holds if d_{1}(3p_{1} − 2) ≤ p_{1}^{d_{1}}. This holds without exception when d_{1} ≥ 3.

If d_{1} = 2, then the required inequality holds for p_{1} > 5. If p_{1} = 5 and d_{1} = 2, we return to the finer estimates. Then p ∈ \{3, 5\}, so that β(p) ≥ 5^{\frac{1}{4}} . We need to show that 13×5^{4} ≤ β(p)^{23}. But 5^{\frac{23}{4}} > 3×5^{5}, and the necessary inequality holds. If p_{1} = 3 and d_{1} = 2, then p_{1}^{2d_{1}}|G_{1}| = 405. But β(3) = 75^{\frac{1}{4}} and p^{d_{1}} − 1 = 8, so p_{1}^{2d_{1}}|G_{1}| ≤ β(p)^{n−1}.

If d_{1} = 1, then p_{1}^{2d_{1}}|G_{1}| ≤ \frac{p_{1}^{2}(p_{1}+1)}{2}. In that case, p_{1} is not the smallest prime divisor of |G|.

In this case, G_{1} is a non-trivial Abelian p′_{1}-subgroup of SL(2, p_{1}) of order dividing \frac{p_{1}+ε}{2} for some sign ε. Now p is a divisor of [G : O_{p_{1}}(G)] = |G_{1}|.

If p = \frac{p_{1}+1}{2}, then β(p)^{p_{1}−1} = h(p)^{p_{1}−1} = \frac{p_{1}^{2}(p_{1}+1)}{2} and n = p_{1}. If p ≤ \frac{p_{1}−1}{2} then

β(p)^{p_{1}−1} ≥ k(p)^{p_{1}−1} ≥ k(\frac{p_{1} − 1}{2})^{p_{1}−1} = \frac{p_{1}^{2}(p_{1} − 1)}{2}.

Hence we may suppose that [G : Z(G)] = p_{1}^{2}(\frac{p_{1}+1}{2}), and that p divides \frac{p_{1}+1}{2}, but p ≠ \frac{p_{1}+1}{2}.Then \frac{p_{1}+1}{2} is certainly odd, so that \frac{p_{1}+1}{2} ≥ p + 2. Now by Lemma 2, we have

α(p)^{p_{1}−1} ≥ k(p)^{p_{1}−1} > h(p + 2)^{p_{1}−1} ≥ h(\frac{p_{1} + 1}{2})^{p_{1}−1} = \frac{p_{1}^{2}(p_{1} + 1)}{2},

so b) part i) is now proved.

To prove part ii) of b), we need to prove that

|G_{1}| ≤ α(p)^{p_{1}^{d_{1}} −1}.

We observed earlier in the proof (using the inductive hypotheses) that |G_{1}| < α(p)^{d_{1}−1}(p_{1}^{2d_{1}} − 1).

Since α(p) ≤ \sqrt{p} ≤ \sqrt{p_{1}}, we obtain |G_{1}| < p1^{\frac{5d_{1}}{2}} .

On the other hand α(p) ≥ p^{\frac{1}{p−1}} ≥ p1^{\frac{1}{p_{1}−1}} , so that

α(p)^{p_{1}^{d_{1}−1}} ≥ p1^{\frac{p{d_{1} \atop 1} −1}{p_{1}−1}} ≥ p_{1}^{d_{1}p \frac{d_{1}−1}{2} ; 1} .

Since p_{1} ≥ 3, we have d_{1} < 3. If d_{1} = 2, we can calculate more carefully. We have

α(p)^{p_{1}^{2}−1} ≥ p1^{\frac{p^{2}_{1}−1}{p_{1}−1}} = p_{1}^{1+p_{1}}

and we know that |G_{1}| < p_{1}^{5}. Thus p_{1} = 3. In that case G_{1} is a completely reducible odd order subgroup of Sp(4, 3). Thus |F (G_{1})| = 5 and in fact |G_{1}| = 5. But p_{1}^{1+p_{1}} = 81 in this case. Hence d_{1} = 1.

But now G_{1} is an odd order completely reducible subgroup of SL(2, p_{1}) so is Abelian of order at most \frac{p_{1}+1}{2}. However, α(p)^{p_{1}−1} ≥ p_{1}, so the proof is complete.

Proof of Corollary: G/F (G) acts completely reducibly on F (G)/Φ(G), and the latter group is Abelian of squarefree exponent. Applying the Fong-Swan theorem to the linear action on each minimal normal subgroup, we see that G/F (G) is isomorphic to a subgroup of GL(n, ℂ) where n is the number of prime factors (including multiplicities) of [F (G) : Φ(G)]. Then certainly n ≤ log_{p}([F (G) : Φ(G)]).

Applying b) ii) of the Theorem to G/F (G), we see that [G : F_{2}(G)] < α(p)^{n} so that

[G : F_{2}(G)] < α(p)^{log_{p}([F (G):Φ(G)])}.

When p ∈ \{3, 5\}, we have α(p) = p^{\frac{1}{p−1}} . When p > 5, we have

log(α(p)) = \frac{log p}{p} + \frac{log 2}{2p} + \frac{log(1 + \frac{1}{2p})}{2p} < \frac{log p}{p} + \frac{log 2}{2p} + \frac{1}{4p^{2}} .

Hence

log([G : F_{2}(G)]) < log([F (G) : Φ(G)])[\frac{1}{p} + \frac{log 2}{2p log p} + \frac{1}{4p^{2} log p}]

and the result follows.

iv) It suffices to consider the case Φ(G) = 1. We have [F_{2}(G) : F (G)] ≤ \frac{|F (G)|−1}{2}, while from part ii) we have [G : F_{2}(G)] < α(p)^{log_{p}(|F (G)|)}. We know that α(p) = p^{\frac{1}{p−1}} when p ∈ \{3, 5\}, while we have seen that α(p) < p^{\frac{3}{2(p−1)}} for p > 5.

Hence we certainly have [G : F (G)] < ^{\frac{|F (G)|^{1+ \frac{3}{2(p−1)}}}{2}} for p > 5 and

[G : F (G)] < ^{\frac{|F (G)|^{1+ \frac{1}{p−1}}}{2}} for p ∈ \{3, 5\}.

References

[1] L.Dornhoff, Jordan’s theorem for solvable groups, Proc. Amer. Math. Soc. 24, (1970), 533–537.

[2] R. Gow, On the number of characters in a p-block of a p-solvable group, J. Algebra 65, (1980), no. 2, 421–426.