arXiv:1610.07253v1 [] 24 Oct 2016

DUAL ORE’S THEOREM FOR DISTRIBUTIVE
INTERVALS OF SMALL INDEX
SEBASTIEN PALCOUX
Abstract. This paper proves a dual version of a theorem of Oystein Ore for every distributive interval of finite groups [H, G] of
index |G : H| < 9720, and for every boolean interval of rank < 7.
It has applications to representation theory for every finite group.

1. Introduction
Oystein Ore has proved that a finite group is cyclic if and only if its
subgroup lattice is distributive [3]. He has extended one side as follows:
Theorem 1.1 ([3]). Let [H, G] be a distributive interval of finite
groups. Then ∃g ∈ G such that hHgi = G.
We have conjectured the following dual version of this theorem:
Conjecture 1.2. Let [H, G] be a distributive interval of finite groups.
Then ∃V irreducible complex representation of G, with G(V H ) = H
(Definition 3.1); this property will be called linearly primitive.
The interval [1, G] is linearly primitive if and only if G is linearly
primitive (i.e. admits a faithful irreducible complex representation).
We will see that Conjecture 1.2 reduces to the boolean case, because a
distributive interval is bottom boolean (i.e. the interval generated by its
atoms is boolean). As application, Conjecture 1.2 leads to a new bridge
between combinatorics and representation theory of finite groups:
Definition 1.3. Let [H, G] be any interval. We define the combinatorial invariant bbℓ(H, G) as the minimal length ℓ for a chain of subgroups
H = H0 < H1 < · · · < Hℓ = G
with [Hi , Hi+1 ] bottom boolean. Then, let bbℓ(G) := bbℓ(1, G).
Application 1.4. Assuming Conjecture 1.2, bbℓ(G) is a non-trivial
upper bound for the minimal number of irreducible complex representations of G generating (for ⊕ and ⊗) the left regular representation.
2010 Mathematics Subject Classification. 20D60, 05E15, 20C15, 06C15.
Key words and phrases. group; representation; lattice; distributive; boolean.
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SEBASTIEN PALCOUX

Remark 1.5. If the normal subgroups of G are also known, note that
cf ℓ(G) := min{bbℓ(H, G) | H core-free}
is a better upper bound. For more details on the applications, see [1, 4].
This paper is dedicated to prove Conjecture 1.2 for [H, G] boolean
of rank < 7, or distributive of index |G : H| < 9720. For so, we will
use the following new result together with two former results:
Theorem 1.6. Let [H, G] be a boolean interval and L a coatom with
|G : L| = 2. If [H, L] is linearly primitive, then so is [H, G].
Theorem 1.7 ([4]). A distributive interval [H, G] with
n
X
1
≤2
|K
:
H|
i
i=1
for K1 , . . . , Kn the minimal overgroups of H, is linearly primitive.
Theorem 1.8 ([1]). A boolean interval [H, G] with a (below) nonzero
dual Euler totient, is linearly primitive.
X
ϕ̂(H, G) :=
(−1)ℓ(H,K) |G : K|
K∈[H,G]

P
Remark 1.9 ([1]). The Euler totient ϕ(H, G) = (−1)ℓ(K,G) |K : H|
is the number of cosets Hg with hHgi = G, so ϕ > 0 by Theorem 1.1;
but in general ϕ̂ 6= ϕ. We extend ϕ to any distributive interval as
ϕ(H, G) = |T : H| · ϕ(T, G)
with [T, G] the top interval of [H, G], so that for n =
Y
Y
pini −1 · (pi − 1)
ϕ(1, Z/n) =
i

Q

i

pni i ,

i

which is the usual Euler totient ϕ(n). Idem for ϕ̂ and bottom interval.
We will also translate our planar algebraic proof of Theorem 1.7 in
the group theoretic framework (one claim excepted).
Contents
1. Introduction
2. Preliminaries on lattice theory
3. A dual version of Ore’s theorem
4. The proof for small index
5. Acknowledgments
References

1
3
4
7
15
15

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 3

2. Preliminaries on lattice theory
Definition 2.1. A lattice (L, ∧, ∨) is a partially ordered set (or poset)
L in which every two elements a, b have a unique supremum (or join)
a ∨ b and a unique infimum (or meet) a ∧ b.
Example 2.2. Let G be a finite group. The set of subgroups K ⊆ G
is a lattice, denoted by L(G), ordered by ⊆, with K1 ∨ K2 = hK1 , K2 i
and K1 ∧ K2 = K1 ∩ K2 .
Definition 2.3. A sublattice of (L, ∧, ∨) is a subset L′ ⊆ L such that
(L′ , ∧, ∨) is also a lattice. Let a, b ∈ L with a ≤ b, then the interval
[a, b] is the sublattice {c ∈ L | a ≤ c ≤ b}.
Definition 2.4. A finite lattice L admits a minimum and a maximum,
called 0̂ and 1̂.
Definition 2.5. An atom is an element a ∈ L such that
∀b ∈ L, 0̂ < b ≤ a ⇒ a = b.
A coatom is an element c ∈ L such that
∀b ∈ L, c ≤ b < 1̂ ⇒ b = c.
Definition 2.6. The top interval of a finite lattice L is the interval
[t, 1̂] with t the meet of all the coatoms. The bottom interval is the
interval [0̂, b] with b the join of all the atoms.
Definition 2.7. The length of a finite lattice L is the greatest length ℓ
of a chain 0 < a1 < a2 < · · · < aℓ = 1 with ai ∈ L.
Definition 2.8. A lattice (L, ∧, ∨) is distributive if ∀a, b, c ∈ L:
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)
(or equivalently, ∀a, b, c ∈ L, a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)).
Lemma 2.9. The reverse lattice and the sublattices of a distributive
lattice are also distributive. Idem for concatenation and direct product.
Definition 2.10. A distributive lattice is called boolean if any element
b admits a unique complement b∁ (i.e. b ∧ b∁ = 0̂ and b ∨ b∁ = 1̂).
Example 2.11. The subset lattice of {1, 2, . . . , n}, for union and intersection, is called the boolean lattice Bn of rank n (see B3 below).

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SEBASTIEN PALCOUX
{1, 2, 3}
{1, 2} {1, 3} {2, 3}
{1}

{2}

{3}

∅

Remark 2.12. Any finite boolean lattice is isomorphic to some Bn .
Theorem 2.13 (Birkhoff’s representation theorem or FTFDL [5]).
Any finite distributive lattice embeds into a finite boolean lattice.
Corollary 2.14. The top and bottom intervals of a distributive lattice
are boolean.
Proof. See [5, items a-i p254-255], together with Lemma 2.9.



3. A dual version of Ore’s theorem
In this section, we will state the dual version of Ore’s theorem, and
prove it for any boolean interval of rank ≤ 4, after Theorem 1.7 proof.
Definition 3.1. Let W be a representation of a group G, K a subgroup
of G, and X a subspace of W . We define the fixed-point subspace
W K := {w ∈ W | kw = w , ∀k ∈ K}
and the pointwise stabilizer subgroup
G(X) := {g ∈ G | gx = x , ∀x ∈ X}
Lemma 3.2. [1, Section 3.2] Let G be a finite group, H, K two subgroups, V a complex representation of G and X, Y two subspaces. Then
(1) H ⊆ K ⇒ V K ⊆ V H
(2) X ⊆ Y ⇒ G(Y ) ⊆ G(X)
(3) V H∨K = V H ∩ V K
(4) H ⊆ G(V H )
(5) V G(V H ) = V H
(6) [H ⊆ K and V K ( V H ] ⇒ K 6⊆ G(V H )
Lemma 3.3. [1] Let V1 , . . . , Vr be the irreducible complex representations of a finite group G (up to equivalence), and H a subgroup. Then
|G : H| =

r
X
i=1

dim(Vi ) dim(ViH ).

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 5

Definition 3.4. An interval of finite groups [H, G] is called linearly
primitive if there is an irreducible complex representation V of G such
that G(V H ) = H.
Remark 3.5. The interval [1, G] is linearly primitive iff G is linearly
primitive (i.e. it admits an irreducible faithful complex representation).
The dual version of Ore’s Theorem 1.1 is the following:
Conjecture 3.6. A distributive interval [H, G] is linearly primitive.
Lemma 3.7. A boolean interval [H, G] of rank 1 is linearly primitive.
Proof. Note that [H, G] is of rank 1 iff H is a maximal subgroup of G.
Let V be a non-trivial irreducible complex representation of G with
V H 6= ∅, by Lemma 3.2 (4), H ⊆ G(V H ) . If G(V H ) = G then V must be
trivial (by irreducibility), so by maximality G(V H ) = H.

Lemma 3.8. [1, Lemma 3.37] An interval [H, G] is linearly primitive
if its bottom interval [H, B] is so (see Definition 2.6).
Proposition 3.9. An interval [H, G] satisfying
n
X
1
≤1
|Ki : H|
i=1
with K1 , . . . , Kn the minimal overgroups of H, is linearly primitive.
Proof. First, by Lemmas 3.7, 3.8, we can assume n > 1. By assumption
Pn |G:H|
Pn
i=1 |Ki :H| ≤ |G : H|, so
i=1 |G : Ki | ≤ |G : H|. Let V1 , . . . , Vr be
the irreducible complex representations of G. By Lemma 3.3
n
n X
r
r
n
X
X
X
X
|G : Ki | =
dim(Vα ) dim(VαKi ) =
dim(Vα )[
dim(VαKi )].
i=1 α=1

i=1

If ∀α,

P

Ki
i Vα

=

α=1

i=1

VαH , then
n
X

dim(VαKi ) ≥ dim(VαH ),

i=1

Pn

Pn
and
Pn so i=1 |G : Ki | ≥ |G : H|, but i=1 |G : Ki | ≤ |G : H|, then
i=1 |G : Ki | = |G : H|. So ∀α,
n
X
dim(VαKi ) = dim(VαH ),
i=1

P
but for V1 trivial, we get that n = ni=1 dim(V1Ki ) = dim(V1H ) = 1,
contradiction with n > 1.
P
Else there is α such that i VαKi ( VαH , then by Lemma 3.2 (6),
Ki 6⊆ G(VαH ) ∀i, which means that G(VαH ) = H by minimality.


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SEBASTIEN PALCOUX

Corollary 3.10. If a subgroup H of G admits at most two minimal
overgroups, then [H, G] is linearly primitive. In particular, a boolean
interval of rank n ≤ 2 is linearly primitive.
P
Proof. i |Ki1:H| ≤ 12 + 12 = 1; the result follows by Proposition 3.9. 
We can upgrade Proposition 3.9 in the distributive case as follows:
Theorem 3.11. A distributive interval [H, G] satisfying
n
X

1
≤2
|Ki : H|

i=1

with K1 , . . . , Kn the minimal overgroups of H, is linearly primitive.
Proof. By Lemma 3.8, Corollaries 2.14 and 3.10, we can assume the
interval to be boolean of rank n > 2.
If ∃α such that
X
(⋆)
VαKi∨Kj ( VαH
i,j,i6=j

then by Lemma 3.2 (6), ∀i, j with i 6= j, Ki ∨Kj 6⊆ G(VαH ) . If G(VαH ) = H
then ok, else by the boolean structure and minimality ∃i such that
G(VαH ) = Ki . Now Li := Ki∁ (see Definition 2.10) is a maximal subgroup
of G, so by Lemma 3.7, there is β such that G(V Li ) = Li .
β
Claim: ∃Vγ ≤ Vα ⊗ Vβ such that Ki ∩ G(VγH ) , G(VγH ) ∩ Li ⊆ Ki ∩ Li .
Proof: See the first part of [4, Theorem 6.8] proof; it exploits (⋆) in a
tricky way (we put this reference because we didn’t find an argument
which avoids the use of planar algebras).

By H ⊆ G(VγH ) , distributivity and Claim, we conclude as follows:
G(VγH ) = G(VγH ) ∨ H = G(VγH ) ∨ (Ki ∧ Li ) = (G(VγH ) ∧ Ki ) ∨ (G(VγH ) ∧ Li )
⊆ (Ki ∧ Li ) ∨ (Ki ∧ Li ) = H ∨ H = H
Else, ∀α,
X

VαKi ∨Kj = VαH .

i,j,i6=j
Ki ∨Kj

∀k, ∀(i, j) with i 6= j, ∃s ∈ {i, j} with s 6= k, but Vα
X
VαKs = VαH .
s6=k

It follows that ∀i, ∀α,
X
j6=i

dim(VαKj ) ≥ dim(VαH ).

⊆ VαKs , so

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 7

Now if ∃α∀i, VαKi ( VαH then (by Lemma 3.2 (6) and minimality)
P
K
G(VαH ) = H. Else ∀α∃i, VαKi = VαH , but j6=i dim(Vα j ) ≥ dim(VαH ), so
X
dim(VαKj ) ≥ 2 dim(VαH )
j

By using Lemma 3.3 and taking V1 trivial, we get
X
XX
X
X
|G : Ki | =
[
dim(Vα ) dim(VαKi )] =
dim(Vα )[
dim(VαKi )]
i

i

≥n+2

α

X

α

i

dim(Vα ) dim(VαH ) = 2|G : H| + (n − 2).

α6=1

It follows that

n
X
i=1

n−2
1
≥2+
|Ki : H|
|G : H|

which contradicts the assumption because n > 2.



Corollary 3.12. A rank n boolean interval [H, G] with |Ki : H| ≥ n/2
for any minimal overgroup Ki of H, is linearly primitive. In particular,
a boolean interval of rank n ≤ 4 is linearly primitive.
P
Proof. i |Ki1:H| ≤ n × n2 = 2; the result follows by Theorem 3.11. 
In the next section, we get a proof at any rank n < 7.
4. The proof for small index
This section will prove dual Ore’s theorem, for any boolean interval of
rank < 7, and then for any distributive interval of index |G : H| < 9720.
Lemma 4.1. Let [H, G] be a boolean interval of rank 2 and let K, L
the atoms. Then (|G : K|, |G : L|) and (|K : H|, |L : H|) 6= (2, 2).
Proof. If |G : K| = |G : L| = 2, then K and L are normal subgroups
of G, and so H = K ∧ L is also normal. So G/H is a group and
[1, G/H] = [H, G] as lattices, but a boolean lattice is distributive, so
by Ore’s theorem, G/H is cyclic; but it has two subgroups of index 2,
contradiction. If |K : H| = |L : H| = 2, then H is a normal subgroup
of K and L, so of G = H ∨ K, contradiction as above.

Note the following immediate generalization:
Remark 4.2. Let [H, G] be boolean of rank 2, with K and L the atoms.
• If H is a normal subgroup of K and L, then |K : H| =
6 |L : H|.
• If K and L are normal subgroups of G then |G : K| =
6 |G : L|.

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SEBASTIEN PALCOUX

Remark 4.3. Let G be a finite group and H, K two subgroups, then
|H| · |K| = |HK| · |H ∩ K| (Product Formula). It follows that
|H| · |K| ≤ |H ∨ K| · |H ∧ K|
Corollary 4.4. Let [H, G] be a boolean interval of finite groups and A
an atom. Any K1 , K2 ∈ [H, A∁ ] with K1 ⊂ K2 satisfy
|K1 ∨ A : K1 | ≤ |K2 ∨ A : K2 |
Moreover if |G : A∁ | = 2 then |K ∨ A : K| = 2, ∀K ∈ [H, A∁ ].
Proof. Suppose that K1 ⊂ K2 . By Remark 4.3,
|K1 ∨ A| · |K2 | ≤ |(K1 ∨ A) ∨ K2 | · |(K1 ∨ A) ∧ K2 |
but K1 ∩ K2 = K1 , K1 ∪ K2 = K2 and A ∧ K2 = H, so by distributivity
|K1 ∨ A| · |K2 | ≤ |K2 ∨ A| · |K1 |
Finally, A∁ ∨ A = G and ∀K ∈ [H, A∁ ], K ⊂ A∁ , so if |G : A∁ | = 2,
then
2 ≤ |K ∨ A : K| ≤ |A∁ ∨ A : A∁ | = 2,
It follows that |K ∨ A : K| = 2.

Lemma 4.5. Let [H, G] rank 2 boolean with K, L the atoms. Then
|K : H| = 2 ⇔ |G : L| = 2.
Proof. If |G : L| = 2 then |K : H| = 2 by Corollary 4.4.
If |K : H| = 2 then H ⊳ K and K = H ⊔ Hτ with τ H = Hτ and
(Hτ )2 = H, so Hτ 2 = H and τ 2 ∈ H. Now L ∈ (H, G) open, then
τ Lτ −1 ∈ (τ Hτ −1 , τ Gτ −1 ) = (H, G), so by assumption τ Lτ −1 ∈ {K, L}.
If τ Lτ −1 = K, then L = τ −1 Kτ = K, contradiction. So τ Lτ −1 = L.
Now H = Hτ 2 ⊂ Lτ 2 , and τ 2 ∈ H ⊂ L, so Lτ 2 = L. It follows that
hL, τ i = L ⊔ Lτ . But by assumption, G = hL, τ i, so |G : L| = 2.

Corollary 4.6. If a boolean interval [H, G] admits a subinterval [K, L]
of index 2, then there is an atom A with L = K ∨ A and |G : A∁ | = 2.
Proof. Let [K, L] be the edge of index |L : K| = 2. By the boolean
structure, there is an atom A ∈ [H, G] such that L = K ∨ A. Let
K = K1 < K2 < · · · < Kr = A∁
be a maximal chain from K to A∁ . Let Li = Ki ∨ A, then the interval
[Ki , Li+1 ] is boolean of rank 2, now |L1 : K1 | = 2, so by Lemma 4.5
2 = |L1 : K1 | = |L2 : K2 | = · · · = |Lr : Kr | = |G : A∁ |.


DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 9

Remark 4.7. Let [H, G] of index |G : H| = 2. Then G = H ⋊ Z/2 if
|H| is odd, but it’s not true in general if |H| even1.
The following theorem was pointed out by Derek Holt2.
Theorem 4.8. Let G be a finite group, N a normal subgroup of prime
index p and π an irreducible complex represenation of N. Exactly one
of the following occurs:
(1) π extends to an irreducible representation of G,
(2) IndG
N (π) is irreducible.
Proof. It is a corollary of Clifford theory, see [2] Corollary 6.19.



Theorem 4.9. Let [H, G] be a boolean interval and L a coatom with
|G : L| = 2. If [H, L] is linearly primitive, then so is [H, G].
Proof. Let the atom A := L∁ . As an immediate corollary of the proofs
of Lemma 4.5 and Corollary 4.6, there is τ ∈ A such that ∀K ∈ [H, L],
Kτ = τ K and τ 2 ∈ H ⊂ K, so K ∨ A = K ⊔ Kτ and G = L ⊔ Lτ . By
assumption, [H, L] is linearly primitive, which means the existence of
an irreducible complex representation V of L such that L(V H ) = H.
Assume that πV extends to an irreducible representation πV+ of G.
Note that G(V+H ) = H ⊔ Sτ with
S = {l ∈ L | πV+ (lτ ) · v = v, ∀v ∈ V H }
If S = ∅ then G(V+H ) = H, ok. Else S 6= ∅ and note that
πV+ (lτ ) · v = v ⇔ πV+ (τ ) · v = πV (l−1 ) · v
but πV+ (τ )(V H ) ⊂ V H and τ 2 ∈ H, so ∀l1 , l2 ∈ S and ∀v ∈ V H ,
πV (l1 l2 )−1 · v = πV+ (τ 2 ) · v = v
It follows that S 2 ⊂ H. Now, HS = S, so HS 2 = (HS)S = S 2 ,
which means that S 2 is a disjoint union of H-coset, then |H| divides
|S 2 |, but S 2 ⊂ H and S 6= ∅, so S 2 = H. Let s0 ∈ S, then the maps
S ∋ s 7→ s0 s ∈ H and H ∋ h 7→ hs0 ∈ S are injective, so |S| = |H|. If
S 6= H, then A = H ⊔Hτ and G(V+H ) = H ⊔Sτ are two different groups
containing H with index 2, contradiction with the boolean structure
by Lemma 4.1. So we can assume that H = S. Now the extension
V+ is completely characterized by πV+ (τ ), and we can make an other
irreducible extension V− characterized by πV− (τ ) = −πV+ (τ ). As above,
G(V−H ) = H ⊔ S ′ τ with
S ′ = {l ∈ L | πV− (lτ ) · v = v, ∀v ∈ V H }.
1http://math.stackexchange.com/a/1609599/84284
2http://math.stackexchange.com/a/1966655/84284

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SEBASTIEN PALCOUX

But πV− (lτ ) = −πV+ (lτ ), so
S ′ = {l ∈ L | πV+ (lτ ) · v = −v, ∀v ∈ V H }.
Then S ∩ S ′ = ∅, but S = H, so S ′ 6= H, contradiction as above.
Next, we can assume that πV does not extend to an irreducible representation of G. So πW := IndG
L (πV ) is irreducible by Theorem 4.8.
We need to check that G(W H ) = H. We can see W as V ⊕ τ V , with
πW (l) · (v1 + τ v2 ) = πV (l) · v1 + τ [πV (τ −1 lτ ) · v2 ],
with l ∈ L, and
πW (τ ) · (v1 + τ v2 ) = πV (τ 2 ) · v2 + τ v2
Then
W H = {v1 +τ v2 ∈ W | πV (h)·v1 = v1 and πV (τ −1 hτ )·v2 = v2 , ∀h ∈ H}
But τ −1 Hτ = H, so W H = V H ⊕ τ V H . Finally, according to πW (l)
and πW (τ ) above, we see that G(W H ) ⊂ L, and then G(W H ) = H.

Remark 4.10. It seems that we can extend Theorem 4.9, replacing
|G : L| = 2 by L⊳G (and so |G : L| = p prime), using Theorem 4.8 and
Remark 4.2. In the proof, we should have K ∨A = K ⊔Kτ ⊔· · ·⊔Kτ p−1 ,
τ p ∈ H, S p = H and πV− (τ ) = e2πi/p πV+ (τ ). We didn’t check the details
because we don’t need this extension.
Corollary 4.11. Let [H, G] be a boolean interval with an atom A satisfying |A : H| = 2. If [H, A∁ ] is linearly primitive, then so is [H, G].
Proof. Immediate by Corollary 4.6 and Theorem 4.9.



One of the main result of the paper is the following:
Theorem 4.12. A boolean interval [H, G] of rank n < 7, is linearly
primitive.
Proof. Let K1 , . . . , Kn be the atoms of [H, G]. By Corollary 4.11, we
can assume that |Ki : H| =
6 2, ∀i. Now n ≤ 6 and |Ki : H| ≥ 3, then
n
X
i=1

1
1
≤ 6 × = 2.
|Ki : H|
3

The result follows by Theorem 3.11.



For the upper bound on the index of distributive interval we will
need a former result (proved group theoretically in [1]):

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 11

Theorem 4.13. [1, Theorem 3.24] A boolean interval [H, G] with a
(below) nonzero dual Euler totient is linearly primitive.
X
ϕ̂(H, G) :=
(−1)ℓ(H,K) |G : K|
K∈[H,G]

with ℓ(H, K) the rank of [H, K].
Conjecture 4.14. A rank n boolean interval has ϕ̂ ≥ 2n−1 .
Remark 4.15. If Conjecture 4.14 is correct, then its lower bound is
optimal, because realized by the interval [1 × S2n , S2 × S3n ].
Lemma
4.16. Let [H, G]Pbe a boolean interval of rank n and index
Q ri
pi with pi prime and i ri = n. Then for any atom A and any
K ∈ [H, A∁ ], |K ∨ A : K| = pi for some i.
W
Proof. Let A1 , · · · , Ar be the atoms of [H, G] such that K = ri=1 Ai ,
let Ar+1 = A and Ar+2 , . . . , An all the other atoms. By considering the
corresponding maximal chain we have that
|G : H| = |A1 : H| · |A1 ∨ A2 : A1 | · · · |K ∨ A : K| · · · |G : A∁n−1 |
It’s a product of n numbers > 1 and the result is composed by n prime
numbers, so by the fundamental theorem of arithmetic, any component
above is prime, then |K ∨ A : K| = pi for some i.

Lemma 4.17. Let [H, G] be a boolean interval of rank n and index pn
with p prime. Then ϕ̂(H, G) = (p − 1)n > 0.

P
Proof. By Lemma 4.16, ϕ̂(H, G) = k (−1)k nk pk = (p − 1)n

Remark 4.18. Lemma 4.17 is coherent with Conjecture 4.14 because
if p = 2 then n = 1 by Lemma 4.1.
Proposition 4.19. Let [H, G] be a boolean interval of rank n and index
pn−1 q, with p, q prime and p ≤ q. Then
q−p
1
ϕ̂(H, G) = (p − 1)n [1 +
(1 −
)] ≥ (p − 1)n > 0.
p
(1 − p)m
with m be the number of coatoms L ∈ [H, G] with |G : L| = q.
Proof. If m = 0, then by Lemma 4.16, Corollary 4.4 and p ≤ q, for any
atom A ∈ [H, G] and ∀K ∈ [H, A∁ ], |K ∨ A : K| = p, so |G : H| = pn
and ϕ̂(H, G) = (p − 1)n by Lemma 4.17, ok.
Else m ≥ 1. We will prove the formula by induction. If n = 1, then
m = 1 and ϕ̂(H, G) = q − 1, ok. Next, assume it is true at rank < n.
Let L be a coatom with |G : L| = q, then for A = L∁ ,
ϕ̂(H, G) = q ϕ̂(H, L) − ϕ̂(A, G)

12

SEBASTIEN PALCOUX

Now |L : H| = pn−1 so by Lemma 4.17, ϕ̂(H, L) = (p − 1)n−1 . But
|A : H| = p or q. If |A : H| = p then |G : A| = pn−2 q and by induction
1
q−p
(1 −
)].
ϕ̂(A, G) = (p − 1)n−1 [1 +
p
(1 − p)m−1
Else |A : H| = q, |G : A| = pn−1 , m = 1 and the same formula works.
Then
q−p
1
ϕ̂(H, G) = (p − 1)n−1 [q − 1 −
(1 −
)]
p
(1 − p)m−1
1
q−1 q−p 1
+
(
−
)]
= (p − 1)n [
p−1
p 1 − p (1 − p)m
1
q−p
1
q−1 q−p
−
(1 +
)+
(1 −
)]
= (p − 1)n [
p−1
p
p−1
p
(1 − p)m
1
q−p
(1 −
)]
= (p − 1)n [1 +
p
(1 − p)m
The result follows.

Definition 4.20. A chain H1 ⊂ · · · ⊂ Hr+1 is of type (k1 , . . . , kr ) if
∃σ ∈ Sr with kσ(i) = |Hi+1 : Hi | (so that we can choose (ki )i increasing).
Remark 4.21. The proof of Proposition 4.19 is working without assuming p, q prime, but assuming type (p, . . . , p, q) for every maximal
chain of [H, G]. For p prime and q = p2 we deduce that at rank n and
index pn+1 , there is 1 ≤ m ≤ n such that
ϕ̂(H, G) = (p − 1)n+1 + (p − 1)n − (−1)m (p − 1)n+1−m ≥ (p − 1)n+1
If there is no edge of index 2, we can also take q = 2p or (p, q) = (3, 4).
Lemma 4.22. A boolean interval [H, G] of index |G : H| = an bc and
rank n+2 with 3 ≤ a ≤ b ≤ c ≤ 12, 1 ≤ n ≤ 6 and every maximal chain
of type (a, . . . , a, b, c), has a dual Euler totient ϕ̂(H, G) ≥ (a − 1)n+2 .
Proof. This is checked by computer calculation using the following iterative method. Let L be a coatom just that |G : L| = c and A = L∁ .
Then ϕ̂(H, G) = cϕ̂(H, L) − ϕ̂(A, G). Now |L : H| = an b so we can use
Propoposition 4.19 formula for ϕ̂(H, L). Next there are three cases:
|A : H| = a, b or c. If |A : H| = c then, by Corollary 4.4, ∀K ∈ [H, L],
|K ∨ A : K| = c, so ϕ̂(H, G) = (c − 1)ϕ̂(H, L). If |A : H| = b, then
|G : A| = an c so we can use Propoposition 4.19 formula for ϕ̂(A, G).
Else |A : H| = a and |G : A| = an−1 bc, so we iterate the method.

Remark 4.23. Let [H, G] be a boolean interval and A an atom such
that ∀K ∈ [H, A∁ ], |K ∨ A : K| = |A : H|. So ϕ̂(H, A∁ ) = ϕ̂(A, G) and
ϕ̂(H, G) = |A : H|ϕ̂(H, A∁ ) − ϕ̂(A, G) = (|A : H| − 1)ϕ̂(A, G).

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 13

Corollary 4.24. Let [H, G] be a boolean interval such that for any
atom A and ∀K ∈ [H, A∁ ], |K ∨ A : K| = |A : H|. Then
n
Y
ϕ̂(H, G) =
(|Ai : H| − 1) > 0.
i=1

with A1 , . . . , An all the atoms of [H, G].


Proof. By Remark 4.23 and induction.

Lemma 4.25. Let [H, G] boolean of rank 2 and index < 32. Let K, L
be the atoms, a = |G : K|, b = |G : L|, c = |L : H| and d = |K : H|.
a

G

b

K

L
d

H

c

If a 6= 7, then (a, b) = (c, d).
If a = 7 and a 6= c then a = b = 7 and c = d ∈ {3, 4}.
Proof. We can check by GAP3 that there are exactly 241 boolean intervals [H, G] of rank 2 and index |G : H| < 32 (up to equivalence).
They all satisfy (a, b) = (c, d), except [D8 , P SL2 (7)] and [S3 , P SL2 (7)],
for which (a, b) = (7, 7) and (c, d) = (3, 3) or (4, 4).

Corollary 4.26. Let [H, G] be a boolean interval having a maximal
chain such that the product of the index of two different edges is < 32,
and no edge has index 7. Then [H, G] satisfies Corollary 4.24.
Proof. Consider such a maximal chain
H = K0 ⊂ K1 ⊂ · · · ⊂ Kn = G
and A1 , . . . , An the atoms of [H, G] such that Ki = Ki−1 ∨ Ai . Now, ∀i
and ∀j < i, [Kj−1, Kj ∨ Ai ] is boolean of rank 2, so by Lemma 4.25,
|Ki : Ki−1 | = |Ki−2 ∨ Ai : Ki−2 | = |Ki−3 ∨ Ai : Ki−2 | = · · · = |Ai : H|
Next, ∀i and ∀j ≥ i, let Lj−1 = Kj ∧ A∁i , then [Lj , Kj+2 ] is boolean of
rank 2 and by Lemma 4.25,
|Ki : Ki−1 | = |Ki+1 : Li | = |Ki+2 : Li+1 | = · · · = |G : A∁i |
Finally, by Corollary 4.4, ∀K ∈ [H, A∁i ],
|Ai : H| ≤ |K ∨ Ai : K| ≤ |G : A∁i |
but |Ai : H| = |Ki : Ki−1 | = |G : A∁i |; the result follows.
3The

GAP Group, http://www.gap-system.org, version 4.8.3, 2016.



14

SEBASTIEN PALCOUX

Remark 4.27. A combinatorial argument could replace the use of
Corollary 4.4 in the proof of Corollary 4.26.
Remark 4.28. Here is the list of all the numbers < 10125 which are
product of at least seven integers ≥ 3; first with exactly seven integers:
2187 = 37
2916 = 36 4
3645 = 36 5
3888 = 35 42
4374 = 36 6

4860 = 35 41 5
5103 = 36 7
5184 = 34 43
5832 = 36 8
6075 = 35 52

6480 = 34 42 5
6561 = 36 9
6804 = 35 41 7
6912 = 33 44
7290 = 36 10

7776 = 35 41 8
8019 = 36 11
8100 = 34 41 52
8505 = 35 51 7
8640 = 33 43 5

8748 = 36 12
9072 = 34 42 7
9216 = 32 45
9477 = 36 13
9720 = 35 41 10;

next with exactly eight integers: 6561 = 38 , 8748 = 374; nothing else.
We can now prove the main theorem of the paper:
Theorem 4.29. A distributive interval [H, G] of index |G : H| < 9720,
is linearly primitive.
Proof. By Lemma 3.8, Corollary 2.14 and Theorem 4.12, we can assume
the interval to be boolean of rank n ≥ 7, and without edge of index
2 by Corollary 4.6 and Theorem 4.9. So by Theorem 4.13, it suffices
to check that for every index (except 9720) in the list of Remark 4.28,
any boolean interval as above with this index has a nonzero dual Euler
totient. We can assume the rank to be 7, because at rank 8, the indices
38 and 37 4 are checked by Lemma 4.17 and Remark 4.21, and there is
nothing else at rank > 8. Now, any maximal chain for such a boolean
interval of index 35 41 5 has type (3, . . . , 3, 4, 5), so it is checked by Corollary 4.26. Idem for index 36 10 with (3, . . . , 3, 10) or (3, . . . , 3, 5, 6). The
index 36 7 is checked by Proposition 4.19. For the index 36 12, if there is
a maximal chain of type (3, . . . , 3, 6, 6), 62 > 32 but using Lemma 4.25
with a, b, c, d ∈ {3, 6} we can deduce that (a, b) = (c, d), so the proof
of Corollary 4.26 is working; else 12 must appears in every maximal
chain, so that the proof of Proposition 4.19 works with q = 12. We can
do the same for every index, except 35 41 7, 35 41 8, 35 51 7, 34 42 7, 35 41 10.
For index 35 41 8, if there is a maximal chain of type (3, . . . , 3, 4, 4, 6),
then ok by Corollary 4.26, else (because there is no edge of index 2)
every maximal chain is of type (3, . . . , 3, 4, 8), so ok by Lemma 4.22.
We can do the same for every remaining index except 35 41 10 = 9720,
the expected upper bound.

Remark 4.30. The tools above don’t check 35 41 10 because the possible maximal chain types are (3, . . . , 3, 4, 5, 6), (3, . . . , 3, 4, 10) and
(3, . . . , 3, 5, 8). The first is ok by Corollary 4.26, but not the two last
because 4 · 10 = 5 · 8 = 40 > 32. So there is not necessarily a unique
maximal chain type, and Lemma 4.22 can’t be applied. Nevertheless,
more intensive computer investigation can probably leads beyond 9720.

DUAL ORE’S THEOREM FOR DISTRIBUTIVE INTERVAL, SMALL INDEX 15

5. Acknowledgments
I would like to thank Derek Holt for showing me a theorem on representation theory used in this paper. This work is supported by the
Institute of Mathematical Sciences, Chennai.
References
[1] Mamta Balodi and Sebastien Palcoux, On boolean intervals of finite groups,
arXiv:1604.06765v5, 25pp. submitted to Trans. Amer. Math. Soc.
[2] I. Martin Isaacs, Character theory of finite groups, Dover Publications, Inc.,
New York, 1994. Corrected reprint of the 1976 original [Academic Press, New
York; MR0460423 (57 #417)]. MR1280461
[3] Oystein Ore, Structures and group theory. II, Duke Math. J. 4 (1938), no. 2,
247–269, DOI 10.1215/S0012-7094-38-00419-3. MR1546048
[4] Sebastien Palcoux, Ore’s theorem for cyclic subfactor planar algebras and applications, arXiv:1505.06649v10, 50pp. submitted to Pacific J. Math.
[5] Richard P. Stanley, Enumerative combinatorics. Volume 1, 2nd ed., Cambridge
Studies in Advanced Mathematics, vol. 49, Cambridge University Press, Cambridge, 2012. MR2868112
Institute of Mathematical Sciences, Chennai, India
E-mail address: palcoux@imsc.res.in

