DEDEKIND S TRANSPOSITION PRINCIPLE

Transcription

1 DEDEKIND S TRANSPOSITION PRINCIPLE AND PERMUTING SUBGROUPS & EQUIVALENCE RELATIONS William DeMeo williamdemeo@gmail.com University of South Carolina Zassenhaus Conference at WCU Asheville, NC May 24 26, 2013 These slides and other resources are available at

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3 INTERVALS IN SUBGROUP LATTICES H, K Let H, K be subgroups of a group G. Recall the set HK = {hk h H, k K } H X H Y XK H 0 = H K Y K is a group if and only if HK = KH = U, H. Let H 0 = H K and define H 0, H := {X H 0 X H}, K, H, K := {X K X H, K }. Define H 0, H K := {X H 0, H XK = KX}. LEMMA If HK = KH, then K, HK = H 0, H K H 0, H.

4 EXAMPLE The group S 4 has permuting subgroups H = D 8 and K = C 3. (neither one normalizes the other) HK = S 4 A 4 S 3 H = D8 C 2 C 2 C 2 K = C3 H 0 = 1 Only four subgroups of H permute with K, including H A 4 = C2 C 2, H S 3 = C2.

5 DEDEKIND S TRANSPOSITION PRINCIPLE FOR MODULAR LATTICES a b Notation Let L = L,, be a lattice with a L. Let ϕ a and ψ a be the perspectivity maps a ψ b (x) y b ϕ a (x) = x a and ψ a(x) = x a x For x, y L, let x, y L = {z L x z y}. ϕ a (y) a b THEOREM (DEDEKIND S TRANSPOSITION PRINCIPLE) L is modular iff for all a, b L the maps ϕ a and ψ b are inverse lattice isomorphisms of a b, a and b, a b.

6 ANOTHER TRANSPOSITION PRINCIPLE FOR LATTICES OF EQUIVALENCE RELATIONS Let X be a set and let Eq X be the lattice of equivalence relations on X. Given α, β Eq X, define the composition of α and β to be the binary relation α β = {(x, y) X 2 ( z X) x α z β y}. For a sublattice L Eq X, with η, θ L, define η, θ L = {γ L η γ θ}, η, θ β L = {γ L η γ θ and γ β = β γ}, i.e., the relations in η, θ L that permute with β. α β x β LEMMA Suppose α and β are permuting relations in L Eq X. Then β, α β L = α β, α β L α β, α L. α x y α α β y β Question: Does this generalize the subgroup lattice lemma?

7 ANSWER: YES! For groups H G, the algebra A = G\H, Ḡ has universe: the right cosets H\G = {Hx x G} operations: Ḡ = {ga : g G}, where g A (Hx) = Hxg. A standard result is Con A = H, G. The isomorphism H, G K θ K Con A is given by θ K = {(Hx, Hy) xy 1 K }. The inverse isomorphism Con A θ K θ H, G is K θ = {g G (H, Hg) θ}. So every lattice property of congruence lattices is also a lattice property of (intervals of) subgroup lattices. Moreover, it s easy to prove: LEMMA In Con G\H, Ḡ, two congruences θ K 1 and θ K2 permute if and only if the corresponding subgroups K 1 and K 2 permute.

8 QUESTIONS Recall that HK = H, K if and only if HK = KH. Question 1. Is it true that HKH = H, K if and only if HKH = KHK? What about HKHK = H, K if and only if HKHK = KHKH? H n K = H, K if and only if H n K = K n H?.

9 QUESTIONS Denote by H n K the n-fold composition of H and K. H 1 K = H, H 2 K = HK, H 3 K = HKH, H 4 K = HKHK,. H n K = H 2 K n 1 H. We say H and K are n-permuting if H n K = K n H. Question 2. Is the following true? If H and K are n-permuting, then interval K, H, K is isomorphic to the lattice of subgroups in H 0, H that n-permute with K.

10 CONNECTION WITH EQUIVALENCE RELATIONS Let A = H\G, Ḡ be the algebra with LEMMA universe: the right cosets H\G = {Hx x G} operations: Ḡ = {ga : g G}, where g A (Hx) = Hxg. The subgroups K 1 and K 2 are n-permuting if and only if their corresponding congruences θ K1 and θ K2 are n-permuting. That is, K 1 n K 2 = K 2 n K 1 θ K1 n θ K2 = θ K2 n θ K1.

11 ANSWER TO QUESTION 1. LEMMA For α, β Eq X, and for every even integer n > 1, TFAE: (I) α n β = α β (II) α n β = β n α (III) α n β β n α For n = 3, but the converse is false. COROLLARY α β α = β α β = α β α = α β For H, K G, and for every even integer n > 1, TFAE: (I) H n K = H, K (II) H n K = K n H (III) H n K K n H For n = 3, HKH = KHK = HKH = H, K but the converse is false. Question 1. What are conditions on G under which the converse is true?

12 ANSWER TO QUESTION 1. CASE n = 5 Question 1. Is it true that H 5 K = H, K if and only if H 5 K = K 5 H? Answer. No. Example. Let G = (C 3 C 3 ) : C 4. This is a group of order 36 with generators f 1, f 2, f 3, f 4. Let H = f 1 = C 2, and K = f 1 f 3 f4 2, f 2 f4 2 = C 4. Then, H K = 1 H, K = K 5 H has order 36 so it is the whole group. The set H 5 K has size 34, so does not generate H, K. H covers 1.

13 ANSWER TO QUESTION 2. No. In general, it is not true that if H and K are n-permuting, then the interval K, H, K is isomorphic to the lattice of those subgroups in H 0, H that n-permute with K. Example. The group A 5 has subgroups H = D 10, and K = C 2 such that H 4 K = K 4 H = A 5, but the map is not one-to-one. K, A 5 J J H 1, H

14 EXAMPLES

15 REVISED QUESTION 2. Question 2. What are conditions on the group G so that if H, K are n-permuting subgroups of G, then K, H, K = H 0, H K n H 0, H?

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