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Basic operations on P -schemes

Chapter VI: Constructing new P -schemes from old ones

6.1 Basic operations on P -schemes

In this section, we introduce some basic operations onP-schemes, including inverse right translation, restriction, and passing to quotient groups. We then use them to prove Lemma 4.12.

Inverse right translation ofP-schemes. LetP be a subgroup system over a finite groupG. For eachH ∈ P, the group Gacts onH\Gby inverse right translation

gHh=Hhg−1. This action induces an action ofGon the set of partitions ofH\G, defined bygP ={gB :B ∈P}for a partitionP ofH\G. ThenGalso acts on the set ofP-collections by inverse right translation:

Definition 6.1. The action ofGon the set ofP-collections by inverse right transla- tion is defined as follows: for aP-collectionC ={CH :H ∈ P}andg ∈G, define

gC ={gCH :H ∈ P}.

Lemma 6.1. For aP-schemeCandg ∈G, theP-collectiongCis also aP-scheme.

Moreover, ifC is antisymmetric (resp. strongly antisymmetric), so isgC.

Proof. This follows in a straightforward manner fromG-equivariance of projections and conjugations (see Lemma 2.2).

SoGalso acts on the set ofP-schemes by inverse right translation, which preserves antisymmetry and strong antisymmetry.

Restriction to a subgroup. We define therestrictionof a subgroup systemPover Gand that ofP-collections to a subgroup ofG.

Definition 6.2(restriction). LetP be a subgroup system over a finite groupG. For a subgroupG0ofG, define

P|G0 :={H ∈ P :H ⊆G0},

which is a subgroup system overG0, called therestrictionofP toG0.

LetC ={CH :H ∈ P}be aP-collection. ForH ∈ P|G0, regardH\G0 as a subset ofH\Gin the obvious way. Then the partition CH ofH\Grestricts to a partition

ofH\G, denoted byCH|G0. Define

C|G0 :={CH|G0 :H ∈ P|G0} which is aP|G0-collection, called therestrictionofC toG0.

Next we show that whenC is P-scheme, its restriction C|G0 to a subgroup G0 is a P|G0-scheme. Moreover, antisymmetry and strong antisymmetry are preserved by restriction.

Lemma 6.2. LetP be a subgroup system over a finite group G. For a subgroup G0 ofGand aP-schemeC, the restrictionC|G0 is aP|G0-scheme. Moreover, ifC is antisymmetric (resp. strongly antisymmetric), so isC|G0.

Proof. We have projections πH,H0 and conjugations cH,g defined between coset spaces H\G for various subgroups H ⊆ G. And we also have projections and conjugations between coset spaces H\G0 for H ⊆ G0. We useπH,H0 0 andc0H,g for the latter maps to distinguish them from the former.

For eachH ∈ P|G0, we have a projectionπH,G0 :H\G→G0\G. This allows us to partitionH\Ginto “fibers” ofπH,G0, i.e., preimages of elements inG0\G:

H\G= a

y∈G0\G

πH,G−1 0(y).

We say x ∈ H\Gis in they-fiber ifπH,G0(x) =y, andyis called the index of x. Note that the subsetH\G0 ⊆H\Gis precisely they-fiber withy=G0e∈G0\G. Consider H, H0 ∈ P|G0 and a map τ : H\G → H0\G that is either a projection πH,H0, or a conjugation cH,g for someg ∈ G0 satisfyingH0 = gHg−1. We claim πH,G0 = πH0,G0 ◦τ, i.e., the mapτ preserves the indices of elements. This can be checked directly: for Hh ∈ H\G, we haveπH,G0(Hh) = G0h. If τ = πH,H0, we haveπH0,G0 ◦τ(Hh) =πH0,G0(H0h) =G0h. And ifτ =cH,g withg ∈G0, we have πH0,G0 ◦τ(Hh) =πH0,G0(H0gh) = G0gh=G0h. So the claim holds.

This means the mapτis also fibered overG0\Gsuch that its “y-fiber”τy :=τ|π−1

H,G0(y)

maps they-fiber ofH\Gto they-fiber ofH0\G. Settingy=G0egives us the map τy :H\G0 →H0\G0that is either the projectionπH,H0 0, or the conjugationc0H,g. From this observation it is easy to see that compatibility, invariance, or regularity ofC|G0 follows from the corresponding property ofC: fixy =G0e. Assume to the

contrary thatC|G0 does not satisfy compatibility. Then some projectionτyH,H0

maps two elements in the same block ofCH|G0 into different blocks ofCH0|G0. But thenτ =πH,H0 also maps these two elements that are in the same block ofCH into different blocks ofCH0, contradicting compatibility ofC. Invariance is proved in the same way except that we consider conjugations instead of projections. For regularity, note that for each projectionπH,H0 0 : H\G0 →H0\G0 and blocksB ∈ CH|G0,B0 ∈ CH0|G0, we haveB = ˜B ∩(H\G0), B0 = ˜B0 ∩(H\G0)whereB˜ ∈CH,B˜0 ∈CH0. And forz ∈B0 we haveπ0−1H,H0(z)∩B =πH,H−1 0(z)∩B˜∩(H\G0) = π−1H,H0(z)∩B˜. Regularity ofC|G0 then follows from regularity ofC.

Now assumeC|G0 is not antisymmetric. Then for someH ∈ P|G0 andg ∈NG0(H), the mapc0H,g restricts to a nontrivial permutation of some blockB ∈ CH|G0. Then we haveg ∈NG(H)andcH,g restricts to a nontrivial permutation ofB˜, whereB˜is the block ofCH satisfyingB˜∩(H\G0) = B. SoC is not antisymmetric.

Finally, assume C|G0 is not strongly antisymmetric. Then there exists a nontrivial permutation τ of a blockB ∈ CH|G0 for some subgroup H ∈ P|G0 such that τ is a composition of mapsσi : Bi−1 → Bi, i = 1. . . , k, where each Bi is a block of CHi|G0,Hi ∈ P|G0, andσiis of the formc0Hi−1,g|Bi−1 (whereg ∈G0),π0Hi−1,H

i|Bi−1, or(πH0

i,Hi−1|Bi)−1 (see Definition 2.7). Each blockBi is of the formB˜i ∩(Hi\G0) for some B˜i ∈ CHi. In the case that σi is of the form (π0H

i,Hi−1|Bi)−1, we know

H0−1

i,Hi−1(z)∩Bi| = |πH−1

i,Hi−1(z)∩B˜i| = 1 for all z ∈ Bi−1. So πHi,Hi−1|Bi is well defined. Then τ = ˜τ|B for the nontrivial permutationτ = σk· · · ◦σ1 of the blockB˜ = ˜B0 ∈ CH, where each map σ˜i is of the formcHi−1,g|B˜i−1, πHi−1,Hi|B˜i−1, or(πHi,Hi−1|B˜i)−1. SoC is not strongly antisymmetric.

Next we describe the analogue of Definition 6.2 form-schemes.

Definition 6.3. Let Π = {P1, . . . , Pm} be an m-scheme on a finite set S. For (x1, . . . , xk)∈S(k) wherek < m, define the(m−k)-collection

Π|x1,...,xk :={P10, . . . , Pm−k0 }

on the setS− {x1, . . . , xk}, wherePi0 is the partition ofS(i)such that two elements (y1, . . . , yi),(y01, . . . , y0i)are in the same block ofS(i)iff(x1, . . . , xk, y1, . . . , yi)and (x1, . . . , xk, y10, . . . , yi0)are in the same block ofS(i+k).

We also have the analogue of Lemma 6.2 form-schemes.

Lemma 6.3. The(m−k)-collectionΠ|x1,...,xkin Definition 6.2 is an(m−k)-scheme.

Moreover, ifΠis antisymmetric (resp. strongly antisymmetric), so isΠ|x1,...,xk. And ifΠdoes not have a matching, neither doesΠ|x1,...,xk.

The proof is straightforward by definition. Indeed, if we viewΠas aP-scheme via Definition 2.12 and Definition 2.13, whereP is the system of stabilizers of depthm with respect to the natural action ofG = Sym(S)on S. ThenΠ|x1,...,xk is simply the restriction of thisP-scheme to the subgroup Gx1,...,xk. We leave the details to the reader.

Passing to quotient groups. LetGbe a finite group and letN be a normal in G.

WriteG¯forG/N andφfor the quotient mapG→G¯.

For a subgroup H ⊆ G¯, the group G acts on H\G¯ by inverse right translation (through its quotient group G¯). The stabilizer of He ∈ H\G¯ is φ−1(H). So by Lemma 2.1, we have an equivalence between the action ofGonH\G¯ and that on φ−1(H)\G, given by the bijection λHe : H\G¯ → φ−1(H)\G sending Hφ(g) to φ−1(H)gforg ∈G.

LetP be a subgroup system over G¯. DefineP˜ = {φ−1(H) : H ∈ P}, which is a subgroup system overG. By identifyingH\G¯withφ−1(H)\GviaλHeforH ∈ P, we see that a P-scheme overG¯ is equivalent to aP˜-scheme overG. This is made formal by the following lemma.

Lemma 6.4. LetP andbe as above. For aP-collection C = {CH : H ∈ P}, define theP-collection˜ C0 ={Cφ0−1(H) :H ∈ P}by choosing

Cφ0−1(H)={λHe(B) :B ∈CH}.

ThenC 7→ C0 is a one-to-one correspondence betweenP-schemes overand- schemes over G. Moreover, C is antisymmetric (resp. strongly antisymmetric) iff C0 is antisymmetric (resp. strongly antisymmetric). And C is homogeneous (resp.

discrete) on a subgroupH ∈ P iffC0 is homogeneous (resp. discrete) onφ−1(H).

Proof. We check that the mapsλHe commute with conjugations and projections:

write πH,H0 and cH,g for conjugations and projections between coset spaces of G¯ and writeπH,H0 0 andc0H,g for those between coset spaces ofG. Then we always have

λH0e◦πH,H0φ0−1(H),φ−1(H0)◦λHe

forH, H ∈ P,H ⊆H, and

λH0e◦cH,φ(g) =c0φ−1(H),g◦λHe.

for H, H0 ∈ P, g ∈ G, H0 = φ(g)Hφ(g)−1. Also note that the maps λHe are bijections. The lemma then follows easily by definition.

We conclude this section by proving Lemma 4.12 using the results developed above.

First we prove the following lemma.

Lemma 6.5. Letk ∈ N+ andGk ⊆ Gk−1 ⊆ · · · ⊆ G1 ⊆ G0 be a chain of finite groups. LetP be a subgroup system overG0. We have:

1. If for alli ∈ [k], all strongly antisymmetric P|Gi−1-schemes are discrete on Gi, then all strongly antisymmetricP-schemes are discrete onGk.

2. If for somei∈[k], all strongly antisymmetricP|Gi−1-schemes are inhomoge- neous onGi, then all strongly antisymmetricP-schemes are inhomogeneous onGk.

The same holds if strong antisymmetry is replaced by antisymmetry.

Proof. Assume that there exists a strongly antisymmetric P-scheme C = {CH : H ∈ P} that is not discrete on Gk. Then there exist two different elements x, x0 ∈ Gk\G lying in the same block of CGk. Pick the greatest integer i ∈ [k]

satisfyingπGk,Gi−1(x) =πGk,Gi−1(x0). Suchiexists asπGk,G0(x) =πGk,G0(x0). Let y = πGk,Gi(x) andy0 = πGk,Gi(x0). Then (1) y 6= y0 by maximality of iand the fact thatx6=x0, (2)y, y0 are in the same block ofCGi by compatibility ofC and the fact thatx, x0are in the same block ofCGk, and (3)πGi,Gi−1(y) =πGi,Gi−1(y0)since πGi,Gi−1(y) =πGk,Gi−1(x)andπGi,Gi−1(y0) = πGk,Gi−1(x0).

SupposeπGi,Gi−1(y) =πGi,Gi−1(y0) =Gi−1g. By replacingC withgC (with respect to the action of Gk on the set of P-schemes by inverse right translation) and applying Lemma 6.1, we may assumeGi−1g =Gi−1e. Then we can writey=Gih and y0 = Gih0 for some h, h0 ∈ Gi−1. By Lemma 6.2, the restriction C|Gi−1 = {CH|Gi−1 : H ∈ P|Gi−1}is a strongly antisymmetricP|Gi−1-scheme. Asy, y0 are in the same block ofCGi, they are also in the same block of CGi|Gi−1. Asy 6= y0, we knowC|Gi−1 is not discrete onGi. This proves the first claim of the lemma.

For the second claim, assume to the contrary that it does not hold. Choosei ∈ [k]

such all strongly antisymmetric P|Gi−1-schemes are inhomogeneous on Gi. Let C = {CH : H ∈ P} be a strongly antisymmetricP-scheme that is homogeneous on Gk. By compatibility, we know C is homogeneous on Gi. Then C|Gi−1 is also homogeneous onGi. It is also strongly antisymmetric by Lemma 6.2, which contradicts the assumption.

The proof for antisymmetry is the same.

Now we are ready to prove Lemma 4.12. For convenience, we restate the lemma.

Lemma 6.6. Let k ∈ N+ and Gk ⊆ Gk−1 ⊆ · · · ⊆ G1 ⊆ G0 be a chain of finite groups. For i ∈ [k], let Ni be a subgroup of Gi that is normal in Gi−1, πi : Gi−1 → Gi−1/Ni be the corresponding quotient map, andPi be a subgroup system overGi−1/Ni that containsGi/Ni. Define

P ={gπi−1(H)g−1 : 1≤i≤k, H ∈ Pi, g ∈G0},

which is a subgroup system overG0 and containsπi−1(Gi/Ni) =Gi for alli∈[k].

Then we have

1. If for alli∈[k], all strongly antisymmetricPi-schemes are discrete onGi/Ni, then all strongly antisymmetricP-schemes are discrete onGk.

2. If for somei∈[k], all strongly antisymmetricPi-schemes are inhomogeneous onGi/Ni, then all strongly antisymmetricP-schemes are inhomogeneous on Gk.

The same holds if strong antisymmetry is replaced by antisymmetry.

Proof. Fixi ∈[k]. By Lemma 6.4 and the definition ofP, if all strongly antisym- metricPi-schemes are discrete (resp. inhomogeneous) onGi/Ni, then all strongly antisymmetric P|Gi−1-schemes are discrete (resp. inhomogeneous) on Gi. The same holds if strong antisymmetry is replaced by antisymmetry. The lemma now follows from Lemma 6.5.