2.6. Seguidores Solares
2.6.1. Tipos de Seguidores Solares
Sometimes, we need to compute the symmetry group of a geometric object like a regular k-gon (k ≥3) or a regular icosahedron. Also, one may ask to compute the symmetry group of a disjoint n copies of a given geometric object when these n copies can be permuted without any or with some rules.
Definition 3.5.1 Let Gbe a finite group and let H be a permutation group on a finite set Ω. We assume that Ω = {1,2, . . . , n}. Let N be the direct product of n copies of G:
N =G| × · · · ×{z G}
n
.
For any h∈H, the map α(h) :N →N defined by
(g1, . . . , gn)α(h) = (g1h−1, . . . , gnh−1), gi ∈G, i= 1, . . . , n
is an automorphism of N. In fact,α(h) permutes the entries g1, . . . , gn
which maps the i-th entry gi to the ih-th entry of (g1, . . . , gn)α(h), and
then for another h1 ∈H, we have
(g1, . . . , gn)α(hh1) = (g1(hh1)−1, . . . , gn(hh1)−1) = (g
1h−11h−1, . . . , gnh−11h−1) = (g1h−1, . . . , gnh−1)α(h1)
= ((g1, . . . , gn)α(h))α(h1).
Hence, α : H → Aut(N) is a homomorphism. Now, the semidirect product N oα H of N by H with respect to α is called the wreath
product of Gby H and denoted byGoH orGwrH. That is,
GoH={(g1, . . . , gn;h)|gi ∈G, h∈H},
and the multiplication is defined by
(g1, . . . , gn;h)(g01, . . . , gn0;h0) = (g1g10h, . . . , gngn0h;hh0).
Note that the wreath product of an abstract group and a permuta- tion group is an abstract group. Obviously we have
Proposition 3.5.1 For any finite group G, we have
Wreath products 131
Remark: (1) In the wreath product GoH, ˜
G={(g,1, . . . ,1; 1)|g ∈G}
and
˜
H ={(1,1, . . . ,1;h)|h∈H}
are subgroups of G oH isomorphic to G and H, respectively. If we identifyGwith ˜GandHwith ˜H, thenG, H can be viewed as subgroups of GoH, and GoH =hG, Hiif H acts on Ω transitively.
(2) In Definition 3.5.1, if G is also a permutation group on a finite set ∆, then the wreath product G oH can be viewed as a permutation group on ∆×Ω in the following way: For any δ∈∆, i∈Ω, we define
(δ, i)(g1,...,gn;h) = (δgi, ih).
It means that for eachδin thei-th entry, first take agi-action and then
shift it to the ih-th entry.
Example 3.5.1 Leta = (1 2 3)(4 5 6) be a permutation in the sym- metric group S6. Then the centralizer CS6(a) of a is isomorphic to Z3oS2. To generalize it, let a= (· · ·)(· · ·) · · · (· · ·) in Sn be a product
of m disjoint cycles of length` and `m=n. (Such a is called aregular permutation on Sn of type (`m).) Then the centralizer CSn(a) of a is
isomorphic to the wreath product Z`oSm.
Solution: Let a= (· · ·)(· · ·) · · · (· · ·) :=a1·a2 · · · am be a regular
permutation on Sn of type (`m), where each ai is a cycle of length `.
Then the symmetric group Sm can be considered as the permutation
group on the cycles a1, a2, . . . , am and the cyclic group Z` acts on each
cycle ai as a cyclic permutation. Then, an element (c1, . . . , cm;ω) ∈
Z`oSm, whereci ∈Z`andω∈Sm,represents a permutation inSnacting
on the set {1, . . . , n} as follows: First identify the ordered sequence 1,2, . . . , n with the orderedm tuples of length `, that is, [1 2 · · · n] = [1 · · · `][`+ 1 · · · 2`] · · · [(m−1)` · · · m`] := α1·α2 · · · αm. Then,
perform a cyclic permutation of the tuple αi by ci ∈ Z` for each i =
1, . . . , m,and then permuteαi’s through the action of the permutation
ωinSm.For example, the element (0,1;ω)∈Z3oS2, whereω 6= 1∈S2, represents a permutation (1 5 2 6 3 4) by actions [1 2 3 4 5 6] =
[1 2 3][4 5 6] Ã [1 2 3][5 6 4] Ã [5 6 4][1 2 3] = [5 6 4 1 2 3]. Under the such representing, all elements (c1, . . . , cm;ω) ∈ Z`oSm belong to
the centralizer CSn(a) of a. To show CSn(a) ≤ Z` oSm, recall that
|Z`oSm|=|Z`|m|Sm|=`m·m!. On the other hand, one can show that
|CSn(a)|=`m·m! by a direct counting. It completes the argument. ¤
The next example is related to the lexicographic product of two graphs.
Given two graphs Γ and Σ with V(Γ) = {1,2, . . . , n}, the lexico- graphic productΓ[Σ] is defined as the graph with vertex setV(Γ)×V(Σ) such that{(i, δ),(i0, δ0)}is an edge if and only if either{i, i0}is an edge
of Γ, or i= i0 and {δ, δ0} is an edge of Σ. Informally Γ[Σ] is obtained
by taking |V(Γ)|copies of Σ, labelling these copies with the vertices of Γ, and, whenever {i, i0}is an edge of Γ, joining each vertex in the copy
of Σ labelledi to each vertex in the copy of Σ labelled i0.
Example 3.5.2 The automorphism group Aut(Γ[Σ]) of Γ[Σ] contains the wreath product Aut(Σ)oAut(Γ), but may be larger. Let Γ = ¯Knbe
the incomplete graph, that is, the graph having onlyn vertices without any edge. Then the lexicographic product Γ[Σ] =nΣ is then copies of Σ and its automorphism group Aut(n[Σ]) contains the wreath product Aut(Σ)oSn as a subgroup.
Solution: LetV(Γ) = {1,2, . . . , n}. For i ∈V(Γ), let Wi :={(i, δ)|
δ∈V(Σ)}. ThenV(Γ)×V(Σ) =∪iWi. Let Σi be the induced subgraph
with vertex set Wi in Γ[Σ], and let
Gi ={g ∈SV(Γ)×V(Σ) | g|Wi ∈Aut(Σi), g|Wj = 1 for j 6=i}.
Then each Σi is a copy of Σ, Gi ∼= Aut(Σ) and G1 × · · · × Gn ≤
Aut(Γ[Σ]). Now, for eachh∈Aut(Γ), we defineα(h) :G1×· · ·×Gn→
G1× · · · ×Gn by
(g1, . . . , gn)α(h) = (g1h−1, . . . , gnh−1), gi ∈G, i= 1, . . . .n
Thenα(h) is an automorphism. Moreover, the element (g1, . . . , gn;h)∈
Aut(Σ)oAut(Γ) = (G1×G2× · · · ×Gn)oαAut(Γ) acts onV(Γ)×V(Σ)
via
Wreath products 133
as an automorphism of the lexicographic product Γ[Σ]. It gives that Aut(Γ[Σ]) ≥Aut(Σ)oAut(Γ).
Finally, we give an example to show that Aut(Σ) oAut(Γ) can be a proper subgroup of Aut(Γ[Σ]). Let Γ = Kn and Σ = Km, where
Ki denote the complete graph with i vertices. Then Aut(Γ) ∼= Sn,
Aut(Σ) ∼= Sm and Aut(Σ)oAut(Γ) =∼ SmoSn. However, since Γ[Σ] ∼=
Knm, we have Snm ∼= Aut(Γ[Σ])>Aut(Σ)oAut(Γ). ¤
The following proposition shows that the wreath product is asso- ciative. The proof is an elementary exercise.
Proposition 3.5.2 Let G, H, K be permutation groups on the sets Ω,
Θ,Υ, respectively. Then (GoH)oK and Go(H oK) can be viewed as permutation groups on the sets(Ω×Θ)×ΥandΩ×(Θ×Υ), respectively. If we identify both (Ω×Θ)×Υ and Ω×(Θ×Υ) withΩ×Θ×Υ, then we have
(GoH)oK =Go(HoK).
The wreath product can be applied to construct a Sylowp-subgroup of the symmetric group Sn. First, note that if we have a Sylow p-
subgroupP ∈Sylp(Sn),then all Sylowp-subgroups ofSnare conjugates
of P. Let P ∈ Sylp(Sn). Then |P| = |Sn|p = (n!)p. Let |P| = ps(n).
Then by elementary number theory, we have
s(n) = · n p ¸ + · n p2 ¸ +· · ·,
where [a] denotes the integral part of a real numbera. If we express n
in the scale of p,
n =arpr+ar−1pr−1+· · ·+a1p+a0,
where 0≤ai ≤p−1, i = 0,1, . . . , r, andar6= 0, then one can see that
s(n) =ars(pr) +ar−1s(pr−1) +· · ·+a1s(p) +a0s(1) and
Clearly, for n =p, the cyclic permutation group Zp acting on Ω =
{1,2, . . . , p} is a Sylow p-subgroup of Sp. Similarly, if n = p2, one can
show that (n!)p = pp+1 and the permutation group Zp oZp acting on
Ω×Ω is a Sylowp-subgroup of Sp2.This can be extended to construct a Sylow p-subgroup of Sn in the general case n=pr.
Theorem 3.5.3 LetZp be a cyclic permutation group of orderpacting
on Ω = {1,2, . . . , p}. Then P =|ZpoZp{zo · · · oZ}p r
, viewed as a permuta- tion group of degree pr acting onΓ = Ω×Ω× · · · ×Ω
| {z }
r
, has order ps(pr)
, and hence is a Sylow p-subgroup of SΓ.
Proof: By Proposition 3.5.2, wreath products of permutation groups have associative law. So P = ZpoZpo · · · oZp makes sense and can be
viewed as a permutation group on Γ. By Proposition 3.5.1, and by a direct computation we get
|P|=ppr−1+pr−2+···+1 =ps(pr).
Since the order of the Sylow p-subgroup ofSΓ is ps(p
r)
, which is equal to|P|. Thus we have P ∈Sylp(SΓ). ¤
Example 3.5.3 Letp be a prime and ∆ ={1,2, . . . , p2}. Find gener- ators for the Sylow p-subgroup of S∆.
Solution: Let Ω ={1,2, . . . , p}. ThenZp =hai, wherea= (1 2· · · p),
is a Sylow p-subgroup of SΩ. Let Γ = Ω ×Ω. By Theorem 3.5.3,
P = Zp oZp is a Sylow p-subgroup of SΓ. By Remark in page 131,
P = hx, yi, where x = (a,1, . . . ,1; 1) and y = (1,1, . . . ,1;a). The actions of xand y on Γ are defined by
(i, j)x = ½
(ia, j), j = 1
(i, j), j 6= 1; (i, j)y = (i, ja),
Wreath products 135
wherei, j = 1,2, . . . , p. Since the map (i, j)7→i+ (j−1)pis a bijection from Γ to ∆, x and y, as permutations of ∆, have the form:
x = (1 2 · · · p),
y = (1 1 +p · · · 1 +p2−p)(2 2 +p · · · 2 +p2−p)· · · (p 2p · · · p2).
¤
Now, for the group Sn, where n=arpr+ar−1pr−1+· · ·+a1p+a0 as before, we divide n letters into ar blocks of pr letters, ar−1 blocks of pr−1 letters, . . . , a
1 of pletters, and a0 single letters. Then for each block Γ one can find a Sylow p-subgroup of SΓ by Theorem 3.5.3, and then the direct product of these Sylow p-subgroups becomes a Sylow
p-subgroup of Sn.
For example, if n = 15 and p = 3, then 15 = 1·32+ 2·3. Hence, (Z3oZ3)×Z3×Z3 is a Sylow 3-subgroup of S15, whereZ3oZ3 acts on
{1,2,3} × {1,2,3} which corresponds the first 9 letters of 15 letters.
Exercises
3.5.1. Let X be a finite set and let R be an equivalence relation on X such that all equivalence classes are of the same size. Describe the automorphism group
Aut(X;R) :={α∈SX |α preserves the equivalence relation}.
3.5.2. Let Γ and Σ be finite graphs with V(Σ) = {y1, . . . , yd} and let dΓ
denote theddisjoint copies of Γ. Then we have a natural embedding of dΓ into the lexicographic product Γ[Σ] where, for 1 ≤ i ≤ d, the i-th copy of Γ is the subgraph induced on the subset {(x, yi) | x ∈
V(Γ)} of vertices of Γ[Σ]. The deleted lexicographic product, denoted by Γ[Σ]−dΓ,is the graph obtained from Γ[Σ] by deleting all the edges of (this natural embedding of)dΓ. Compute the automorphism group of the deleted lexicographic product Γ[Σ]−dΓ when Γ = Kn and
Σ =Km are complete graphs.
3.5.3. Let G be a finite group, and H a permutation group on Ω. Let Gp∈Sylp(G),Hp∈Sylp(H). Then GpoHp∈Sylp(GoH).
3.5.4. Find the generators of a Sylow 3-subgroupP of the symmetric group S27.
3.5.5. Continuing Exercise 3.5.4, prove that P can be generated by three elements of order 3, butP has elements of order 27.
3.5.6. Prove that the automorphism group of the complete graphKn isSn. 3.5.7. Prove that the automorphism group of the bipartite complete graph
Kn,n isSnoZ2.
3.5.8. Construct a p-Sylow subgroupP of the symmetric groupSp3. Find a generating set for the groupP.
Chapter 4
Nilpotent groups and
solvable groups
Nilpotent and solvable groups are two important classes of groups. In this chapter, we shall give an exposition of some basic concepts and knowledge about these groups, mainly about finite nilpotent and solv- able groups. To give a formal definition of a nilpotent group we first need to study commutators and commutator formulas.