Análisis del proceso de preparado de la harina y bebida de Chonta Duro

The groupsA4, S4 andA5 have presentations

A4 ' ha, b|a3 =b3 = (ab)2 = 1i

S4 ' ha, b|a4 =b3 = (ab)2 = 1i

A5 ' ha, b|a5 =b3 = (ab)2 = 1i.

(42)

We verified this presentation of A5 above and the others can be done in the same way. The triples

(2,3,3),(2,3,4),(2,3,5)also arise from the Platonic solids, as follows.

LetGbe a finite group acting on a setX, with the following two properties: 1. For allx∈Xthe stabilizerGxis nontrivial.

2. Any nonidentity element ofGhas exactly two fixed-points inX.

What can we say aboutG? Let|G| = nand let there be r-orbitsO1, . . . ,Or inX, and letmi be the

order of the stabilizer of a point inOi. By the Burnside Counting formula we have r X i=1 1 = 1 n X g∈G |Xg_|= 1 n(|X|+ 2(n−1)) = 1 n " _r X i=1 n mi + 2(n−1) # ,

so r X i=1 1− 1 mi = 2− 2 n. (43) We haver >1, lest 1> 1 mi = 2− 2 n ≥1.

And since1−1/mi ≥1/2, we have

r 2 ≤2 1− 1 n , implyingr≤3. Hencer = 2orr= 3. Ifr= 2we have 1 m1 + 1 m2 = 1 n + 1 n,

which means thatm1 =m2 =n, soXhas just two elements andGacts trivially onX.

We arrive atr= 3, and 1 m1 + 1 m2 + 1 m3 = 1 + 2 n.

Index so thatm1 ≤m2 ≤m3. We cannot havem1 ≥3, lest the left side be≤1. Som1 = 2and

1 m2 + 1 m3 = 1 2+ 2 n.

We cannot havem2 ≥4, lest the left side be≤1/2. Ifm2 = 2thenn = 2m3. Ifm2 = 3we have

1 m3 = 1 6+ 2 n,

so thatm3 = 3,4,5withn = 12,24,60, respectively. To summarize, we have the following possibili-

ties: r m1 m2 m3 n 2 n n − n 3 2 2 m 2m 3 2 3 3 12 3 2 3 4 24 3 2 3 5 60

Now assume that G is a finite group acting by by rotations on the two-dimensional sphere S2. Let

X = {x ∈ S2 _{: G}

x 6= 1}. Then X consists of antipodal pairs {x,−x} on the axes of rotation of

the non-trivial elements ofG. For the first two rows of the table above we haveGcyclic or dihedral, respectively.

For (m1, m2, m3) = (2,3,3), G has 3 elements of order 2 and 8 elements of order three. Since a

2-Sylow subgroup of G has order four and G has no elements of order four, the 2-Sylow must be

K4, and is unique, hence normal in G. But the3-Sylows are not unique, hence are not normal. It

follows thatG ' A4. A 3-Sylow is the stabilizer Gx of a G-orbit {x, y, z, w} in S2 and permutes

{y, z, w}transitively. Hence each of y, z, w have the same distance from x. Likewise Gy permutes

x, z, wtransitively, so these points all have the same distance fromy. It follows thatx, y, z, w are the vertices of a tetrahedron, whose symmetry group isG.

For (m1, m2, m3) = (2,3,4), G has 6 + 3 elements of order two, 8 elements of order three and 6

elements of order four. HenceGhas four3-Sylow subgroups and we have a homomorphismπ :G→ S4. IfP andQare distinct3-Sylows 1hen |N(P)∩N(Q)| ≤ 2, so|kerπ| ≤ 2. If|kerπ| = 2then

kerπ is central in G andimπ = A4. Since there are eight involutions outsidekerπ we would have

at least4involutions inA4, which is not the case. Sokerπ = 1andπ : G → S4 is an isomorphism.

As above, one can show that the six points inS2 with stablizerC4 form the vertices of an octahedron

whose symmetry group isG.

For(m1, m2, m3) = (2,3,5), we have|G| = 60. We show that Gis simple. There are24elements of

order five, henceGhas six Sylow 5-subgroups. LetN be a non-trivial normal subgroup ofG. If|N|

is divisible by5then all six5-Sylows are inN, so |N| ≥ 1 + 24, so|N| ≥30. ThereforeN contains an element of order two. ButGhas fifteen conjugate elements of order two, so|N| ≥25 + 15 > 30, henceN =G, and we have proved thatGis simple. By Cor. 5.19it follows thatG' A5. There is a

G-orbit of12points inS2 _{whose stabilizers have order 5. One can show that these are the vertices of}

an icosahedron.

9

Building new groups from old

9.1

Automorphisms

Recall that an automorphism of a group Gis an isomorphismf : G → GfromG to itself. The set

Aut(G)of automorphisms ofGforms a group under composition, with identity elementIG, given by

IG(g) =g for allg ∈G.

There are various kinds of automorphisms; some automorphisms come fromGitself: For eachg ∈G, letcg :G→Gbe the function given bycg(x) = gxg−1. It is easy to check thatcg ∈Aut(G)and that

c:G→Aut(G)

is a group homomorphism. In general, the homomorphism c is neither injective nor surjective (see examples below). The imageInn(G) = {cg : g ∈ G}ofcis the groupinner automorphisms ofG.

In general, the kernel ofcis the centerZ(G)ofG, andcinduces an isomorphism

G/Z(G)'Inn(G)⊂Aut(G).

You can check that ifα∈Aut(G), then

ThereforeInn(G)is a normal subgroup ofAut(G); the quotient

Out(G) := Aut(G)/Inn(G)

is theouter automorphism groupofG. All of these groups fit into the exact sequence

1−→Z(G)−→G−→c Aut(G)−→Out(G)−→1.

Typically, outer automorphisms ofGarise from conjugation in a larger groupGe, in whichG /Ge. This is why

Aut(An) =Sn,

forn 6= 6(see table below). Often one needs to know which subgroups of G are normalized by Ge. This leads to the notion of characteristic subgroup: We say that a subgroupH ≤ Gis characteristic inGifα(H) =H for everyα ∈ Aut(G). The centerZ(G)and commutator[G, G]are examples of characteristic subgroups in any groupG.

IfH is characteristic inG thenH is normal in G, but not conversely. For example, if G = C2 ×C2

then every subgroup is normal, butAut(G) = GL2(2)moves the subgroups of order two G, so these

are not characteristic. InG = Q8 the subgroupshii, hji, hkihave index two, hence are normal in G,

but there is an automorphismα ∈ Aut(Q8)of order three, sendingi 7→ j 7→ k 7→ i. Hence none of

these subgroups are characteristic.

Examples of Automorphism groups

G Z(G) Aut(G) Inn(G) Out(G)

Z Z C2 1 C2 Zn Zn GLn(Z) 1 GLn(Z) Cn Cn (Z/nZ)× 1 (Z/nZ)× Cn p Cpn GLn(p) 1 GLn(p) Sn,n6= 2,6 1 Sn Sn 1 S6 1 S6·2 S6 C2 An,n6= 2,3,6 1 Sn An C2 A6 1 S6·2 A6 C2 ×C2 D4 C2 D4 D2 C2 Q8 C2 S4 D2 S3

The notationAut(S6)'S6·2means thatAut(S6)fits into an exact sequence

1−→S6

c

and similarly forA6. We will examine this exceptional case in the next section.

9.1.1 Automorphisms ofSn

An automorphism of a group G permutes the conjugacy classes inG, and the inner automorphisms preserve each conjugacy class. Ifα ∈Aut(G)andX, Y are conjugacy classes inGsuch thatα(X) =

Y, then|X|=|Y|and the elements inY have the same order as the elements inX.

Supposeα is an automorphism of the symmetric groupSn, for n ≥ 2. Then αsends the classX of

2-cycles in Sn to another classY of elements of order two such that|Y| =|X| =n(n−1)/2. There

is1≤k≤n/2such that the elements inY have cycle type[2k₁n−2k_{]. One possibility isk = 1, which}

meansY =X, as occurs for the inner automorphisms.

Suppose that k ≥ 2. There are n!/(k!2k(n−2k)!) elements in Sn with cycle type[2k1n−2k], so we

must have

n(n−1)

2 =|X|=|Y|=

n!

k!2k_(n−2k)!.

We rewrite this equation as

k(2k−2)(2k−3)· · ·2·1 = (n−2)(n−3)·(n−4)(n−5)· · ·(n−2k+ 2)(n−2k+ 1).

As2k−2≤n−2and2k−4≤n−4etc, we must haven = 2kand find that

n= 2(n−3)(n−5)· · ·3·1≥2(n−3),

which implies thatn = 6 andk = 3. We note that the classes [21111]and [222]in S6 both have15

elements. We have proved the following.

Lemma 9.1 IfSn has an automorphismα which does not preserve the class of2-cycles, thenn = 6 andαsends the class of2-cycles to the class of222-cycles.

We next investigate the casek= 1.

Lemma 9.2 Supposeα ∈Snpreserves the class of2-cycles. Thenαis inner.

Proof: 24 _{For each 2≤ r ≤ nthere area}

r, br ∈ {1, . . . , n}such thatα(1r) = (ar br). The order of

arandbris not determined, and we will exploit this ambiguity.

We single outr = 2, and seta = a2, b = b2, so that α(1 2) = (a b). Let r ≥ 3. Since (1r)(2 r) =

(1 2r)has order three, we must have

α(1r)·α(1 2) = (ar br)(a b)

also of order three. This means the intersection{ar, br} ∩ {a, b}consists of a single element. Hence

either ar ∈ {a, b} and br ∈ {a, b}/ or vice-versa. Let us switch ar and br if necessary so that

ar ∈ {a, b}andbr ∈ {a, b}/ for allr≥3.

It appears that whetherar =aorar = bcould depend on r. Suppose that for somer, s≥ 3we have

ar =aandas=b. Then

α(1 2r) = α(1r)·α(1 2) = (a br)·(a b) = (a b br),

and

α(1 2s) = α(1s)·α(1 2) = (b bs)·(a b) = (a bsb).

Now(1 2r)(1 2s) = (1r)(2s)has order two, so(a b br)(a bsb)must also have order two. But this is

impossible, for ifbr = bs then(a b br)(a bs b) = e, while ifbr 6= bs then(a b br)(a bs b) = (a bs br)

has order three.

This contradiction shows that eitherar = a for all r ≥ 3 orar = b for allr ≥ 3. We now switch a

andb, if necessary, so thatar = a for allr ≥ 3. Nowα is conjugation by the permutationσ sending

17→a,27→b, andr7→brfor allr≥3.