Ideas in finite group theory

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A = (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (234), (243), (134), (143)}.

4 Note that although 6 divides 12, A has no subgroups of order 6.

4

Although it is not true that there exist subgroups of G of order any given divisor

|G|,

of Cauchy’s theorem implies that they exist if the given divisor is a prime.

|G|

The next natural step is asking what happens with prime-powers. Suppose

IDEAS IN FINITE GROUP THEORY 7

k ≤

is divisible by a prime-power p . Can we always find a subgroup H G with

k

|H| = p ? The answer is yes. n

|G|

Theorem 5 (Sylow (1832 - 1918)). Let G be a finite group and write = mp

where p is a prime and m is not divisible by p.

n

• G contains a subgroup P of order p . P is called “Sylow p-subgroup” of G.

k

• ≤ ≤

G contains a subgroup of order p for every 0 k n.

• If P, Q are two Sylow p-subgroups of G then they are conjugated: there

−1

∈

exists g G such that g P g = Q.

• The number of Sylow p-subgroups of G is congruent to 1 mod p.

• |H|

If H is a subgroup of G such that is a power of p then there exists a

≤

Sylow p-subgroup P of G such that H P .

Consider the following example. Let F = and let

Z/5Z

a 0

{ ∈ 6

G := : a, b, c F, a, c = 0},

b c

a 0

{ ∈

H := G : b = 0},

b c

a 0

{ ∈

K := G : a = c = 1}.

b c 2 4

• · ·

G is a group (with respect to multiplication) of order 4 5 = 2 5,

2 4

• |H| ⇒

= 4 = 2 H is a Sylow 2-subgroup of G and

• |K| ⇒

= 5 K is a Sylow 5-subgroup of G.

Finally, Cayley theorem says that every finite group can be found inside some

symmetric group.

Theorem 6 (Cayley (1821 - 1895)). Let G be a group. Then the map

→ 7→ 7→

G Sym(G), g (x gx)

is an injective homomorphism.

In particular, G is isomorphic with a subgroup of Sym(G).

Corollary 1. Let G be a finite group. There exists a positive integer n such that

G is isomorphic with a subgroup of Sym(n). |G|.

Cayley’s theorem says that we may choose n = But sometimes we can

choose a smaller n (cf. [7]). ∈

For example, if G is the Galois group of the polynomial f (X) with n

Q[X]

f

distinct roots, then the permutation action of G on the n roots gives an injective

f

→

homomorphism G Sym(n).

f 4. Simple groups, solvable groups {1}

A group G is said to be “simple” if the only normal subgroups of G are and

G. Since every subgroup of an abelian group is normal, it is easy to show that

Remark 3 (Abelian simple groups). Abelian simple groups are the cyclic groups

of prime order, ∼

2 p−1 p

{g, {1, −

C = g , . . . , g , g = 1} (Z/pZ, +) = 2, . . . , p 1, p = 0}.

=

p

8 MARTINO GARONZI

The following results provide infinite families of non-abelian simple groups.

≥

Theorem 7 (Alternating Groups). If n 5 is an integer, Alt(n) is a non-abelian

simple group.

Theorem 8 (Projective Linear Groups). Let F be a field, and let GL(n, F ) be the

group of invertible matrices over F . Let SL(n, F ) be the subgroup of GL(n, F ) con-

sisting of matrices of determinant 1. Let Z be the subgroup of GL(n, F ) consisting

≥ |F | ≥

of scalar matrices. If n 2 and 4, the quotient

∩

P SL(n, F ) := SL(n, F )/Z SL(n, F )

(projective linear group) is a non-abelian simple group.

Given a finite group G, we can costruct longest possible chains of subgroups of

the form

{1} · · ·

= G G G G = G.

0 1 2 k

Maximality of k implies that the factor groups G /G are all simple groups.

i i−1

Such chain is called “composition series” and its factors G /G are called

i i−1

“composition factors”.

Theorem 9 (Jordan-Holder). Any two composition series of a given finite group

have the same length and the same composition factors (up to reordering and iso-

morphism).

For example the composition factors of the cyclic group C correspond to the

n

2 · ·

prime divisors of n, counted with multiplicity. If n = 60 = 2 3 5,

∼ ∼ ∼

30 15 5

hg i( hg i( hg i( hgi

1 C ) C ) C ) = C .

= = =

2 4 12 60

Definition 2 (Solvable groups). If the composition factors of the finite group G

are all abelian (hence cyclic of prime order) then G is said to be solvable.

∈

Evariste Galois proved that the zeros of a polynomial f (X) can be

Q[X]

expressed by starting from the elements of and performing sums, differences,

Q

products, divisions, and root extractions if and only if the Galois group G is

f

solvable. In this case f (X) is said to be “solvable by radicals”. Let us give some

examples. 4 − ∈

The Galois group of f (X) = X 4X + 2 is S , so f (X) is solvable by

Z[X] 4

radicals. Indeed, S is solvable:

4 / / /

/ C C C

C 2 3 2

2 h(12)(34)i

{1} O (S ) A S

2 4 4 4

Arrows are inclusions. O (S ) denotes the intersection of the Sylow 2-subgroups of

2 4 ×

S : it is a normal subgroup of S of order 4 isomorphic to the Klein group C C .

4 4 2 2

3

|S | ·

The composition factors of S are C (three times) and C . = 24 = 2 3.

4 2 3 4

More generally, all polynomials of degree 2, 3, 4 are solvable by radicals. Indeed,

all subgroups of Sym(4) are solvable. On the other hand, the symmetric group S

n

≥

is not solvable when n 5: / /

A C 2

n

{1} A S

n n

The composition factors of S are A (not abelian) and C .

n n 2

IDEAS IN FINITE GROUP THEORY 9

2

Let a, b, c be indeterminates over The roots of the polynomial P (X) = aX +

Q.

bX + c are given by the well-known formula

√ 2

−b ± −

b 4ac

x = .

1,2 2a

It follows that P (X) is solvable by radicals over b, c). I want to consider now

Q(a,

degrees larger than 2.

Let a , . . . , a be indeterminates over It is interesting to ask when the

Q.

1 n−1

generic polynomial of degree n

n n−1 · · ·

P (X) = X + a X + + a X + a

n−1 1 0

is solvable by radicals over the field generated by its coefficients, , . . . , a ).

Q(a

1 n−1

In other words, we ask when the roots of P (X) can be expressed by starting from

the coefficients a , . . . , a and performing sums, differences, products, divisions

0 n−1

and root extractions. It turns out that P (X) is irreducible in , . . . , a )[X],

Q(a

1 n−1

it has distinct roots (as does any irreducible polynomial in characteristic zero) and

its Galois group over , . . . , a ) is Sym(n). Since, as we have seen, Sym(n)

Q(a

1 n−1

≥

is not a solvable group if n 5, it follows that P (X) is solvable by radicals if and

≤

only if n 4.

As you might have noticed, above I used the expression “over the field generated

by its coefficients”. Let us clarify this. If F/K (to be read “F over K”) is any field

extension (meaning that F and K are fields and F contains K) then the Galois

G(F/K),

group of F/K, denoted is defined to be the set of all field automorphisms

∈

g of F such that (*) g(a) = a for every a K. Note that if K = then condition

Q

⊆

(*) is automathic. The inclusion K F gives F a canonical structure of K-vector

space. The extension F/K is said to be “finite” if F has finite dimension as K-vector

space. The “degree” of the finite field extension F/K, usually denoted [F : K], is

the dimension dim (F ). So for example is a finite field extension of degree 2

C/R

K

with Galois group C (its two elements are the identity and the complex conjugation

2

7→ −

a + ib a ib). A finite extension F/K is said to be a Galois extension if

{a ∈ ∀g ∈ G(F/K)}

F : g(a) = a = K.

It turns out that an extension F/K is a Galois extension if and only if [F : K] =

|G(F/K)|, that is, the degree equals the size of the Galois group. For example,

∈

whenever f (X) is a polynomial in with roots a , . . . , a the extension

Q[X], C,

1 n 2 ∈

, . . . , a )/Q is a Galois extension. For example consider f (X) = X +1

Q(a R[X]:

1 n −i),

since = = the extension is Galois.

C R(i) R(i, C/R √

3 2)/Q.

Here is an example of an extension that is not Galois: F/K = Q(

√

3

→

Indeed, the only K-automorphism of F is the identity, id : F F ( 2 is the only

F

3 − |G(F/K)|

root of X 2 that belongs to F !), so = 1, while the degree [F : K] is 3:

√ √

3 3

a K-basis is given by 1, 2, 4. However, F is contained in a Galois extension of

3 −

K: b, c)/Q, where a, b, c are the three roots of X 2 in It is an extension

Q(a, C.

of degree 6 with Galois group isomorphic to Sym(3).

Here comes the main property of Galois extensions, which makes them very nice.

If F/K is a finite Galois extension then the correspondences

7→ {a ∈ ∀h ∈

H F : h(a) = a H}

10 MARTINO GARONZI

7→ {g ∈ G(F/K) ∀a ∈

L : g(a) = a L}

provide inclusion-reversing bijections, inverses of each other, between the family of

G(F/K) ⊆ ⊆

subgroups of and the family of fields L such that K L F (intermediate

fields of F/K). So, if you want to see how the intermediate field lattice of a Galois

extension looks like just take the subgroup lattice of its Galois group and turn it

upside-down (remember that inclusions are reversed).

I now spend some words on the classification of the finite simple groups. The

starting point for the classification was the following beautiful result, which is known

a