The General Linear Group Over the ﬁeld Fq, Is the Group of Invertible N × N Matrices with Coeﬃcients in Fq

Home , Center (group theory), Cyclic group, Direct product of groups, Finite field, Finite group, General linear group

ALGEBRA - LECTURE II

1. General Linear Group

Let Fq be a finite field of order q. Then GLn(q), the general linear group over the field Fq, is the group of invertible n × n matrices with coefficients in Fq. We shall now compute n the order of this group. Note that columns of an invertible matrix give a basis of V = Fq . Conversely, if v1, v2, . . . vn is a basis of V then there is an invertible matrix with these vectors as columns. Thus we need to count the bases of V . This is easy, as v1 can be any non-zero vector, then v2 any vector in V not on the line spanned by v1, then v3 any vector in V not n n on the plane spanned by v1 and v2, etc... Thus, there are q − 1 choices for v1, then q − q n 2 choices for v2, then q − q choices for v3, etc. Hence the order of GLn(q) is

n−1 Y n i |GLn(q)| = (q − q ). i=0

2. Group action Let X be a set and G a group. A group action of G on X is a map G × X → X (g, x) 7→ gx such that (1) 1 · x = x for every x in X. (Here 1 is the identity in G.) (2) (gh)x = g(hx) for every x in X and g, h in G. Examples: n (1) G = GLn(q) and X = Fq . The action is given by matrix multiplication. (2) X = G and the action is the conjugation action of G. (3) X is the set of all subgroups in G and the action is the conjugation action. An action of G on a set X deﬁnes an equivalence relation by x ∼ y if there exists an element g in G such that y = gx. The equivalence classes are called G-orbits. If x is in X, the equivalence class of x (or its G-orbit) is denoted by xG = {gx|g ∈ G}. The orbits in the three examples respectively are n (1) Two orbits. One is {0} and the other Fq \{0}. (2) The orbits are conjugacy classes of elements of G. (3) The orbits are conjugacy classes of subgroups of G. Let x be in X. The stabilizer of x in G is deﬁned by

SG(x) = {g ∈ G|gx = x}. 1 2 ALGEBRA - LECTURE II

Note that the group action axioms guarantee that the stabilizer SG(x) is a subgroup of G. For example, if X = G and the action is the conjugation action, then the stabilizer of x in G is the centralizer CG(x) of x in G: −1 SG(x) = CG(x) = {g ∈ G|gxg = x}. If X is the set of all subgroups of G and the action is the conjugation action, then the stabilizer of a subgroup H is the normalizer NG(H) of H in G: −1 SG(H) = NG(H) = {g ∈ G|gHg = H}.

3. Class formula Assume now that G is finite. Then the orbits are finite also. I claim that the number of element in the orbit of x is equal to [G : SG(x)], the index of the subgroup SG(x) in G. This is easy. Let g1, g2,... be a list of all elements of G. Then g1x, g2x, . . . are all elements G −1 −1 of x . Now gix = gjx is equivalent to x = (gi gj)x which means that gi gj is in SG(x) G or gj ∈ giSG(x). It follows that the elements of x correspond to SG(x)-cosets in G. This proves the claim. Assume further that X is finite. Let x1, . . . , xn representatives of all G-orbits in X. The the following class formula is obvious: n X |X| = [G : SG(xi)]. i=1 We discuss some applications of the class formula. The first is well known: Proposition 1. Let p be a prime and G be a group of order pm. Then p divides the order of the center Z(G) of G. In particular, the center of G is not trivial. Proof. Consider the special case of the class formula where G is acting on itself by conjugation. Note that the conjugation class of x consists of only one element (so the class is {x}) if and only ix x is in the center Z(G) of the group G. If x1, . . . , xn are representatives of non-central conjugacy classes then the class formula takes the form n X |G| = |Z(G)| + [G : CG(xi)]. i=1

Since the index [G : CG(xi)] is a non-trivial factor of G, it is a proper power of p. It follows that p divides the left side of the class equation and the terms [G : CG(xi)] on the right side. It follows that p divides the order of Z(G), as claimed. The proposition is proved.

Exercise: Let p be a prime and U be the subgroup of GL3(p) consisting of all matrices of the form  1 x z   0 1 y  . 0 0 1 Then |U| = p3, and the center is non-trivial, according to the previous proposition. Find that center. The second application of the class formula is more exciting. ALGEBRA - LECTURE II 3

Proposition 2. Let p be a prime and G a group whose order is divisible by p. Then there is an element in G of order p. Proof. Let X be the subset of Gp defined by p X = {(x1, x2, . . . , xp) ∈ G |x1x2 ··· xp = 1}. p−1 Note that |X| = |G| since x1 through xp−1 can be freely picked and then xp is fixed. The group acting on X will not be G but a cyclic group Cp = hσi of order p where σ is defined by

σ(x1, x2, . . . , xp) = (xp, x1, . . . , xp−1).

Since the order of Cp is the prime p, an orbit in X has either one or p elements. An orbit has one element if and only if x1 = x2 = ··· = xp. Thus single element orbits consist of (x, x, . . . , x) where x is in G such that xp = 1. Of course, one such x is the identity element. The class formula now implies that the number of orbits with one element is divisible by p. p This proves that there exists x 6= 1 such that x = 1. The proposition is proved. 4. Semi-direct product of groups Let T and U be two subgroups of G. Assume that T is contained in the normalizer of U in G. Then UT , the set of all possible products ut where u ∈ U and t ∈ T is a group. Indeed, −1 (u1t1)(u2t2) = (u1t1u2t1 )(t1t2) = (u1u3)(t1t2) −1 where u3 = t1u2t1 ∈ U which shows that the set TU is closed under multiplication. Since (ut)−1 = (t−1u−1 t)t−1 the set TU is closed under taking inverses as well. It is also convenient to consider U × T , a semi-direct product of T and U. As a set it is equal to the direct product of U and T . However, the multiplication is defined by −1 (u1, t1) · (u2, t2) = (u1u3, t1t2) where u3 = t1u2t1 ∈ U. Note that there is a natural homomorphism from U × T onto UT defined by (u, t) 7→ ut. The kernel consists of elements (u, u−1) where u is in U ∩ T . If the groups are finite, it follows that the order of UT is |U||T | |UT | = . |U ∩ T | Example: Let p be a prime and G = GL3(p). Let U be the group of upper triangular matrices with 1 on the diagonal (so called unipotent matrices), and let T be the group of all diagonal matrices. Then T normalizes U. Since U ∩ T = {1} the group UT is isomorphic to the semi-direct product of U and T .

5. Sylow Theorems

Let G be a ﬁnite group and p a prime number. By |G|p we shall denote the biggest power of p dividing |G|. A subgroup P of G of order pr is called a p-subgroup of G. If the order of P is equal to |G|p then P is called a Sylow p-subgroup of G.

Theorem 3. Assume that |G|p 6= 1. Let X be the set of Sylow p-subgroups. Then (1) The set X is non-empty. (2) |X| ≡ 1 (mod p). (3) All Sylow p-subgroups are G-conjugated. 4 ALGEBRA - LECTURE II

(4) Any p-subgroup is contained in a Sylow p-subgroup. Proof. Let X be the set of all p-subgroups maximal with respect to inclusions. Note that this set is non-trivial since G has elements of order p. (In the last step of the proof we will show that elements of X have the order |G|p, so it OK to use the letter X.) We show first: Lemma 4. (1) |X| ≡ 1 (mod p). (2) The set X consists of only one conjugacy class. Proof. Let P be in X. Then P acts on X by conjugation. I claim that P fixes P and no other elements in X. Indeed, if P fixes Q, another maximal p-subgroup, that means that P is contained in NG(Q). It follows that QP is a subgroup of G. Its order is |Q||P |/|Q ∩ P |, so QP is also a p-subgroup. But QP contains both Q and P . This is a contradiction since P and Q are two different maximal p-subgroups. It follows that the P -orbit of Q 6= P is non-trivial and, therefore, has the order equal to pr for some r ≥ 1. This shows the first part of Lemma. The second is now easy to check. Assume that X = Y ∪ Z where both Y and Z are conjugation invariant. Let P be in Y (so it is not in Z). Since the order of any P -orbit in X different from {P } is divisible by p, it follows that |Y | ≡ 1 (mod p) and |Z| ≡ 0 (mod p). Now let Q be in Z. Since the order of any Q-orbit in X different from {Q} is divisible by p, it follows that |Y | ≡ 0 (mod p) and |Z| ≡ 1 (mod p). Thus we have derived a contradiction to the assumption that X does not consist of only one G-orbit. So far we have shown that the set X of maximal p-subgroups satisfies |X| ≡ 1 (mod p) and that it consist of one G-conjugacy class. In particular, all maximal p-subgroups have the same cardinality. Thus, to finish the proof we need to show that elements of X have the order |G|p. This is shown by induction on the order of G. Let P be in X. If P is normal in G, then we can consider the quotient G/P . The inverse image of a Sylow p-subgroup in G/P (which exists by induction assumption) is a Sylow p-subgroup in G. Since P is a maximal p-subgroup, it has to be a Sylow p-subgroup in G. If P is not normal subgroup in G then the normalizer NG(P ) is a proper subgroup of G. By the induction assumption on NG(P ), the group P has to be the Sylow p-subgroup of NG(P ). (Here we use that P is a maximal p-subgroup in G and hence also in NG(P ).) Since |G| |X| = |NG(P )| and |X| is not divisible by p, it follows that |G|p = |NG(P )|p and P is a Sylow p-subgroup in G as well. The theorem is proved. 6. Application to groups of order pq Let p and q be two primes. Let G be a group of order pq. Let P be a Sylow p-subgroup. ∼ Since |P | = p, we must have P = Cp. The number of Sylow p-subgroups is congruent to 1 modulo p, one one hand, and equal to the index of the normalizer of P in G, on the other ALGEBRA - LECTURE II 5 hand. Since NG(P ) contains P , the theorem of Lagrange implies that the normalizer is either P or G. If the normalizer is P , then there are q Sylow p-subgroups. This is, of course possible, if q ≡ 1 (mod p). Proposition 5. Let G be a group of order pq where p and q are two different primes such that p does not divide q − 1 and q does not divide p − 1. Then G is cyclic. Proof. Let P and Q be the Sylow p and q-subgroups of G, respectively. Assumptions on the ∼ ∼ primes imply that that P = Cp and Q = Cq are normal. Let x be a generator of P and y a generator of Q. Then the commutator [x, y] = xyx−1y−1 = (xyx−1)y = x(yx−1y−1) clearly sits in both P and Q, by normality of both subgroups. Since P ∩ Q = {1} by the ∼ ∼ theorem of Lagrange, we see that x and y commute. It follows that P × Q = Cp × Cq = Cpq is a subgroup of G. The proposition is proved. Of course, one example is pq = 5 · 7 = 35.

Exercises

1) Let p be a prime and U be the subgroup of GL3(p) consisting of all matrices of the form  1 x z   0 1 y  . 0 0 1

Compute the center of U and the normalizer of U in GL3(p). What is the number of Sylow p-subgroups in GL3(p)?

2) Let G be a group and Z(G) its center. Show that G is commutative if G/Z(G) is cyclic. Let p be a prime. Show that a group G of order p2 is commutative.

3) An action of a group G on a set X deﬁnes a relation on X by x ∼ y if there exists an element g in G such that y = gx. Show that this is an equivalence relation on X.

4) Let p > q be two primes. Let G be a group of order pq2. Show that G has a normal Sylow p-subgroup unless |G| = 12. Find a group of order 12 which does not have a normal Sylow 3-subgroup.

5) A group G is called simple if it has no no-trivial normal subgroup. Show that there is no simple group of order 992 = 31 · 32. Hint: if a Sylow 31-subgroup is not normal, how many elements of order 31 does G have?