Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
7 Index of a subgroup and order of an element 7.1 Definition. Let H be a subgroup of G. Then [G : H] := the number of distinct left cosets of H in G This number is called the index of H in G. 7.2 Note. 1) The number of left cosets of H in G is the same as the number of right cosets (check!), so also [G : H] = the number of distinct right cosets of H in G 2) If H is a normal subgroup of G then [G : H] = |G/H|. 7.3 Lemma. If H is a subgroup of G then every left (and right) coset of H in G has the same number of elements as H. Proof. If a ∈ G then the map of sets f : H → aH, f (h) = ah is a bijection (check!). 7.4 Theorem (Lagrange). If H is a subgroup of G then |G| = [G : H]|H| Proof. Recall: 1) G is a disjoint union of left cosets of H (5.5) 2) each left coset has as many elements as H (7.3) 20 This gives: |G| = (number of left cosets of H) · (number of elements of H) = [G : H] · |H| 7.5 Definition. If a ∈ G then the order |a| of a is the order of the subgroup �a� ⊆ G generated by a. 7.6 Proposition. |a| = n if n is the smallest positive integer such that an = e, and |a| = ∞ if such n does not exist. Proof. Take the homomorphism f : Z → �a�, f (k) = ak Note that f is onto. If an �= e for all n > 0 then Ker(f ) = {0}. Then f is an isomorphism and so |�a�| = |Z| = ∞. If an = e for some n > 0 and n is the smallest positive number with this property then Ker(f ) = nZ (check!). Therefore �a� ∼ = Z/nZ and so |�a�| = |Z/nZ| = n. 7.7 Proposition. If G is a finite group and a ∈ G then |a| divides |G|. Proof. Follows from Lagrange’s theorem (7.4). 7.8 Note. If a number n divides |G| then G need not contain an element of order n. E.g. |GT | = 6 but GT does not have an element of order 6: |S1 | = |S2 | = |S3 | = 2, |R1 | = |R2 | = 3, |I| = 1 We will see later that if p is a prime number that divides |G| then G contains an element of order p. 21 7.9 Definition. An element a ∈ G is a torsion element if |a| < ∞. A group G is a torsion group if all its elements are torsion elements. A group is torsion free if it has not torsion elements. 7.10 Examples. 1) Every finite group is a torsion group. 2) Q/Z is also a torsion group. 3) Z, Q, R, C are torsion free. 4) Q∗ is neither torsion nor torsion free: 2 ∈ Q∗ has infinite order, −1 ∈ Q∗ has order 2. 22 8 Free groups and presentations of groups 8.1. Let S be a set. A word in S is a finite sequence of the form w = xλ1 1 xλ2 2 · · · · · xλk k where xi ∈ S and λi = ±1 for i = 1, 2, . . . , k. Also, take e := “the empty word” (i.e. the word corresponding to the sequence of length 0). We identify two words if one can be obtained from the other by a series of “cancellations” and “insertions” of subwords of the form xx−1 and x−1 x, e.g.: −1 −1 −1 x1 x2 x3 x−1 3 ∼ x1 x2 ∼ x1 x4 x4 x2 ∼ x1 x4 x1 x1 x4 x2 x1 x−1 1 ∼ e Let F (S) be the set of equivalence classes of words under this equivalence relation. Note. A word is reduced if it does not contain any subwords of the form xx−1 or x−1 x. Every equivalence class in F (S) is represented by a unique reduced word. Define multiplication in F (S) by concatenation of words, e.g.: −1 −1 −1 (x1 x2 x−1 1 ) · (x2 x3 ) = x1 x2 x1 x2 x3 F(S) with this multiplication becomes a group: • the identity element in F (S): e −1 −1 −1 • inverses in F (S): (x1 x2 x−1 = x−1 3 x2 ) 2 x3 x2 x1 23 8.2 Definition. F (S) is called the free group generated by the set S. In general, a group G is free if G ∼ = F (S) for some set S. 8.3 Note. • If S = ∅ then F (S) = {e} is the trivial group. • If S consists of a single element, S = {x} then F (S) is an infinite cyclic group, so F (S) ∼ = Z. 8.4 Note. We have a map of sets i : S → F (S), i(x) = x 8.5 Theorem (The universal property of free groups). Let S be a set and G be a group. For any map of sets f : S → G there exists a unique homomorphism f¯: F (S) → G such that the following diagram commutes: f S � i � � � � � � �G �� � � f¯ F (S) Proof. f¯ is defined by f¯(xλ1 1 xλ2 2 · · · · · xλk k ) := f (x1 )λ1 f (x2 )λ2 · · · · · f (xk )λk 8.6 Corollary. Every group is the homeomorphic image of a free group. Proof. Let G be a group. Take the set S := {xg | g ∈ G} 24 We have a map of sets f : S → G, f (xg ) := g This gives a homomorphism f¯: F (S) → G. Since f is onto thus also f¯ is onto, i.e. G = f¯(F (S)). 8.7 Note. By corollary 8.6 and the First Isomorphism Theorem (6.1) we have G∼ = F (S)/ Ker(f¯) One can show that any subgroup of a free group is free. In particular Ker(f¯) is free. This shows that any group is isomorphic to a quotient of two free groups. 8.8 Definition. Let S be a set and let R be a subset of F (S). Then �S | R� := F (S)/H where H is the smallest normal subgroup of F (S) such that R ⊆ H. We say then that • elements of S are generators of �S | R� • elements of R are relations (or relators) in �S | R� 8.9 Definition. If G is a group and G ∼ = �S | R� for some set S and some subset R ⊆ F (S) then we say that �S | R� is a presentation of G. We say that a group G is finitely presentable if it has a presentation such that both S and R are finite sets. 8.10 Examples. 1) F (S) ∼ = �S | ∅� 25 2) Z/nZ ∼ = �x | xn � Note: in particular Z/6Z ∼ = �x | x6 �. Here is a different presentation of Z/6Z: Z/6Z ∼ = �x, y | x2 , y 3 , xyxy 2 � 3) GT ∼ = �x, y | x2 , y 3 , xyxy� (isomorphsm: x �→ S1 , y �→ R1 ) 4) Recall: Sn = the symmetric group on n letters (1.8) Sn ∼ = �x1 , . . . xn−1 | x2i , (xi xi+1 )3 , (xi xj )2 for |i − j| > 1� (isomorphism: xi �→ σi where σi : {1, . . . , n} → {1, . . . , n}, σi (i) = i + 1, σi (i + 1) = i and σi (j) = j for j �= i, i + 1). 26