Download 7 Index of a subgroup and order of an element

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