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image of some free group F . Indeed, let F be the free group on the set G;
the map F ’! G is then that induced by 1G : G ’! G. More generally (and
economically), if G = S , and if F is free on S, then the homomorphism
F ’! G induced by the inclusion S ’! G is surjective.
1.8. FREE GROUPS; GENERATORS AND RELATIONS 37
The following notation, though not standard, will prove useful. Let H be
a group, and let R be a subset of H. Denote by R the smallest normal
subgroup of H which contains R. This normal subgroup R is called
the normal closure of R. The reader should note carefully the difference
between R and R . Now assume that G = S is a group, and that
F is free on S, with the obviously induced homomorphism F ’! G. Let
K = ker(F ’! G), and assume that K = R . Then it is customary to say
that G has generators S and relations R, or that G has presentation
G = S| r = e, r " R .
A simple example is in order here. Let D be a dihedral group of order
2k; thus D is generated by elements n, k " D such that nk = h2 = e, hnh =
n-1. Let F be the free group on the set S = {x, y}. The kernel of the
homomorphism F ’! D determined by x ’! n, y ’! h can be shown to be
xk, y2, (yx)2 (we ll prove this below). Thus D has presentation
D = x, y| xk = y2 = (xy)2 = e .
One need not always write each relation in the form r = e. Indeed, the
above presentation might just as well have been written as
D = x, y| xk = y2 = e, yxy = x-1 .
The concept of generators and relations is meaningful in isolation, i.e.,
without reference to a given group G. Thus, if one were to write Consider
the group
G = x, y| x4 = y2 = (xy)2 = e ,
then one means the following. Let F be the free group on the set S = {X, Y }
2
and let K = X4, Y , (XY )2 . Then G is the quotient group F/K, and
the elements x, y are simply the cosets XK, Y K " G/K.
Presented groups, i.e., groups of the form S|R are nice in the sense
that if H is any group and if Æ : S ’! H is any function, then Æ determines
a uniquely defined homomorphism S|R ’! H precisely when Æ kills all
elements of R. This fact is worth displaying conspicuously.
Theorem 1.8.3 Let S| R be a presented group, and let Æ : S ’! H be a
function, where H is a group. Then Æ extends uniquely to a homomorphism
S| R ’! H if and only if
Æ(s2)Æ(s2) · · · Æ(sr) = e
whenever s1s2 · · · sr " R.
38 CHAPTER 1. GROUP THEORY
To see this in practice, consider the presented group G = x, y|x2 = y3 =
(xy)3 = e , and let H = A4, the alternating group on 4 letters. Let Æ be
the assignment
Æ : x ’! (1 2)(3 4); y ’! (1 2 3);
since ((1 2)(3 4))2 = (1 2 3)3 = ((1 2)(3 4)(1 2 3))3 = e, the above theorem
guarantees that Æ extends to a homorphism Æ : x, y|x2 = y3 = (xy)3 =
e ’! A4. Furthermore, since A4 = (1 2)(3 4), (1 2 3) , we conclude that
Æ is onto. That Æ is actually an isomorphism is a little more difficult (see
Exercise 11); we turn now to issues of this type.
Consider again the dihedral group D = n, h of order 2k, where nk =
h2 = e, hnh = n-1. Set G = x, y| xk = y2 = (xy)2 = e . We have
immediately that the map x ’! n, y ’! h determines a surjective homo-
morphism G ’! D. Since |D| = 2k, we will get an isomorphism as soon
as we learn that |G| d" 2k. This isn t too hard to show. Indeed, note that
the relation (xy)2 = e implies that yx = x-1y. From this it follows easily
that any element of G can be written in the form xayb. Furthermore, as
xk = e = y2, we see also that every element of G can be written as xayb,
where 0 d" a d" k - 1, 0 d" b d" 1. Thus it follows immediately that |G| d" 2k,
and we are done.
The general question of calculating the order of a group given by gen-
erators and relations is not only difficult, but, in certain instances, can be
shown to be impossible. (This is a consequence of the unsolvability of the
so-called word problem in group theory.) Consider the following fairly simple
example: G = x, y| xy = y2x, yx = x2y . We get
y-1xy = y-1y2x = yx = x2y = xxy,
so that y-1 = x. But then
e = xy = y2x = y(yx) = y,
so y = e, implying that x = e. In other words, the relations imposed on
the generating elements of G are so destructive that the group defined is
actually the trivial group.
Exercises 1.8
1. Show that if F is free on the set S via the map Æ : S ’! F , then
1.8. FREE GROUPS; GENERATORS AND RELATIONS 39
(a) Æ is injective.
(b) F = Æ(S) .
2. Let |S| = 1, and let F be free on S. Prove that F (Z, +).
=
3. Let |S| e" 2, and let F be free on S. Prove that F is not abelian.
4. Let F be free on the set S, and let F0 be the subgroup of F generated
by S0 †" S. Prove that F0 is free on S0.
5. Prove that x, y, z| yxy2z4 = e is a free group. (Hint: it is free on
{y, z}. )
6. Prove that x, y| yx = x2y, xy3 = y2x = {e}.
7. Prove that x, y| xy2 = y3x, x2y = yx3 = {e}.
8. Let G be a free group on a set of more than one element. Prove that
G/G is infinite.
9. Compute the structure of G/G for each finitely presented group below.
(i) x, y| x6 = y4 = e, x3 = y2 ,
(ii) x, y| x3 = y2 = e ,
(iii) x, y| x2 = y3 = (xy)3 = e ,
(iv) x, y| x2 = y3 = (xy)4 = e ,
(v) x, y| x2 = y3 = (xy)5 = e .
10. Prove that
x, y| x4 = e, y2 = x2, yxy-1 = x-1 r, s, t|r2 = s2 = t2 = rst .
=
11. Show that | x, y| x2 = y3 = (xy)3 = e | d" 12. Conclude that A4
=
x, y| x2 = y3 = (xy)3 = e .
12. (a) Show that | x, y| x2 = y3 = (xy)4 = e | = 24.
(b) Show that | x, y| x2 = y3 = (xy)5 = e | = 60.
13. Let D = D8, the dihedral group of order 8. Prove that Aut(D) D8.
=
40 CHAPTER 1. GROUP THEORY
14. Let k, l, m be positive integers and set
D = D(k, l, m) = ±, ²| ±k = ²l = (±²)m = e ,
" = "(k, l, m) = a, b, c| a2 = b2 = c2 = (ab)l = (bc)l = (ac)m = e .
Prove that D is isomorphic with a subgroup of index 2 in ".
15. Let Q2n+1 be the generalized quaternion group of order 2n+1. Show
that Q2n+1 has presentation
a a-1
Q2n+1 = x, y| x2 = e, y4 = x2 , yxy-1 = x-1 .
(Hint: see Exercise 11.)
16. The Free Product of Groups. Let Ai, i " I be a family of groups. A
free product of the groups Ai, i " I is a group P , together with a
family of homomorphisms µi : Ai ’! P , such that if ¸i : Ai ’! G is any
family of homomorphisms of the groups Ai into a group G, then there
exists a unique homomorphism f : P ’! G making the diagram below
commute for each i " I:
µi
Ai P
i f
G
Prove that the free product of the groups Ai, i " I exists and is unique
up to isomorphism. (Hint: the uniqueness is just the usual categorical [ Pobierz całość w formacie PDF ]