Download Introduction to Group Theory (cont.) 1. Generic Constructions of

Introduction to Group Theory (cont.)
1. Generic Constructions of Groups
Definition: Let G be a group and S an arbitrary set. M(S,G) is defined to be the
set of all maps from S into G. Assume that the product of two elements of G
is denoted by · . Then the product of two maps f,g: S → G is the map
f ⋅ g : S → G; x → f (x ) ⋅ g (x ).
Note: If S is the empty set, then there is exactly one map from S into G. In
this case, M(S,G) with · is a group with exactly one element and the product of
the map with itself is itself (by abuse of notation). Don’t consider this case when
proving the following proposition.
Proposition: Let (G,·) be a group and S an arbitrary set. Then (M(S,G), · ) is a
group.
Definition: Let S ≠ ∅ be a set. (∅ is the empty set, i.e. the unique set without
any elements.) A permutation of S is a map from S → S which is bijective. (A
map f : S → S is bijective if there is a map g : S → S such that g ° f = f ° g =
idS, where idS : S → S; s → s is the “identity map”. The composite g ° f of two
maps f: A → B and g: B → C is the map g ° f : A → C; a → g(f(a)). )
Proposition: Let S be a non-empty set. The set P(S) of permutations together
with the composition as the multiplication is a group.
Grothendiek Group
Let (S, +) be a set with an operation that has an identity element 0 and is
associative and commutative. Assume that the cancellation law holds in S, that
is, that
∀x , y , z ∈ S : x + z = y + z ⇒ x = y .
Define a relation on S×S = { (s,t) | s,t ∈ S} (the set of all pairs of elements in S)
by (a,b) ~ (c,d) iff a+d = b+c.
Lemma: ~ is an “equivalence relation”, that is, the following are true for all a, b,
c, d, e, f ∈ S :
(a,b) ~ (a,b).
(i)
(a,b) ~ (c,d) ⇒ (c,d) ~ (a,b)
(ii)
(a,b) ~ (c,d) and (c,d) ~ (e,f) ⇒ (a,b) ~ (e,f).
(iii)
For (a,b) ∈ S × S define [a,b] = { (x,y) ∈ S × S | (x,y) ~ (a,b) }. This is called the
equivalence class of a and b. Denote G(S) = {[a,b] | a,b ∈ S × S}.
Lemma: The binary operation G(S) × G(S) → G(S); ([a,b],[c,d]) → [a+c,b+d] is
well-defined.
(Basically, the definition of the product depends on the particular choice of
elements representing it. Thus, you have to show that if (a,b)~(a’,b’) and
(c,d)~(c’,d’) that then (a+c,b+d) ~ (a’+c’,b’+d’). )
Proposition: In this case:
( G(S), + ) is a group.
(i)
ι: S → G(S); s → [s,0] is an embedding, that is, if for s, t, ∈ S ι(s) =
(ii)
ι(t) is true, then necessarily s = t.