Download Definition of a quotient group. Let N ¢ G and consider as before the

Definition of a quotient group. Let N G and consider as before the equivalence relation
a ∼ b iff ab−1 ∈ N.
We define a new group G/N , called the quotient of G by N as follows:
(1) The elements of G/N are the equivalence classes, that is the right cosets (or the left cosets
or just cosets as these are equal).
(2) The group multiplication is derived from G by using the set multiplication. Let a, b ∈ G. As
N G we have aN = N a. Hence
(∗)
N a · N b = N · N · a · b = N ab.
Expressed differently, we have
[a] · [b] = [ab].
It is now easy to see that G/N is a group:
(i) The associative law holds as it holds in G. Thus ([a] · [b]) · [c] = [ab] · [c] = [(ab)c] =
[a(bc)] = [a] · [bc] = [a] · ([b] · [c]).
(ii) The identity is [e] = N as [a] · [e] = [a · e] = [a] and [e] · [a] = [e · a] = [a].
(iii) The inverse of [a] = N a is [a−1 ] = N a−1 as [a] · [a−1 ] = [a · a−1 ] = [e] and [a−1 ] · [a] = [e].
Remark. The number of elements in G/N is [G : N ].
Definition. Let φ : G → H be a homomorphism
(1) The image of φ is the set
im φ = {h ∈ H : h = φ(g) for some g ∈ G}.
The kernel of φ is the set
ker φ = {g ∈ G : φ(g) = eH }.
On exercise sheet 4 we see that im φ ≤ H and ker ≤ G.
Lemma 3.3 A homomorphism φ : G → H is injective if and only if ker φ = {eG }.
Proof (⇒) If φ(g) = eH then φ(g) = φ(eG ) and injectivity gives us g = eG .
(⇐) We have
φ(a) = φ(b) ⇒ eH = φ(a) · φ(b)−1 = φ(a) · φ(b−1 ) = φ(ab−1 ).
By our assumption, it follows that eG = ab−1 and therefore a = b. This proves that φ is injective.
2.
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Remark. If φ : G → H is an injective homomorphism then we have a bijective homomorphism from G to im φ. So G is then isomorphic to the subgroup im φ of H.
Theorem 3.4 (The Fundamental Isomorphism Theorem). Let φ : G → H be a homomorphism.
Then
G/Ker φ ≃ Im φ.
Proof Let N = Ker φ and I = Im φ. Consider the map
Φ : G/N → I, N a 7→ φ(a).
(1) Φ is well defined: If [a] = [b] then ab−1 ∈ N and therefore eH = φ(ab−1 ) = φ(a)φ(b)−1 and
thus φ(a) = φ(b).
(2) Φ is a homomorphism: Φ([a] · [b]) = Φ([ab]) = φ(ab) = φ(a)φ(b) = Φ([a])Φ([b]).
(3) Φ is bijective: Φ is clearly surjective. To show that it is injective notice that if Φ([a]) = Φ([b])
then φ(a) = φ(b) and thus eH = φ(a)φ(b)−1 = φ(ab−1 ). It follows that ab−1 ∈ N and thus
[a] = [b]. 2
Remark. Let N be a normal subgroup of G and consider the map
φ : G → G/N, a 7→ [a] = aN.
Then φ is a surjective homomorphism as φ(ab) = [ab] = [a] · [b] = φ(a) · φ(b). We also have that
a ∈ ker φ iff [e] = φ(a) = [a] iff a ∈ N . So any normal subgroup N of G is a kernel of some
homomorphism from G to another group.
Example Let G = hai be a cyclic group. Consider the homomorphism
φ : Z → G, n 7→ an .
If G is finite of order n the Ker φ = nZ and if G is infinite then Ker φ = {0} = 0Z. So by the
Isomorphism Theorem we have that all cyclic groups of order n are isomorphic to Z/nZ and all
infinite cyclic groups are isomorphic to Z. Something we had seen before.
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4
Permutations
I. Groups as permutations. Cayley’s Theorem.
Let X be a set. By a permutation of X we mean a bijective map from X to X. The set,
Sym (X), of all permutations of X is a group with respect to composition called the symmetric
group on X.
We are going to be particularly interested in the case when X = {1, 2, . . . , n}. In this case
we will often use Sn for Sym (X). This group is often referred to as the symmetric group on n
letters.
We are now going to see that any group can be thought of as a group of permutations.
Theorem 4.1 (Cayley). Any group G is isomorphic to a subgroup of Sym (G).
Proof For a ∈ G, let Ta ∈ Sym (G) be the permutation of G that arises from left multiplication
by a. So Ta (x) = ax. Consider the map
Φ : G → Sym (G), a 7→ Ta .
As Ta (Tb (x)) = Ta (bx) = abx = Tab (x) we have that Ta ◦ Tb = Tab . In other words
Φ(ab) = Φ(a) ◦ Φ(b)
and Φ is a homomorphism. It is injective since
Φ(a) = Φ(b) ⇒ Ta (e) = Tb (e) ⇒ a = b.
Thus G ≃ Im Φ ≤ Sym (G). 2
Remark. In particular, if |G| = n then G is isomorphic to a subgroup of Sn . Notice that
if r ≤ n then we can think of Sr as being a subgroup of Sn (any permutation α in Sr corresponds to a permutation in Sn that permutes the elements 1, . . . , r like α and fixes the remaining
numbers r + 1, . . . , n). Thus any finite groups of order ≤ n is isomorphic to a subgroup of Sn .
Thus the subgroup structure of Sn becomes very rich as n grows larger.
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