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INTRODUCTION TO GROUPS AND SYMMETRY (MTH 203)
Lecture-15, (05-09-2019)
Let G be a group and N be a normal subgroup of G. Note that the normality of N implies that aN = N a. So the set of left cosets of N is same as the set of right cosets of N. Also by normality of N we know that the set of left (right) cosets of N forms a group with respect to the binary operation
aN.bN = abN = N ab = N a.N b
where a, b ∈ G.
We denote this by (^) NG and call it the quotient group of G by N. Example-15.1: Z/mZ is the quotient group of Z by mZ.
Examples of group homomorphism:
(1) Let G be a group and N be a subgroup. Define
φ : G →
G
N
g 7 → gN. The φ is a group homomorphism and Ker(φ) = N. (2) Let G be any group and H be any subgroup of G. Then φ : G → (^) HG is a group homomorphism and H = Ker(φ). (3) φ : Z → Z 10 , n 7 → ¯n, Ker(φ) = 10Z.
Definition 0.1. We say two groups G 1 and G 2 are isomorphic if there exists an isomor- phism φ : G 1 → G 2. We denote it by G 1 ∼= G 2.
Examples of Isomorphisms:
(1) Z 4 ∼= U 5. Define a map φ : Z 4 → U 5 such that φ(¯0) = ¯ 1 φ(¯1) = ¯ 2 φ(¯2) = ¯ 4 φ(¯3) = ¯ 3. Check that φ is an isomorphism and hence Z 4 ∼= U 5.
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2 MTH 203
(2) Zp− 1 ∼= Up, where p is a prime number. Hint: Construct an isomorphism as above. To construct a map you only need image of ¯ 1 ∈ Zp− 1 (why?). We know under an isomorphism order is preserve so find an element in Up which is generator of Up and send ¯ 1 ∈ Zp− 1 to that element.
(3) Show that there exists no isomorphism from S 4 to Z 24.
First Isomorphism Theorem.
Theorem 0.2. Let φ : G 1 → G 2 be a group homomorphism. Then
G 1 Ker(φ)
∼= Im(φ).
Proof. For simplicity let K = Ker(φ). Define the map
ψ : G/K → Im(φ))
by
ψ(gK) =: φ(g).
Claim: ψ is a group isomorphism.
- Step-1: First, we need to check that ψ is well-defined. To show ψ is well defined, we have to check that if g 1 K = g 2 K then ψ(g 1 K) = ψ(g 2 K). Note that g 1 K = g 2 K implies that g 1 = g 2 k for some k ∈ K. Now φ(g 1 g 2 − 1 ) = e 2 since (g 1 g− 2 1 ∈ K). By homomorphism property of φ we get φ(g 1 )φ(g 2 )−^1 = e 2 and hence ψ(g 1 K) = ψ(g 2 K).
- Step-2: ψ is a group homorphism. ψ(g 1 Kg 2 K) = ψ(g 1 g 2 K) = φ(g 1 g 2 )
= φ(g 1 )φ(g 2 ) = ψ(g 1 K)ψ(g 2 K).
- Step-3: ψ is one-one. For g 1 , g 2 ∈ G, let ψ(g 1 K) = ψ(g 2 K), ⇒ φ(g 1 ) = φ(g 2 ), ⇒ φ(g 1 )φ(g 2 )−^1 = e 2 , ⇒ g 1 g− 2 1 ∈ K, and hence g 1 K = g 2 K.
- Step-4: ψ is onto. Let x ∈ Im(φ) i.e. there exists g ∈ G 1 such that φ(g) = x. It is clear that ψ(gK) = φ(g) = x. So ψ is onto.
Home Work 15.1: Can you give a group structure on a set of group homomorphisms from G 1 to G 2.
Home Work 15.2: Show that there exists only one group of order 2 upto ismomor- phism.
