Math 402B-6

# Math 402B-6 - Math 402B Solutions to Homework 6 Chapter 2...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 402B Solutions to Homework 6 Chapter 2 Section 5-7 Show that φ : G → G , where φ is a homomorphism, is a monomorphism if and only if Ker φ = ( e ) . Proof. By definition, a monomorphism is an injective homomorphism. Ker φ = { g ∈ G | φ ( g ) = e } , i.e. Ker φ is the preimage of e . ” ⇒ ” For a homomorphism φ we always have φ ( e ) = e . If a ∈ G is an element such that φ ( a ) = e , since φ is injective so a = e . Therefore the only element in Ker φ is e . ” ⇐ ” If Ker φ = ( e ), we show the homomorphism φ is injective. Suppose φ ( a ) = φ ( b ) for some a,b ∈ G , then φ ( ab- 1 ) = φ ( a ) φ ( b )- 1 = e , so ab- 1 ∈ Ker φ , which implies ab- 1 = e , so a = b and φ is an injective homomorphism. Remark : (1)The proof is only a restatement of definitions. Here is a suggestion: before trying to prove anything, make sure you know clearly the definition of each word. In mathematics definitions are always important. (2)This conclusion is so frequently used that from now on you can use it without a proof, that is, to show a homomorphism is injective, it suffices to check whether the kernel of this homomorphism contains only the unit element. To state it another way, the kernel of a homomorphism can be used to test how far this homomorphism is from an injective homomorphism. Section 5-21 Let S be any set having more than two elements and A ( S ) the set of all 1-1 mappings of S onto itself. If s ∈ S , we define H ( s ) = { f ∈ A ( S ) | f ( s ) = s } . Prove that H ( s ) cannot be a normal subgroup of A ( S ) Proof. A normal subgroup H of a group G is a subgroup which is normal, i.e., which satisfies ∀ g ∈ G , g- 1 Hg = H . In this problem H ( s ) is indeed a subgroup. To show H ( s ) is not a normal subgroup, it suffices to find a g ∈ A ( S ) and an h ∈ H ( s ) such that g- 1 hg / ∈ H ( s ). Most of you gave the following proof, which is concise and clear. Since S has more than two elements, we can pick 3 distinct elements a,b and s . Now let g = ( as ) ∈ A ( S ) (i.e. g exchanges a and s and fixes any other element), h = ( ab ) ∈ H ( s ), then g- 1 hg = ( as )( ab )( as ) = ( bs ), which is not in H ( s ). Hence H ( s ) is not a normal subgroup. 1 Remark : (1)For a subgroup H of a group G , the following are equivalent: (a) H is normal; (b) g- 1 Hg ⊂ H for any g ∈ G ; (c) g- 1 Hg = H for any g ∈ G ; Here is a point I need to make clear: in (c) g- 1 Hg = H means as two SETS they are equal, i.e. they contain exactly the same elements. This does not mean for any h ∈ H , g- 1 hg = h ; the correct interpretation is: for any h ∈ H , g- 1 hg = h for some h ∈ H , here h may be different from h itself....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

Math 402B-6 - Math 402B Solutions to Homework 6 Chapter 2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online