HW6_Math 113 - MATH 113 HW#6 KEVIN JORGENSEN 1 Prove the...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 113 HW #6 KEVIN JORGENSEN 1) Prove the Third Isomorphism Theorem: Let G be a group and let H and K be normal subgroups of G such that H is a subgroup of K . Then K/H is a normal subgroup of G/H and ( G/H ) / ( K/H ) is isomorphic to G/K . Suppose H,K E G and H ≤ K . It is trivial to see that K/H is a subgroup of G/H . Let gH ∈ G/H and kH ∈ K/H . Then gH ◦ kH ◦ ( gH )- 1 = gH ◦ kH ◦ g- 1 H = ( gk ) H ◦ g- 1 H = ( gkg- 1 ) H ∈ G/H. So K/H is a normal subgroup of G/H . Let ϕ : G/H → G/K , where gH 7→ gK . (Since H ≤ K , this is well-defined. Further, this map is clearly a surjection, as the choice of g is arbitrary.) This mapping is also very clearly a homomorphism: ϕ ( aH ◦ bH ) = ϕ (( ab ) H ) = ( ab ) K . ϕ ( aH ) ◦ ϕ ( bH ) = aK ◦ bK = ( ab ) K . Finally, ker ( ϕ ) = { gH ∈ G/H | ϕ ( gH ) = K } = { gH ∈ G/H | gK = K } = { gH ∈ G/H | g ∈ K } = K/H And by the First Isomorphism Theorem, ( G/H ) /ker ( ϕ ) ∼ = ϕ ( G/H ), so ( G/H ) / ( K/H ) ∼ = G/K , since ϕ is surjective. Q.E.D. 1 2 KEVIN JORGENSEN 2) Let G be a group and let N be a normal subgroup. Let π N : G → G/N be the canonical projection and let H < G/N . Show that π- 1 ( H ) is a subgroup of G ....
View Full Document

This note was uploaded on 10/15/2011 for the course CHEM 1 taught by Professor Gelfand during the Summer '11 term at Solano Community College.

Page1 / 5

HW6_Math 113 - MATH 113 HW#6 KEVIN JORGENSEN 1 Prove the...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online