Flashcards on Normal Subgroups, Quotient Groups, and Isomorphism Theorems

Normal Subgroups, Quotient Groups & Isomorphism Theorems

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What is a normal subgroup (H ◁ G) in a group G?

A subgroup H of G is normal (H ◁ G) if ghg^{-1} ∈ H for all g ∈ G and all h ∈ H. Equivalently gH = Hg for all g ∈ G.

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Group Theory

71 cards

Card 1

Question: What is a normal subgroup (H ◁ G) in a group G?

Answer: A subgroup H of G is normal (H ◁ G) if ghg^{-1} ∈ H for all g ∈ G and all h ∈ H. Equivalently gH = Hg for all g ∈ G.

Card 2

Question: Why do we require normal subgroups to form a quotient group G/H?

Answer: To make coset multiplication well-defined: for Hg1·Hg2 := H(g1g2) we need that writing representatives differently doesn't change the product, which r

Card 3

Question: State the three equivalent conditions for a subgroup H ≤ G to be normal (Theorem 5.1).

Answer: (1) H ◁ G. (2) gHg^{-1} ⊆ H for all g ∈ G. (3) gHg^{-1} = H for all g ∈ G.

Card 4

Question: What are the trivial normal subgroups always present in any group?

Answer: The trivial normal subgroups are {e} and G; both are always normal in G.

Card 5

Question: When is every subgroup of a group normal?

Answer: Every subgroup is normal when the group is abelian.

Card 6

Question: What does the Index-2 Theorem state (Theorem 5.2)?

Answer: If H ≤ G and the index [G : H] = 2, then H is normal in G.

Card 7

Question: Give a quick reason why a subgroup of index 2 is always normal.

Answer: There are exactly two cosets (H and G\H); for any g, the left and right coset must be either H or G\H, so gH = Hg for all g, hence H ◁ G.

Card 8

Question: Provide an example of an index-2 normal subgroup in the symmetric group context.

Answer: A_n is normal in S_n because [S_n : A_n] = 2.

Card 9

Question: Give an example of a normal and a non-normal subgroup in S_3 from the content.

Answer: Normal: H = ⟨(123)⟩ = {e,(123),(132)} is normal in S_3 (index 2). Non-normal: K = {e,(12)} is not normal since (123)K ≠ K(123).

Card 10

Question: What is the kernel of the sign homomorphism φ: S_3 → Z_2, and why is it normal?

Answer: ker(φ) = A_3 = {e,(123),(132)}. The kernel of any homomorphism is always a normal subgroup of the domain.