(1)ALGEBRA MIDTERM No credit will be given for an answer without reasoning

ALGEBRA MIDTERM

No credit will be given for an answer without reasoning.

1.

(1) [5%] a⁻¹∗ (a ∗ b ∗ c) = a⁻¹∗ e = a⁻¹, so b ∗ c = a⁻¹. Hence (b ∗ c) ∗ a = e.

(2) [5%] ρ₁µ₁ 6= µ₁ρ₁. 2.

(1) [5%] (a ∗ b) ∗ c 6= a ∗ (b ∗ c) i.e., the associativity does not hold.

(2) [5%] (G : H) = 2.

3.

(1) [5%] (1, 5, 3, 6, 9, 2, 8, 7, 10, 4) (2) [5%] (Z × Z)/h(1, 3)i ' Z

4.

(1) [5%] Let φ : Z₄ → Z be a homomorphism. Suppose that φ(1) = a ∈ Z. Then 0 = φ(0) = φ(1 + 1 + 1 + 1) = φ(1) + φ(1) + φ(1) + φ(1) = 4a. Hence, a = 0. Therefore, φ must be trivial.

(2) [5%] φ : Z₄× 2Z → Z × Z given by φ((a, b)) = (0, b) is a nontrivial homomorphism.

5.Let GL(n, R) be the general linear group of degree n.

(1) [5%]

(2) [5%] Suppose AA^t = BB^t = I. Then (AB)(AB)^t = ABB^tA^t = I. II^t = I. And A⁻¹(A⁻¹)^t= A^t(A^t)^t= A^tA = (AA^t)^t= I^t= I. So O(n, R) is a subgroup.

6.φ(1) = (13)(24)(517) = (24)(5317).

(1) [5%] Find φ(14) = ((24)(5317))¹⁴= (5317)² = (51)(37).

(2) [5%] Ker(φ) = {4, 8, 12, 16, 0}.

7.[10%] Let X = { (a, b) ∈ Z × Z | b 6= 0 }, and define a relation “∼” on X by (a, b) ∼ (c, d) if ad = bc.

Show that “∼” is an equivalence relation.

8.[10%] Let φ : U × R⁺' C^∗be given by φ((u, r)) = ru. Check that φ is an isomorphism.

9. [10%] Let φ : G → G⁰ be a group homomorphism. Show that φ[G] is abelian if and only if for all x, y ∈ G, we have xyx⁻¹y⁻¹∈ Ker(φ).

10.Let ig(x) = gxg⁻¹.

(1) [5%] i_g◦ i_g⁰ = i_gg⁰, the identity map is i_e ∈ H, and i⁻¹_g = i_g⁻¹. Hence, H is a subgroup of Aut(G).

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2 ALGEBRA MIDTERM

(2) [5%] Let σ be an element in Aut(G). Then σ ◦ i_g◦σ⁻¹= i_σ(g). Hence, H is a normal subgroup of Aut(G).