Algebra I by Randall R. Holmes

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By Randall R. Holmes

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4, −1, 2, 5, 8, . }, 3 + H = 0 + H, 4 + H = 1 + H, 5 + H = 2 + H,. . 3. If H = τ = {1, τ }, then ρH = {ρ, ρτ } = {ρ, ρ3 τ } = {ρ, τ ρ} = Hρ, which shows that left and right cosets determined by an element need not coincide. 4 Equality of cosets Let G be a group, let H be a subgroup of G, and let a, b ∈ G. 2 shows, it is possible to have aH = bH with a = b. Here is a useful criterion for equality of cosets: Theorem. (i) aH = bH if and only if a−1 b ∈ H, (ii) Ha = Hb if and only if ab−1 ∈ H. Proof.

I) ϕ(e) = e . (ii) ϕ(x−1 ) = ϕ(x)−1 for each x ∈ G. (iii) If H ≤ G, then ϕ(H) ≤ G , where ϕ(H) = {ϕ(h) | h ∈ H}. (iv) If H ≤ G , then ϕ−1 (H ) ≤ G, where ϕ−1 (H ) = {x ∈ G | ϕ(x) ∈ H }. Proof. (i) We have ϕ(e)ϕ(e) = ϕ(ee) = ϕ(e), so cancellation gives ϕ(e) = e . (ii) Let x ∈ G. Using part (i) we have ϕ(x)ϕ(x−1 ) = ϕ(xx−1 ) = ϕ(e) = e , so ϕ(x−1 ) = ϕ(x)−1 . (iii) Let H ≤ G. By (i), e = ϕ(e) ∈ ϕ(H). For h, k ∈ H, we have ϕ(h)ϕ(k) = ϕ(hk) ∈ ϕ(H) and ϕ(h)−1 = ϕ(h−1 ) ∈ ϕ(H) (using (ii)), so ϕ(H) is closed under multiplication and under inversion.

C) Give an example to show that without the assumption of surjectivity in the previous part, ϕ(N ) need not be a normal subgroup of G . 8–4 Let G and G be groups with G finite, let ϕ : G → G be an epimorphism, let p be a prime number, and let g be an element of G of order pn for some nonnegative integer n. Prove that there exists an element g of G of order pm for some nonnegative integer m such that ϕ(g) = g . 1 Isomorphism Theorems First Isomorphism Theorem Let ϕ : G → G be a homomorphism. 5, ker ϕ is a normal subgroup of G so the quotient group G/ ker ϕ is defined.

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