Abstract Algebra Questions

Anonymous
timer Asked: Feb 21st, 2019
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Question Description

Hello,

In the attachments three pictures each one has a questions,


Please solve all the three problems,


Thanks for your hard work in advance

Abstract Algebra Questions
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Abstract Algebra Questions
dafb0053_3e5b_4c6e_99d7_eb1ccf287cde.jpeg
Abstract Algebra Questions
d12f5751_7cd0_4d25_acd4_899900da1a73.jpeg

Tutor Answer

Borys_S
School: UCLA

The solutions are ready, please ask if something is unclear.

1. Homomorphisms.
πœ‘: ℝ β†’ ℝ, πœ‘(π‘₯) = 3π‘₯ βˆ’ 5 is NOT a homomorphism.
Indeed, a homomorphism must preserve neutral element, i.e. πœ‘(𝑒) = 𝑒.
For the (additive) group ℝ, the neutral element is 𝑒 = 0.
But πœ‘(0) = βˆ’5 β‰  0. It is simple to check that πœ‘ does not preserve addition, too.

2. Isomorphisms.

πœ‘: (ℝ, +) β†’ (ℝ+ ,Γ—), πœ‘(π‘₯) = 𝑒 π‘₯ .
(a) It is a homomorphism because it preserves neutral element, the group operation and
inversion:
πœ‘(𝑒) = πœ‘(0) = 𝑒 0 = 1 = 𝑒 β€² ,
πœ‘(π‘₯ + 𝑦) = 𝑒 π‘₯+𝑦 = 𝑒 π‘₯ Γ— 𝑒 𝑦 ,
βˆ’1

πœ‘(βˆ’π‘₯) = 𝑒 βˆ’π‘₯ = (𝑒 π‘₯ )βˆ’1 = (πœ‘(π‘₯)) .
(b) It is one-to-one (injective) because if πœ‘(π‘₯) = 𝑒 π‘₯ = 𝑒 𝑦 = πœ‘(𝑦) implies π‘₯ = 𝑦.
(c) It is onto because for each 𝑦 ∈ ℝ+ there is π‘₯: πœ‘(π‘₯) = 𝑦 (we call it π‘₯ = ln 𝑦).
(d) It is an isomorphism by definition: a bijective homomorphism.

3. A Special Normal Subgroup.
...

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Anonymous
Thanks, good work

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