# Abstract Algebra Questions

Anonymous
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### Question Description

Hello,

In the attachments three pictures each one has a questions,

Please solve all the three problems,

7c2d80af_1962_4420_bd82_b6bba7be2981.jpeg
dafb0053_3e5b_4c6e_99d7_eb1ccf287cde.jpeg
d12f5751_7cd0_4d25_acd4_899900da1a73.jpeg

Borys_S
School: UCLA

1. Homomorphisms.
π: β β β, π(π₯) = 3π₯ β 5 is NOT a homomorphism.
Indeed, a homomorphism must preserve neutral element, i.e. π(π) = π.
For the (additive) group β, the neutral element is π = 0.
But π(0) = β5 β  0. It is simple to check that π does not preserve addition, too.

2. Isomorphisms.

π: (β, +) β (β+ ,Γ), π(π₯) = π π₯ .
(a) It is a homomorphism because it preserves neutral element, the group operation and
inversion:
π(π) = π(0) = π 0 = 1 = π β² ,
π(π₯ + π¦) = π π₯+π¦ = π π₯ Γ π π¦ ,
β1

π(βπ₯) = π βπ₯ = (π π₯ )β1 = (π(π₯)) .
(b) It is one-to-one (injective) because if π(π₯) = π π₯ = π π¦ = π(π¦) implies π₯ = π¦.
(c) It is onto because for each π¦ β β+ there is π₯: π(π₯) = π¦ (we call it π₯ = ln π¦).
(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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