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Consider the set $S = \{1, ω, ω^2\}$, where $ω$ and $ω^2$ are cube roots of unity. If $*$ denotes the multiplication operation, the structure $(S, *)$ forms

1. A Group
2. A Ring
3. An integral domain
4. A field
edited | 1.9k views
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This might help ...

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@srestha ma'am

is ring , field in syllabus ?

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no

ring , field out of syllabus

 1 ω ω2 1 1 ω ω2 ω ω ω2 1 ω2 ω2 1 ω

The structure (S,*) satisfies closure property, associativity, commutativity. The structure also has an identity element (i.e. 1) and an inverse for each element. So, the structure is an abelian group.

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Can anyone explain what are ring, field, and integral domain and why the given structure is not one of them?
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Rajarshi Sarkar can you please explain why it is not a ring ?

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It can't be ring as set under addition doesn't contain identity element , read more here.

https://en.wikipedia.org/wiki/Ring_(mathematics)

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In the question, it is only asked for multiplication so why even do we need to check for ring or field because for both of them two binary operations should be defined. right? Thus it clearly not a ring or a field.
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Tuhin , ring is defined over two operations (Multiplication and addition)and they are fixed so when someone talk about ring , we can consider the binary operation operation + and · ourselves.

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use this while designing "cayley table"

1+W+W^2 =0

W^3=1

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Also, it is a cyclic group with generators $\omega,\omega^2$

In this question, only one binary operation is given so it cannot be a Ring or Integral domain or field.For Ring or Integral domain or for field there must be Two binary operations are required and the algebraic structure looks like (S ,+ ,⨉) .

So ,option b ,c and d are False.

Now check the properties of groups.It satisfies all the conditions of groups.

Hence, it is a Group.

## The correct answer is,(A) A Group

Its satisfies 1.closure property 2.associativity 3.identity element exists i.e 1 4.inverse element exist for each element So it is a group to attack these type of questions draw the composition table then it wl be easier ...

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