I’m trying to see if I understand the notation. I think that

(C,+)=⟨1,i⟩(\mathbb{C},+)=\langle1,i\rangle

Is this true?

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1

Looks right to me. EDIT: Apologies, misread the problem. No, you can’t generate the complex numbers in general, but that does generate the complex numbers with integer components.

– Ispil

2 days ago

1

No, ⟨1,i⟩⊂Z[i]\langle 1,i\rangle \subset \mathbb{Z}[i] since the latter is a subgroup of C\mathbb{C} that contains both 11 and ii.

– Prahlad Vaidyanathan

2 days ago

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2 Answers

2

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Since the group operation in question is addition, ⟨1,i⟩\langle 1,i \rangle generates only the complex numbers with integer real and imaginary parts: Z[i]\mathbb{Z}[i]. In fact, the additive group C\mathbb{C} is not finitely-generated: it cannot be generated by any finite subset.

If we think of C\mathbb{C} as a vector space over R\mathbb{R}, then it is generated by the set {1,i}\{1,i\}.

If we think of C\mathbb{C} as an unital algebra over R\mathbb{R}, then it is generated solely by ii.

Great answer! Thanks!

– Skeleton Bow

2 days ago

No. The group generated by 11 and ii is the group {a+bi∣a,b∈Z}\{a+bi\mid a,b\in\mathbb{Z}\}.