# Is the group (C,+)=⟨1,i⟩(\mathbb{C},+)=\langle1,i\rangle?

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

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

Is this true?

=================

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.
2 days ago

=================

2

=================

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.