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The context I'm seeing it in says it's isomorphic to the automorphisms on $\mathbb{C}$. Thanks for your help.

fuqnuts
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2 Answers2

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Given a field $\mathbb F$, the notation $\mathbb F^{\times}$ denotes the multiplicative group of the field. That is, it is the group where the space is the non-zero elements of $\mathbb F$, and the binary operator is what, in the field, is the multiplicative operator (zero has to be excluded because zero does not have a multiplicative inverse). The multiplicative group of a field is always isomorphic to a (not necessarily proper) subgroup of the automorphism group of the field; this can be shown by the fact that $\phi: a \rightarrow (f_a:b \rightarrow ab)$ is a homomorphism (note that $\phi$ is a function whose input is an element of $\mathbb F$, and output is itself a function from $\mathbb F$ to $\mathbb F$).

I'm not clear on how $\mathbb C^{\times}$ is isomorphic to the set of automorphisms; it seems to me that $\{f_a:b \rightarrow ab \}$ and $\{a \rightarrow a, a \rightarrow \bar a \}$ are both subgroups of $\text {Aut}(\mathbb C)$. Perhaps the claim is that $\mathbb C^{\times}$ is isomorphic to the orientation-preserving automorphisms?

Acccumulation
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  • The question one should always ask in these situations is, automorphism in what category? The automorphisms of $\mathbb{C}$ as a one-dimensional vector space is certainly isomorphic to $\mathbb{C}^\times$. – Zhen Lin Sep 20 '21 at 09:27
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This refers to the complex numbers without $0$. The automorphism is given by multiplication.

quarague
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