What group is the Galois group $G(\mathbb{C}/\mathbb{R})$ isomorphic to?

Question

Recently in my abstract algebra II course we have been studying Galois theory. I have been doing a lot of reading and came across the Galois group $G(\mathbb{C}/\mathbb{R})$ (The set of all automorphisms of $\mathbb{C}$ that fix $\mathbb{R}$ pointwise) and naturally being an abstract algebra student I immediately asked the question:

To what group is $G(\mathbb{C}/\mathbb{R})$ isomorphic to?

I believe $G(\mathbb{C}/\mathbb{R})$ has two elements: the complex conjugation automorphism and the identity automorphism. Would it maybe be isomorphic to $\mathbb{Z_2}$? I can’t find any suppporting evidence of this in my book so if there are any relevant theorems that could supplement my question, please feel free to add.

Answer 1

Without deep machinery, any field automorphism $\phi$ of $\Bbb C$ that leaves $\Bbb R$ pointwise fix is determined by $\phi(i)$ as we must have $\phi(x+iy)=x+\phi(i)y$. Also $$0=\phi(0)=\phi(1+i^2)=1+\phi(i)^2$$ implies that $\phi(i)\in\{i,-i\}$. Hence indeed there are exactly two such automorphisms: the identity and complex conjugation. So $G(\Bbb C/\Bbb R)$ is the group of order $2$.

Answer 2

As outlined in the comments, $\mathbb C = \mathbb R(i)$ makes it a degree 2 extension. It is normal but the fundamental theorem of algebra and separable because the characteristic is 0, hence Galois. So the size of the Galois group is the degree, and clearly complex conjugation works as the nonzero element generating that Galois group.

Answer 3

It is $\mathbb{C} = \mathbb{R}(i) $ with $ [\mathbb{C}: \mathbb{R}] = 2 $ since the minimal polynomial of $ i $ is $ X^2+1$. So the extension $\mathbb{C}/\mathbb{R}$ is a galois extension with galois group of order 2. Hence it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.