Is $C_2$ the correct Galois Group of $f(x)= x^3+x^2+x+1$? Let $\operatorname{f} \in \mathbb{Q}[x]$ where $\operatorname{f}(x) = x^3+x^2+x+1$. This is, of course, a cyclotomic polynomial. The roots are the fourth roots of unity, except $1$ itself. I get $\mathbb{Q}[x]/(\operatorname{f}) \cong \mathbb{Q}(\pm 1, \pm i) \cong \mathbb{Q}(i) = \{a+bi : a,b \in \mathbb{Q}\}.$
Let $\alpha : \mathbb{Q}(i) \to \mathbb{Q}(i)$ be a $\mathbb{Q}$-automorphism. We have: 
$$\alpha(a+bi) = \alpha(a)+\alpha(bi) = \alpha(a)+\alpha(b)\alpha(a)i = a+b\alpha(i).$$
Since $\alpha(i)^2 = \alpha(i)\alpha(i) = \alpha(ii) = \alpha(-1)=-1$ we have $\alpha(i) = \pm\sqrt{-1} = \pm i$. There are then two $\mathbb{Q}$-automorphisms: the identity with $\alpha(z)=z$ and the conjugate $\alpha(z)=\overline{z}$.
This tells me that the Galois Group is $S_2=\langle(12)\rangle.$ I've been using GAP software, and it says that the Galois Group is $\langle(13)\rangle$. I can see that $\langle(12)\rangle \cong \langle(13)\rangle$. However, $\langle(13)\rangle < S_3$. My suspision is that because $x^3+x^2+x+1$ is reducible over $\mathbb{Q}$: $x^3+x^2+x+1 \equiv (x+1)(x^2+1)$.
Is GAP telling me that the Galois Group of $x^3+x^2+x+1$ is $C_1\times C_2$?
How should I think about the Galois Group of $x^3+x^2+x+1$? Is it $C_2$, is it a subgroup of $S_3$ which is isomorphic to $C_2$, or is it the product $C_1 \times C_2$. I realise that these are all isomorphic, but what's the best way to think of it?

The Galois group is the group of authomorphisms of the splitting field. It acts on the roots of any splitting polynomial (such as $f$) by permuting the roots. In your case, there are three roots, $-1, i, -i$ and the automorphisms must leave $-1$ fixed. Since the action is also free, you can view $G$ (via this action) as a subgroup of $\operatorname{Sym}(\{-1,i,-i\})$ and of cours it as only one nontrivial element $(1)(i\ {-i})$.
