Group of Units a Subgroup of some ${S}_n$

Question

Consider the polynomial $f(x) \in \mathbb{F}[x]$ where $\mathbb{F}$ is some field of characteristic 0. Let $\mathbb{E}$ be the splitting field of $f$. We know that if $f$ is separable of degree $n$, the Galois Group $\operatorname{Gal}(\mathbb{E}/\mathbb{F})$ is isomorphic to some subgroup of $S_n$ where $S_n$ is the symmetric group on $n$ letters. Moreover, we know that the order of the Galois Group is equal to the degree of the field extension. So, for example, since $[\mathbb{C} : \mathbb{R}] = 2$, we have that $|\operatorname{Gal}(\mathbb{C}/\mathbb{R})| = 2$.

I'm having a hard time understanding this definition in the context of Cyclotomic fields, for what I assume is a relatively trivial reason. Let $\zeta$ be a primitive $p$-th root of unity adjoined to $\mathbb{Q}$ where $p$ is prime. We then have that $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^\times$ (the group of units in the integers modulo $p$).

This group is simply a set of integers (specifically equivalence classes modulo $p$). A subgroup of $S_n$ might be some collection of cycles, transposition, etc... For example, the alternating (sub)group of $S_3$ is, $$A_3 = \{id, (123), (321)\} = \{ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \}.$$ It's tricky for me to understand how a group of units could be a subgroup of this group, at least in terms of the different representations. Could someone clarify?

For instance, if we have a cyclotomic polynomial of degree 3 over $\mathbb{Q}$, we know that the Galois group is isomorphic to a subgroup of $S_3$. Moreover, we know that $|\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})| = 3$. So, $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/3\mathbb{Z})^\times = \mathbb{Z}/3\mathbb{Z} = \{ 0, 1, 2 \}$. But what does it mean for $\mathbb{Z}/3\mathbb{Z}$ to be a subgroup of $S_3$? Apologies for any lack of coherence –– I'm just having a hard time picturing it.

Answer 1

Every finite group $G$ is isomorphic to a subgroup of some $\mathfrak S_n$ by Cayley's theorem. This is done by considering the action of $G$ on itself by left multiplication.

Answer 2

You write that:

"...the Galois Group $\text{Gal}(\mathbb{E}/\mathbb{F})$ is isomorphic to some subgroup of $\mathfrak{S}_n$.

and later

"We then have that $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^\times$."

In both cases you make statements about the Galois group being isomorphic to some other group.

So your particular example of a cyclotomic field shows that $(\mathbb{Z}/p\mathbb{Z})^\times$ is isomorphic to some subgroup of $\mathfrak{S}_n$, where $n=\varphi(p)=p-1$. This should not be surprising at all; the group $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic of order $p-1$ and the symmetric group $\mathfrak{S}_{p-1}$ contains a $p-1$-cycle.

In general, two completely unrelated groups can be isomorphic without there being any 'visual' or 'intuitive' reason. This is especially true for smaller groups, as there are many ways to construct small groups, but only very few small groups to construct.

A classic example is given by the group $(\Bbb{Z}/8\Bbb{Z})^{\times}$ and the group of symmetries of a diamond-shape in the plane. The two groups are easily verified to be isomorphic, though I cannot think of any particular visual or intuitive reason for this to be true. But they are both groups of order $4$, and up to isomorphism there are only two groups of order $4$, so it's not at all unlikely for these two two 'random' groups of order $4$ to be isomorphic.