# Notes - Galois Theory HT25, Group actions

### Flashcards

Suppose $G$ is a group that acts on a set $X$. @Define what it means for $Y \subseteq X$ to be $G$-stable.

For all $y \in Y$, $g \cdot y \in Y$.

(Hence also $G$ acts on $Y$ by restriction).

Suppose:

- $K/F$ is a field extension
- $G = \text{Gal}(K/F)$

In what way does $G$ act on $K$?

$G$ is the set of all $F$-linear automorphisms of $K$, so it acts on $K$ via

\[g \cdot k = g(k)\]Suppose:

- $K/F$ is a field extension
- $G = \text{Gal}(K/F)$
- $f \in F[t]$
- $V(f) := \{\alpha \in K \mid f(\alpha) = 0\}$

@Prove that $V(f)$ is a $G$-stable subset of $K$ and therefore that $G$ also acts on $V(f)$.

Let $\sigma \in G$ and $\alpha \in V(f)$. Suppose

\[f = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\]Then:

\[\begin{aligned} f(\sigma(\alpha)) &= \sum^n_{i = 0} a_i \sigma(\alpha)^i \\\\ &= \sigma \left(\sum^n_{i = 1} a_i \alpha^i\right) &&(\star)\\\\ &= \sigma(f(\alpha)) \\\\ &= \sigma(0) \\\\ &= 0 \end{aligned}\]where $(\star)$ is justified by the fact that $\sigma$ is an $F$-linear ring homomorphism. Hence $\sigma(\alpha) \in V(F)$ for all $\alpha \in V(f)$, so $V(f)$ is $G$-stable and hence $G$ also acts on $V(f)$.

Suppose:

- $G$ is a group
- $X$ is a set
- $x \in X$

@Define the orbit map $\pi _ x$ and connect it to $\text{Stab} _ G(x)$.

Then

\[\text{Stab}_G(x) = \pi_x^{-1}(x)\]Suppose:

- $G$ is a group
- $X$ is a set
- $x \in X$
- $g \in G$
- $\pi _ x$ is the orbit map given by $\pi _ x(g) = g \cdot x$

@Prove that:

\[\pi_x^{-1}(g \cdot x) = g \text{Stab}_G(x)\]

For any $h \in G$,

\[\begin{aligned} &h \in \pi_x^{-1}(g \cdot x) \\\\ \iff &\pi_x(h) = g \cdot x \\\\ \iff &h \cdot x = g \cdot x \\\\ \iff &g^{-1} h \in \text{Stab}_G(x) &&(\star1)\\\\ \iff &h \in g\text{Stab}_G(x) &&(\star2) \end{aligned}\]where:

- $(\star 1)$ is justified by considering $g^{-1} h \cdot x = g^{-1} \cdot (h \cdot x) = g^{-1} \cdot (g \cdot x) = g^{-1} g \cdot x = x$.
- $(\star 2)$ is justified by expanding definitions.

Hence the two sets are equal.