Galois Theory HT25, Groups

\[\begin{aligned} \text{sgn} &: S _ n \to \pm 1 \\\\ \sigma &\mapsto \begin{cases} 1 &\text{if } \sigma \in A _ n \\\\ -1 &\text{if } \sigma \notin A _ n \end{cases} \end{aligned}\]

A finite abelian group $G$ is cyclic iff there is one subgroup for each divisor of $ \vert G \vert $.

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1. Count the number of divisors, each of which corresponds to a subgroup / subfield
2. Show that there is exactly one subgroup for each divisor
3. Show that there are two distinct subgroups of the same order

\[\text{Aut}(\mathbb Z / n \mathbb Z) \cong (\mathbb Z / n \mathbb Z)^\times\]

They are equivalent modulo $ \vert G \vert $.

For any $x \in X$, $G \cdot x = X$.

Equivalently, for any $x$ and $y$, there is some $g$ such that $g(x) = y$.

• $S _ 3$ (order 6)
• $C _ 3$ (order 3)

• $S _ 4$ (order 24)
• $A _ 4$ (order 12)
• $D _ 8$ (order 8)
• $V _ 4$ (order 4)
• $C _ 4$ (order 4)

• $S _ 5$ (order 120)
• $A _ 5$ (order 60)
• $F _ {20} = \mathbb Z/5\mathbb Z \rtimes (\mathbb Z/5\mathbb Z)^\times$ (order 20)
• $D _ {10}$ (order 10)
• $C _ 5$ (order 5)

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

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

$G$ is the set of all $F$-linear automorphisms of $K$, so it acts on $K$ via \(g \cdot k = g(k)\)

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 $V(f)$.

\(\begin{aligned} \pi _ x &: G \to G \cdot x \\\\ g &\mapsto g\cdot x \end{aligned}\) Then \(\text{Stab} _ G(x) = \pi _ x^{-1}(x)\)

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.

There does not exist a non-trivial $g \in G \setminus \{1\}$ such that $g \cdot x = x$ for all $x$, i.e. if some element fixes everything, that element is the identity.

There exists $N \trianglelefteq H \le G$ such that \(Q \cong H/N\)

Suppose:

• $G$ is a finite group
• $p$ is a prime which divides $ \vert G \vert $

Then:

• $G$ has an element of order $p$

• $G$ contains a subgroup $H$ such that $ \vert H \vert = p^n$
• If $H, H’ \le G$ are two subgroups of size $p^n$, then they are conjugate to each other, i.e. there exists $g \in G$ such that $g^{-1} H g = H’$.

Suppose:

• $p$ is a prime
• $a$ is an integer coprime to $p$

Then:

\[a^{p-1} \equiv 1 \pmod p\]

Suppose:

• $n \ge 2$
• $a$ is an integer coprime to $p$
• $\varphi$ is Euler’s totient function

Then:

\[a^{\varphi(n)} \equiv 1 \pmod n\]