Cayley Theorem

2020-2학기, 대학에서 ‘현대대수1’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)

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군에 대한 일반적인 성질을 기술하기에 명제 자체는 놀랍지만, 별 쓸모는 없는 명제, “Cayley’s Theorem”이다!

Theorem.

Every group is isomorphic to a subgroup of a suitable permutation group.

\[G \le S_G\]

proof.

Let $G$ be a group, and $S_G$ be a set of permutations on $G$.

We will show that $G$ is isomorphic to some subgroup of $S_G$.

Chose one of $a \in G$, and consider a simple mapping $\lambda_a$ s.t.

\[\begin{aligned} \lambda_a: G & \longrightarrow G \\ g & \longmapsto ag \end{aligned}\]

We claim that $\lambda_a$ is 1-1 & onto, i.e. $\lambda_a \in S_G$.

(1) $\lambda_a$: 1-1

Supp. $ag = ag’$, THEN

$\implies a^{-1}(ag) = a^{-1}(ag’) \implies g = g’$

따라서 $\lambda_a$ is 1-1.

(2) $\lambda_a$: onto

For $g’ \in G$, find $g$ s.t. $g’ = ag$, THEN

$a^{-1}g’ = g \in G$ is that inverse image of $g’$.

따라서 $\lambda_a$ is onto.

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Let $G’ := \{ \lambda_a \mid a \in G\} \subseteq S_G$.

We claim $G’$ is a subgroup of $S_G$; $G’ \le S_G$

(1) $G’$ is closed under opr.

$\lambda_a \cdot \lambda_{a’} = \lambda_{aa’} \in G’$

(2) Identity

$\lambda_e: g \mapsto g = \textrm{id}_{G’}$

(3) Inverse

If $\lambda_a \in G’$. THEN there exist an inv. map.

$(\lambda_a)^{-1} = \lambda_{a’} \in G’$

That inv. map $\lambda_{a’}$ is about $a’ = a^{-1}$.

따라서 $G’ \le S_G$이다!

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Finally, we will show $G \cong G’$.

Define a group iso- $\psi$ as

\[\begin{aligned} \psi: G & \longrightarrow G' \\ a & \longmapsto \lambda_a \end{aligned}\]

(1) $\psi$ is 1-1

Supp. $\lambda_a = \lambda_b$, THEN

$\lambda_a (e) = a = b = \lambda_b (e)$

(2) $\psi$ is onto

Clear! $\lambda_a$ has inv. image $a \in G$ under $\psi$.

(3) $\psi$ is homo-

\[\begin{aligned} \psi(ab) &= \lambda_{ab} \\ &= \lambda_a \cdot \lambda_b = \psi(a) \cdot \psi(b) \\ \end{aligned}\]

따라서 $G \cong G’$이고, $G’ \le S_G$이므로

\[G \cong G' \le S_G\]

Every group is isomorphic to a subgroup of a suitable permutation group. $\blacksquare$