📂Abstract Algebra

Cayley's Theorem Proof

Theorem ¹

Every group is isomorphic to some subgroup of the symmetric group.

Explanation

This concise yet significant theorem conveys the message that by studying symmetric groups, one can grasp all groups.

Proof

Although the proof may seem tedious at first glance, the techniques involved are quite interesting, so it is recommended to try and follow it at least once.

---

Part 1. $f : G \to G'$ is injective then $G \simeq f (G)$

For group $G$ and $G'$, let’s show that if a homomorphism $f : G \to G'$ is injective, then $G \simeq f (G)$.

Definition of a group: A group is a binary operation structure that satisfies the following properties.

• (i): The operation is associative.
• (ii): There is an identity element for all elements.
• (iii): There is an inverse for all elements.

$f$, when viewed as a function $f : G \to f(G)$ with the image of $G$ under $f$, $f(G) \subset G'$, as its codomain, is naturally surjective. Assuming that $f$ is injective, showing that $f(G)$ is a group leads to $G \simeq f(G)$.

If we let $x,y \in G$ and $x', y ' \in G'$, then $$ f(xy) = f(x) f(y) = x ' y’ $$ so $f(G)$ is closed under the operation of $G'$. Also, since $f(G) \subset G'$ and $G'$ is a group, it satisfies the associative law.

If we designate $e$ as the identity of $G$, and $e'$ as the identity of $G'$, $$ e' f(e) = f(e) = f(ee) = f(e) f(e) $$ then $f(G)$ has the identity element $e'$.

Lastly, $$ e' = f(e) = f(x x^{-1}) = f(x) f(x^{-1}) = x ' f(x^{-1}) $$ this means every $x' \in f(G)$ has an inverse $f(x^{-1})$, so $f(G)$ is a group.

---

Part 2. $\exists \phi : G \hookrightarrow S_{G}$

Now, proving the existence of a monomorphism (a homomorphism that is also injective) $\phi : G \to S_{G}$ concludes the proof.

Defining $\lambda_{x} : G \to G$ for $x \in G$ as $\lambda_{x} (g) := xg$, $$ \lambda_{x} (a) = \lambda_{x} (b) \implies xa = xb \implies a = b $$ shows that $\lambda_{x}$ is injective. Also, for every $c \in G$, $$ \lambda_{x} (x^{-1} c ) = x x^{-1} c = c $$ thus $\lambda_{x}$ is surjective, and therefore $\lambda_{x} : G \to G$ is a permutation of $G$.

So, now it is possible to define $\phi : G \to S_{G}$ for $x \in G$ as $\phi (x) = \lambda_{x}$. If we define $\phi (x) = \phi (y)$, then $\lambda_{x} = \lambda_{y}$ and $\lambda_{x} (e) = \lambda_{y} (e)$, therefore $$ x e = y e \implies x = y $$ meaning, $\phi$ is injective. Meanwhile, since $\phi (xy) = \lambda_{xy}$, thus $\lambda_{xy}(g) = (x y) g$, and since $\lambda_{x} , \lambda_{y}$ is a permutation, $$ ( \lambda_{x} \lambda_{y} ) (g) = \lambda_{x} (\lambda_{y} (g)) = \lambda_{x} (yg) = x (yg) $$ To summarize, $$\phi (xy) = \lambda_{xy} = \lambda_{x} \lambda_{y} = \phi (x) \phi (y)$$ thus, $\phi$ is a homomorphism.

---

Therefore, $G$ is isomorphic to some subgroup of $S_{G}$.

■