📂Abstract Algebra

Proof of Cayley's Theorem

Theorem ¹

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

Explanation

This short but weighty theorem carries the message that by studying symmetric groups, one can grasp all groups.

Proof

The proof may seem tedious at first glance, but reading through it reveals that the technique is quite interesting, so it is recommended to follow along yourself at least once.

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

For groups $G$ and $G'$, let us show that if the homomorphism $f : G \to G'$ is injective, then $G \simeq f (G)$.

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

• (i): The associative law holds for the operation.
• (ii): An identity element exists for all elements.
• (iii): An inverse element exists for all elements.

When $f$ is viewed as a function $f : G \to f(G)$ having the image of $G$ under $f$ $f(G) \subset G'$ as its codomain, it is naturally surjective. If we assume that $f$ is injective, then we can obtain $G \simeq f(G)$ by showing that $f(G)$ is a group.

For $x,y \in G$ and $x', y ' \in G'$, if $f(x) = x '$, $f(y) = y '$, 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, the associative law is satisfied.

Letting $e$ be the identity element of $G$ and $e'$ be the identity element of $G'$, $$ e' f(e) = f(e) = f(ee) = f(e) f(e) $$ so $f(G)$ has the identity element $e'$.

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

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

Now the proof is complete once we show that a monomorphism (a function that is both a homomorphism and injective) $\phi : G \to S_{G}$ exists.

For $x \in G$, defining $\lambda_{x} : G \to G$ by $\lambda_{x} (g) := xg$, $$ \lambda_{x} (a) = \lambda_{x} (b) \implies xa = xb \implies a = b $$ so $\lambda_{x}$ is injective. Also, for all $c \in G$, $$ \lambda_{x} (x^{-1} c ) = x x^{-1} c = c $$ so $\lambda_{x}$ is surjective, and therefore $\lambda_{x} : G \to G$ is a permutation of $G$.

Then, for $x \in G$, we may define $\phi : G \to S_{G}$ by $\phi (x) = \lambda_{x}$. If $\phi (x) = \phi (y)$, then $\lambda_{x} = \lambda_{y}$ and $\lambda_{x} (e) = \lambda_{y} (e)$, so $$ x e = y e \implies x = y $$ that is, $\phi$ is injective. Meanwhile, since $\phi (xy) = \lambda_{xy}$, we have $\lambda_{xy}(g) = (x y) g$, and since $\lambda_{x} , \lambda_{y}$ are permutations, $$ ( \lambda_{x} \lambda_{y} ) (g) = \lambda_{x} (\lambda_{y} (g)) = \lambda_{x} (yg) = x (yg) $$ holds. In summary, $$\phi (xy) = \lambda_{xy} = \lambda_{x} \lambda_{y} = \phi (x) \phi (y)$$ that is, $\phi$ is a homomorphism.

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

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