# Tour:Cayley's theorem

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WHAT YOU NEED TO DO:
• Read and understand the statement of Cayley's theorem, both in terms of group actions and in terms of homomorphisms.
• Understand the proof of the theorem.

## Statement

### In terms of group actions

Let $G$ be a group. The group multiplication $G \times G \to G$, defines a group action of $G$ on itself. In other words, the left multiplication gives an action of $G$ on itself, with the rule $g.h = gh$. This action is termed the left-regular group action.

This group action is faithful -- no non-identity element of $G$ acts trivially.

### In terms of homomorphisms

Let $G$ be a group. There is a homomorphism from $G$ to $\operatorname{Sym}(G)$ (the symmetric group, i.e., the group of all permutations, on the underlying set of $G$). Moreover, this homomorphism is injective. Thus, every group can be realized as a subgroup of a symmetric group.

## Proof

### In terms of group actions

Given: A group $G$.

To prove: $G$ acts on itself by left multiplication, and this gives an injective homomorphism from $G$ to the symmetric group on $G$.

Proof: Define the left-regular group action of $G$ on itself by $g.h = gh$.

1. This is a group action: $e.s = s$ follows from the fact that $e$ is the identity element, while $g.(h.s) = (gh).s$ follows from associativity.
2. The action is faithful; every non-identity element of the group gives a non-identity permutation: Assume that there are $g, h \in G$ such that their action by left multiplication is identical. But then $ge = he$ so $g = h$. Therefore, the action is faithful.

Thus, we get a homomorphism from $G$ to $\operatorname{Sym}(G)$. Since the action is faithful, distinct elements of $G$ go to distinct elements of $\operatorname{Sym}(G)$, so the map is injective. In particular, $G$ is isomorphic to a subgroup of $\operatorname{Sym}(G)$.