# Lecture - Groups TT23, Cayley’s theorem

> Source: https://ollybritton.com/notes/uni/prelims/ht23/groups/lectures/cayleys-theorem/ · Updated: 2023-05-25 · Tags: uni, notes

- [Course - Groups and Group Actions TT23](https://ollybritton.com/notes/uni/prelims/tt23/groups/)

### Flashcards
<!-- flash-start -->
What theorem establishes a correspondence between actions and homomorphisms?::
Given a left action of a group $G$ on $S$, then there is an associated homomorphism
$$
\rho : G \to \mathrm{Sym}(S)
$$
and likewise for each homomorphism $\rho : G \to \mathrm{Sym}(S)$ there is an associated left action of $G$ on $S$.

^action-homomorphism-correspondence
<!-- flash-end -->

<!-- flash-start -->
If you have a homomorphism $\rho : G \to \mathrm{Sym}(S)$, how can you construct a group action (there’s no need to prove this is a group action)?::
$$
g \cdot s = (\rho(g))(s)
$$

^homomorphism-to-action
<!-- flash-end -->

<!-- flash-start -->
If you have a left action $\langle\cdot\rangle$ can you give an associated homomorphism (there’s no need to prove this is a homomorphism)?::
$$
\rho(g) = (x \mapsto g \cdot x)
$$

^action-to-homomorphism
<!-- flash-end -->

<!-- flash-start -->
@State Cayley’s theorem.::
Every finite group is isomorphic to a subgroup of a permutation group $S_n$ for some $n$.

^cayleys-theorem
<!-- flash-end -->

<!-- flash-start -->
Quickly justify that the map given by $G \to \mathrm{Sym}(S)$ given by
$$
\rho (g) = (x \mapsto g \cdot x)
$$
is a homomorphism.::
$$
\rho(gh) = (x \mapsto gh \cdot x) = (y \mapsto g \cdot y) \circ (x \mapsto h \cdot x) =  \rho(g) \circ \rho(h)
$$

^action-homomorphism-is-homomorphism
<!-- flash-end -->

<!-- flash-start -->
Quickly justify that there’s a valid left action given by $g \cdot x = \rho(g)(x)$ where $\rho$ is a homomorphism $\rho : G \to \mathrm{Sym}(S)$.::
$$
e \cdot x = \rho(e)(x) = \mathrm{id}(x) = x
$$
and
$$
g \cdot (h \cdot x) = \rho(g)(h \cdot x) = \rho(g)(\rho(h)(x)) = (\rho(g) \circ \rho(h) )(x) = \rho(gh)(x) = gh \cdot x
$$

^homomorphism-action-is-action
<!-- flash-end -->

### Proofs

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