Tour:Cayley's theorem
This article adapts material from the main article: 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.
Contents
Statement
In terms of group actions
Let be a group. The group multiplication
, defines a group action of
on itself. In other words, the left multiplication gives an action of
on itself, with the rule
. This action is termed the left-regular group action.
This group action is faithful -- no non-identity element of acts trivially.
In terms of homomorphisms
Let be a group. There is a homomorphism from
to
(the symmetric group, i.e., the group of all permutations, on the underlying set of
). 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 .
To prove: acts on itself by left multiplication, and this gives an injective homomorphism from
to the symmetric group on
.
Proof: Define the left-regular group action of on itself by
.
- This is a group action:
follows from the fact that
is the identity element, while
follows from associativity.
- The action is faithful; every non-identity element of the group gives a non-identity permutation: Assume that there are
such that their action by left multiplication is identical. But then
so
. Therefore, the action is faithful.
Thus, we get a homomorphism from to
. Since the action is faithful, distinct elements of
go to distinct elements of
, so the map is injective. In particular,
is isomorphic to a subgroup of
.
This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
PREVIOUS: Equivalence of definitions of group action| UP: Introduction five (beginners)| NEXT: External direct product
General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part