Can someone explain to me how Isomorphism is linked to cayley's theorem?
Using cayley's theorem, it is stated that ' every group is isomorphic to a group of permutations'
Step 1: Let G be a given group and set G' of permutations form a grp isomorphic to G. Let Sg be the grp of all permutations of G. For a in G, let Pa be the mapping of G into G given by xPa = xa for x in G.
We then proceed by proving that Pa is one- to - one and onto.
May I know why there is a need to prove that Pa is one to one and onto?
Step 2: Claiming that G' is a subgroup of Sg, we then show that it is closed under permutation mulitplication, has identity permutation and an inverse.
This shows that G' is a subgroup of G but is this needed to prove the theorem?
Step 3: lastly, defining a mapping Ø: G -> G' and show that Ø is an isomorphism of G with G'.
define Ø: G -> G' by aØ = Pa for a in G
aØ = bØ
then Pa and Pb must be in the same permutations of G.
ePa = ePb
so a = b. thus Ø is one to one.
why do we have to prove that Ø is one to one when we have earlier proved that Pa is one to one?
my notes then continue to state that :
for the proof of the theorem, we consider the permutations xλa = xa for x in G
these permutations would have formed a subgroup G'' of Sg, again isomorphic to G but under the map ψ: G -> G'' defined by
aψ = λa-1
what does this remaining part of the proof mean?
Can someone explain what does right and left regular representations of a group G mean and the purpose of it?
Originally Posted by alexandrabel90
Let be a group. Define to be the symmetry group on , that is, consists of all bijections and under the binary operation of function composition. As you learned is the symmetric group (by definition). We are going to define a function . So for every we need to define so that becomes a permutation on . That is done by defining to be the function that is defined by . Notice that is a permutation of . Therefore, we will define . Notice, that is an element of but is a function on ! Be careful, this sometimes confuses people. The next thing to notice is that if then it must mean that so is one-to-one. Finally, is a homomorphism because . From this point you use the result from group theory that under an injective homomorphism the group is isomorphic to its homomorphic image, so, , but is a subgroup of . Since is a subgroup of a symmetric group it itself is a permutation group. Thus, all groups are isomorphic to permutation groups.