Sep1-09, 08:20 AM
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?
Sep6-09, 08:07 PM
I am confused by your definition of G'. Could you clarify it a bit?
The gist of the proof is simple: each element a in G gives rise to a permutation Pa:G->G which sends x to ax. Pa is a permutation because, as a function, its inverse is Pa-1. In other words, Pa lives in Sg. Now consider the map F:G->Sg sending a to Pa. This map is an injective homomorphism. So G is isomorphic to F(G), and F(G) is a group of permutation. QED.
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