Register to reply

Cayley's theorem

by alexandrabel
Tags: cayley, theorem
Share this thread:
alexandrabel
#1
Sep1-09, 08:20 AM
P: 1
Hello!

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'

Proof:

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 : G -> G' and show that is an isomorphism of G with G'.

define : G -> G' by a = Pa for a in G

a = b
then Pa and Pb must be in the same permutations of G.
ePa = ePb
so a = b. thus is one to one.


why do we have to prove that 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?

thanks!
Phys.Org News Partner Science news on Phys.org
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
morphism
#2
Sep6-09, 08:07 PM
Sci Advisor
HW Helper
P: 2,020
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.


Register to reply

Related Discussions
Rolles Theorem/ Mean Value Theorem + First Derivative Test Calculus & Beyond Homework 6
Extreme Value Theorem & MVT/Rolles Theorem Calculus & Beyond Homework 2
Categorical extension of Cayley's Theorem General Math 3
Greens theorem and cauchy theorem help Calculus & Beyond Homework 4
Gauss's Divergance Theorem and Stokes's Theorem Classical Physics 1