# Cayley's theorem together with isomorphic groups

• Sep 1st 2009, 03:29 AM
alexandrabel90
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 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?

thanks!

Can someone explain what does right and left regular representations of a group G mean and the purpose of it?

thanks!
• Sep 1st 2009, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by alexandrabel90
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 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?

thanks!

Can someone explain what does right and left regular representations of a group G mean and the purpose of it?

thanks!

Let $G$ be a group. Define $\text{sym}(G)$ to be the symmetry group on $G$, that is, $\text{sym}(G)$ consists of all bijections $f:G\to G$ and under the binary operation of function composition. As you learned $\text{sym}(G)$ is the symmetric group (by definition). We are going to define a function $\phi: G\to \text{sym}(G)$. So for every $g\in G$ we need to define $\phi(g)$ so that $\phi(g)$ becomes a permutation on $G$. That is done by defining $\phi(g)$ to be the function $p_g: G\to G$ that is defined by $p_g(x) = gx$. Notice that $p_g$ is a permutation of $G$. Therefore, we will define $\phi(g) = p_g$. Notice, that $g$ is an element of $G$ but $\phi(g)$ is a function on $G$! Be careful, this sometimes confuses people. The next thing to notice is that if $p_{g_1} = p_{g_2}$ then it must mean that $g_1 = g_2$ so $\phi$ is one-to-one. Finally, $\phi$ is a homomorphism because $\phi(g_1g_2) = \phi(p_1)\phi(g_2)$. From this point you use the result from group theory that under an injective homomorphism the group is isomorphic to its homomorphic image, so, $G\simeq \phi[G]$, but $\phi[G]$ is a subgroup of $\text{sym}(G)$. Since $\phi[G]$ is a subgroup of a symmetric group it itself is a permutation group. Thus, all groups are isomorphic to permutation groups.