# Math Help - Cayley's theorem together with isomorphic groups

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 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!

2. 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.