Group Theory inner automorphism

Lee33
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Homework Statement

How do I prove that the inner automorphisms is isomorphic to ##S_3##?


The attempt at a solution

I know ##S_3 = \{f: \{ 1,2,3 \}\to\{ 1,2,3 \}\mid f\text{ is a permutation}\}## and I know for every group there is a map whose center is its kernel so the center of of ##S_3## is trivial therefore ##S_3/Z(S_3) = 6##.

So an inner automorphism of a group ##G## is an automorphism of the form ##ρ_g :x↦gxg^{-1}. ## What I am having trouble with is verifying that distinct elements of ##S_3## give distinct inner automorphism. How can I prove this problem directly?
 
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The inner automorphisms of what are supposed to be isomorphic to S3?
 
Lee33 said:
Homework Statement

How do I prove that the inner automorphisms is isomorphic to ##S_3##?


The attempt at a solution

I know ##S_3 = \{f: \{ 1,2,3 \}\to\{ 1,2,3 \}\mid f\text{ is a permutation}\}## and I know for every group there is a map whose center is its kernel so the center of of ##S_3## is trivial therefore ##S_3/Z(S_3) = 6##.

So an inner automorphism of a group ##G## is an automorphism of the form ##ρ_g :x↦gxg^{-1}. ## What I am having trouble with is verifying that distinct elements of ##S_3## give distinct inner automorphism. How can I prove this problem directly?

Can you prove that if \rho_g = \rho_h then g^{-1}h \in Z(G)?
 
Opps, I forgot to add the inner automorphism of ##S_3## is isomorphic to ##S_3.## Sorry for the confusion!

Pasmith - Can you elaborate please?
 
Lee33 said:
Opps, I forgot to add the inner automorphism of ##S_3## is isomorphic to ##S_3.## Sorry for the confusion!

Pasmith - Can you elaborate please?

There's no need to elaborate. pasmith pretty much spelled it out. Try to prove the hint he gave you. Once you've done that figure out what the center of ##S_3## is. It should click easily after that.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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