Groups and Inner Automorphisms

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Homework Statement


Let G be a group. Show that G/Z(G) \cong Inn(G)


The Attempt at a Solution


G/Z(G) = gnZ(G) for some g ε G and for any n ε N
choose some g-1 such that
g(g-1h) = g(hg-1)
and the same can be done switching the g and g-1

This doesn't feel right at all...
 
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Locoism said:

The Attempt at a Solution


G/Z(G) = gnZ(G) for some g ε G and for any n ε N
choose some g-1 such that
g(g-1h) = g(hg-1)
and the same can be done switching the g and g-1

This doesn't feel right at all...

I don't see what this has to do with the problem??

Can you find a surjective homomorphism

f:G\rightarrow Inn(G)

and then apply the first isomorphism theorem?
 
Locoism said:

Homework Statement


Let G be a group. Show that G/Z(G) \cong Inn(G)


The Attempt at a Solution


G/Z(G) = gnZ(G) for some g ε G and for any n ε N
choose some g-1 such that
g(g-1h) = g(hg-1)
and the same can be done switching the g and g-1

This doesn't feel right at all...

No, not right. Given an element g of G can you name an inner automorphism of G corresponding to g? When do two different elements of G, g1 and g2 give define the same automorphism?
 
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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