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gentsagree
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From Wikipedia:
Consider the map [itex]f: G \rightarrow Aut(G)[/itex] from G to the automorphism group of G defined by [itex]f(g)=\phi_{g}[/itex], where [itex]\phi_{g}[/itex] is the automorphism of G defined by
[tex]\phi_{G}(h)=ghg^{-1}[/tex]
The function f is a group homomorphism, and its kernel is precisely the center of G, and its image is called the inner automorphism group of G, denoted Inn(G). By the first isomorphism theorem we get
[tex]G/Z(G) \simeq Inn(G)[/tex]
My understanding is this: the centre of G is the kernel of the automorphism as its elements give the identity mapping under automorphisms (conjugation leaves the element unchanged)?
But then, why isn't the quotient [itex]G/Z(G)[/itex] isomorphic to Aut(G) rather than Inn(G)? The first isomorphism theorem states that the quotient group [itex]G/Ker(\phi)[/itex] is isomorphic to the image of the homomorphism, and here the homomorphism is the automorphisms, not the inner automorphisms.
Where am I going wrong?
Consider the map [itex]f: G \rightarrow Aut(G)[/itex] from G to the automorphism group of G defined by [itex]f(g)=\phi_{g}[/itex], where [itex]\phi_{g}[/itex] is the automorphism of G defined by
[tex]\phi_{G}(h)=ghg^{-1}[/tex]
The function f is a group homomorphism, and its kernel is precisely the center of G, and its image is called the inner automorphism group of G, denoted Inn(G). By the first isomorphism theorem we get
[tex]G/Z(G) \simeq Inn(G)[/tex]
My understanding is this: the centre of G is the kernel of the automorphism as its elements give the identity mapping under automorphisms (conjugation leaves the element unchanged)?
But then, why isn't the quotient [itex]G/Z(G)[/itex] isomorphic to Aut(G) rather than Inn(G)? The first isomorphism theorem states that the quotient group [itex]G/Ker(\phi)[/itex] is isomorphic to the image of the homomorphism, and here the homomorphism is the automorphisms, not the inner automorphisms.
Where am I going wrong?
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