Exhibiting a homomorphism f: G-> Aut(G)

[AdrianZ]

I'm watching Harvard's abstract algebra course online and the professor says that the map f(g)(h)=ghg^-1 is a homomorphism between the groups G and Aut(G). the thing that I don't understand is that whether ghg^-1 is an element of Aut(G) or not. It must be an element of Aut(G) because It's in the image of f but I can't figure out how it is an element of Aut(G). Can someone clarify what f(g)(h) means at the first place? if h is the input, then what is f(g)? Is f(g)(h) the same thing as fog(h)? Shouldn't our input be from G and our output in Aut(G)? so if h is in G, Is ghg^-1 an automorphism on the group G under its operation? I'm confused

 

[HallsofIvy]

Do you agree that $f_g(h)= ghg^{-1}$ (I prefer that notation so we don't get "g" and "h" confused) is "auto"- that is, that it maps amember of the group to a member of the group? That should be clear because the group is closed under the group operation.

So the only thing left to prove is that it is a homomorphism: that is, that $f_g(hk)= f_g(h)f_g(k)$.

Okay, $f_g(h)f_g(k)= (ghg^{-1})(gkg^{-1})$. What does that reduce to?

 

[AdrianZ]

well, I think the main confusion was about the notation he had used. so, ghg^-1 is indeed an automorphism therefore it's in Aut(G). now It's clear that f is a group homomorphism. Thanks for the help HallsofIvy.

 

[micromass]

so, ghg^-1 is indeed an automorphism therefore it's in Aut(G).

No, it's not. $ghg^{-1}$ is an element of G, not of Aut(G).
The map

$$f_g:G\rightarrow G:h\rightarrow ghg^{-1}$$

is in Aut(G). And the map

$$f:G\rightarrow Aut(G):g\rightarrow f_g$$

is the homomorphism.

 

[AdrianZ]

No, it's not. $ghg^{-1}$ is an element of G, not of Aut(G).
The map

$$f_g:G\rightarrow G:h\rightarrow ghg^{-1}$$

is in Aut(G). And the map

$$f:G\rightarrow Aut(G):g\rightarrow f_g$$

is the homomorphism.

Yea, I meant the same thing. Thanks for pointing out the slight difference in the way I had phrased my sentence. I meant ghg^-1 defines a set theoretic automorphism on G (It's one-to-one and onto) and f_g (the function defined by that relation) satisfies f_g(h.h')=f_g(h) . f_g(h').
Now as you said, the map $$f:G\rightarrow Aut(G):g\rightarrow f_g$$ is a homomorphism because we have: f_g.g'(h)=(g.g').h.(g.g')^-1=(g.g').h.(g')^-1g^-1=g(g'h(g')^-1)g^-1=g(f_g'(h))g^-1=f_g(f_g'(h))=f_gof_g'(h).

the thing is that Benedict Gross first misinterpreted the problem, so I too got confused what he was talking about and lost the path we were following.