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

1. Sep 18, 2011

### 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

2. Sep 18, 2011

### 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?

Last edited by a moderator: Sep 18, 2011
3. Sep 18, 2011

### 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.

4. Sep 18, 2011

### micromass

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.

5. Sep 18, 2011

### AdrianZ

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 fg (the function defined by that relation) satisfies fg(h.h')=fg(h) . fg(h').
Now as you said, the map $$f:G\rightarrow Aut(G):g\rightarrow f_g$$ is a homomorphism because we have: fg.g'(h)=(g.g').h.(g.g')-1=(g.g').h.(g')-1g-1=g(g'h(g')-1)g-1=g(fg'(h))g-1=fg(fg'(h))=fgofg'(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.

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