Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows:
If Aut(G) and is conjugation by g prove that = .
Deduce that Inn(G) is a normal subgroup of Aut(G).
Can anyone help me get started on this problem?
(1) I have attached Dummit and Foote section 4.4 in case readers need to view the terminology and notation used. This upload includes the text of the exercise.
(2) I am assuming that the mapping is h gh
(3) I am not sure what Dummit and Foote mean by - unless they mean the set of automorphisms that map g into g???
The example was very helpful in getting an understanding of the problem.
Basically he equivalence between your notation and D&Fs is as follows:
where (as is usual) is the map of g under
I am still struggling with the proof of = and would appreciate some further help.
Regarding Inn(G) Aut(G) it appears to me that is essentially conjugation of G by g and hence is an inner automorphism of G and hence Inn(G).
But where to from here to show Inn(G) Aut(G)
two maps are equal if they take on equal values for every element in their domain.
let's say we have any old automorphism of G, which we call . this is a bijective homomorphism, and let's say . clearly, .
now by , we mean the particular automorphism that is conjugation by g: for any x. so:
(since is a homomorphism)
(again, since is a homomorphism, and by our definition of y)
so when we conjugate the inner automorphism corresponding to conjugation by g, by another automorphism , we get the same map as conjugation by the element .
this shows that for any automorphism , that