Hi, I've got this question which I'm sure should be easy but I just don't seem to be able to get started...
I have a centre of a group G defined as
Z(G) = {x is an element of G : for all y contained in G xy=yx}.
Also k: G is mapped to G is an automorphism of G and automorphisms form a group Aut(G) under composition.
I have to show that the function f: G is mapped to Aut(G) defined by
[f(g)](h) = ghg^-1 is a group homomorphism.
I know that I have to show that f(g1,g2) = f(g1)f(g2) but I don't know how to go about it when the function is defined the way it is.
Any hints or advice would be very much apprecaited.
THANKS!
Yes. I understood what you meant by it, but I did not like how it was written, so I wrote it in a way I found nicer.
If define as then is an automorphism by a straightforward computation. Now define as (which was just defined above), this is a valid mapping because it takes elements in an turns them into automorphisms.
Thanks again for your help. I've managed to get a lot further now. I have just got to a point where I have shown that Ker(f) = Z(G) but now I need to prove that Im(f) is a normal subgroup of Aut(G). I know the following:
- By a proposition in my notes im(f) is a subgroup of Aut(G)
- G/Z(G) = Inn(G) by the 1st isomorphism theorem (I'm not sure if I can use this as we haven't defined the inner automorphism group in class)
- N is normal in G implies that there is some homomorphism of G for which N is the Kernel.
I don't know if I'm on track with any of those but it does seem like the question has led me straight to the First Isomorphism Theorem by getting me to find the Kernel. I'm not sure how to use it though! Its so frustrating because I feel like I'm close but I don't quite understand all this stuff well enough yet to make any connections or have a brainwave!
Any help/advice would be brilliant!
Thanks,
Sooz
Okay, thus we established that defined as is a homomorphism. Now by the fundamental isomorphism theorem is means . Now where is the group of all inner automomorphism*. The kernel is the set of all such that gets mapped into the identity of , i.e. the identity automorphism. Thus, we require that to leave all element unpremuteted meaning if and only if if and only if (the center of the group) by definition.
Hence, we have shown that, .
*)The automorphism is called an inner automorphism. It can be shown that the set of all inner automorphisms forms a normal subgroup of the automorphism group .
Let . First we need to prove it is a subgroup. It is closed because if is one inner automorphism and is another inner automorphism then their product (which is function composition) is which is an inner automorphism. Also, is the identity automorphism. It is associative because in general function compositions are associative. And finally it has an inverse. Thus, the set of inner automorphism form a subgroup. To show that we need to show that for all and we have . Can you do this last part? If not post what you can do.
I can't seem to do very much at all! I started off with
but do I just look at the whole thing in one go or just look at first? By doing this I get that this is equal to but I don't see how that helps or if it is even right! I understand the subgroup bit though!
Thanks again,
Sooz