You have to prove that : it's a consequence of associativity and the fact that .
Can someone help me with proving this proposition:
Let a group. For we define:
Proof is an automorphism of (We call it an inside automorphism of determined by ).
If I wan't to proof is an automorphism I've to demonstrate that is an isomorphism or an bijective homomorphism.
1) is an homorphism
But to prove if is an homorphism in my inion I need a composite law for the group .
I guess I've to proof for two arbitrary chosen numbers in that:
But how do I now: ? Or doesn't that matter? ...
2) is bijective:
I've to prove that:
In this case:
This is true because in a group a is the symmetric element of and so:
So the only thing I haven't proved yet is the homorphism. Does someone have a hint?
Thanks in advance! .
(Sorry If my English wouldn't be correct, but English mathematical terminology is not easy for me as a Belgian).
No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).
Incidentally, it's good to get into the habit of thinking of an isomorphism as a homomorphism with homomorphic inverse, rather than a bijective homomorphism. In the category of groups, the latter is a consequence of the former, and either may be taken as the definition, but in many categories it's not true. For example, in the category of topological spaces, the equivalent notions to homomorphism and isomorphism are continuous map and homeomorphism--and a bijective continuous map need not be a homeomorphism. The correct definition is a continuous map with continuous inverse.
You have already shown that is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).
It suffices to show that has an inverse. When you look further into this then you will notice that is indeed the inverse of . You can see this by noting that if then it follows that for any element (because for any element ).
Does this make sense?
(For example, if we take the function , then if we define then , but clearly isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))