>>10132505Okay.
Imagine an abstract set X and a set S of bijections from X to X that is closed under composition. If the set is closed, f=f^n guarantees the presence of the identity function, otherwise we have to specify it and also inverses are in the set. S is a group.
If I have two groups, R and S a homomorphism maps elements of R to elements of S, but we can also interpret it as a function from X to X. This might sound weird, but imagine X is the Natural numbers, and we have the five symmetric group swapping (1, 2, 3, 4, 5), and fixing the rest of the naturals and the seven symmetric group on (6, 7, 8, 9, 10, 11, 12). If a function ? from X to X maps 6 to 1, 7 to 2, etc, 11 to 1, and 12 to 1, then to each element of the seven group is associated an element of the five group. ?(6, 7)=(1, 2), and (11, 12) naturally is the kernel. Of course, we can still interpret the homomorphism as a function from one group to another, and its cleaner to define it innths old way.
Now I take every element of a group G and associate to each of them an automorphism, writing a=(f, ?), b=(g, ?) and c=(h, ?). We define sum as a+b=fog, and axb=fo?=go?. Why we intepret a multiplication as an automorphism is really simple: distribution. If a homomorphism gives ?o(aob)=(?oa)o(?ob), then classical multiplication is a(c+d)=ac+ad!
I've only started formulating this, so sorry if it's a bit confusing or even wrong.
