§ Semidirect product is equivalent to splitting of exact sequence
Consider the exact sequence
- We want to show that if there exists a map such that (ie, ), then G . So the splitting of the exact sequence decomposes into a semidirect product.
- The idea is that elements of have an part and a part. We can get the part by first pushing into using and then pulling back using . So define . This gives us the "K" part. To get the part, invert the "k part" to annihiliate it from . So define a map .
- See that the image of lies entirely in the kernel of , or the image of indeed lies in . This is a check:
- Hence, the image of is entirely in the kernel of . But the kernel of is isomorphic to , and hence the image of is isomorphic to . So we've managed to decompose an element of into a part and an part.
- Write as , by the map . Let's discover the composition law.
We need the second to be , so that composes in an
entirely straightforward fashion. For the other component, we need:
So we need the of to be twisted by the component of by a conjugation. So we define the semidirect
We've checked that this works with the group structure. So we now have a morphism .
we need to check that it's an isomorphism, so we need to make sure that this has full image and trivial kernel.
- Full image: Let . Create the element . We get . We get