2008 February 08 « The Lord of The Definite Integrals

Well-Definededness

In Algebra, we just finished chapter 10 of Gallian. One of the big theorems in the chapter is the First Isomorphism Theorem – and since I’m a lazy typist I’ll call it FIT from now on. (Isomorphism theorems 2 and 3 are left as exercises.) Skipping over what a group is, what a homomorphism is, what the kernel of said homomorphism is, and what normal subgroups and factors groups are (see, FIT talks about all sorts of cool stuff!), FIT says:

Suppose $\theta :G \longrightarrow G^\prime$ is a homomorphism. Then $G/\text{ker}\theta \cong G^\prime\theta$ ; moreover, the map $\theta : G/\text{ker}\theta \longrightarrow G^\prime\theta$ by $g\text{ker}\theta \mapsto g\theta$ is an isomorphism.

One of the points of the proof, since $\theta$ is defined on the cosets of $G/\text{ker}\theta$ by specifying the image of a representative, is whether or not the map $\theta$ is even well defined for some $g\text{ker}\theta \in G/\text{ker}\theta.$ In particular, will taking different representatives of the coset $g\text{ker}\theta$ always have the same image? It turns out that the map is indeed well defined – it’s a rather straightforward argument – but this just brings up another question: