Definition of the Isomorphism theorem

Question

I have a question about the definition of the Isomorphism theorem given my notes.

The Isomorphism Theorem: Let $\phi : G \rightarrow G'$ be a group homomorphism with kernel $K$. Then $\phi$ induces an isomorpism $\bar{\phi}: G/K \rightarrow \phi (G')$ given by $\bar{\phi}(Kg) = \phi(g)$ for all $Kg$ in $G/K$ ...

However shouldn't the part be "for all $g\in G$"? Is this the same thing?

Answer 1

Formally, the statement is more correct. If you define a function $f:A\longrightarrow B$, you must specify the action of $f$ over a generic element of $A$.

In your case, a more pompous statement (but formally more precise) should be: $\bar{\phi}:G/K\longrightarrow \phi(G)$ is defined in the following way: for every coset $\alpha \in G/K$, let be $g\in G$ a representative of $\alpha$, id est such that $\alpha=Kg$, let be $$\bar{\phi}(\alpha)=\bar{\phi}(Kg)=\phi(g)$$

The solution you provide is perfectly correct, since $G/K=\{Kg\mid g\in G\}$.

Answer 2

It is the same thing because $Kg$ runs over $G/K$ when $g$ runs over $G$.

However, the key point is that $\bar{\phi}(Kg) = \phi(g)$ is well defined, that is, does not depend on the choice of the representative $g$ for the coset $Kg$. This follows because $K = \ker \phi$, of course.

Answer 3

Do you understand the line " induces an isomorphism...." ?

Your statement of the 1st Isomorphism Theorem does not mention the canonical homomorphism G to G/K , nor the factorisation of the homomorphism G to G'.