Proving $\phi$ is well-defined Let $H$ and $K$ be normal subgroups of a group $G$, with $H \subseteq K$. Define $\phi: G/H \rightarrow G/K$ by $\phi(Ha)=Ka.$
Prove that $\phi$ is a well-defined function (i.e., if $Ha=Hb$, then $\phi(Ha)= \phi(Hb)$.

Honestly I have never understood well-defined and how to prove it.
 A: First, if $J$ is any subgroup of $G$ note that $Ja=Jb$ if and only if $ab^{-1}\in J$ (Using the definition of right cosets) then:
$Ha=Hb \implies ab^{-1}\in H\subset K$ then $ab^{-1}\in K \implies Ka=Kb$
So we have proved $Ha=Hb$ implies $\phi(Ha)=\phi(Hb)$. Then $\phi$ is a well defined function since for each way to write the same right coset (element in domain of $\phi$) we have only one image (element in codomain of $\phi$).
A: The concept is indeed a bit subtle. Let’s restate the definition of $\phi$, which is supposed to take an $H$-coset and assign to it a $K$-coset. The definition can be read this way: “Given an $H$-coset $S$, $\phi(S)$ is defined to be a $K$-coset as follows. Choose any element $a$ of $S$, and then $\phi(S)$ will be the $K$-coset that contains $a$.”
Now it becomes clear that the resulting $K$-coset apparently depends on your choice of $a$: if you chose a different element of $S$, would you get the same $K$-coset by applying the recipe to that? Showing that which element of $S$ you chose doesn’t affect the final resulting $K$-coset is just what is being asked for.
