# 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.

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When you are proving something is well defined, you are making sure that it is actually a function. This means that if two things are equal, they must be sent to the same thing, which is the only real requirement for a function. The most common time that you have to check if a function is well defined is when the domain is a set of equivalence classes (Like a quotient group). You have to make sure that the value of the function doesn't depend on the representative you choose from the equivalence class, because different representatives from the same class need to evaluate to the same thing. – Thomas Credeur May 2 '14 at 3:57
Also, sorry for the wall of text. – Thomas Credeur May 2 '14 at 3:58
Ah thank you so much for putting that into plain English for me! That makes so much sense – allie May 5 '14 at 20:23
No problem. Until my first semester of graduate algebra I never really knew when to check for that either. I had a great professor who made it easy to understand though. – Thomas Credeur May 5 '14 at 20:58

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 $\phi$).

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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.

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