Proving Map is Well-defined Okay so I have this question:

Let $G$ be a group and suppose that $N$ and $K$ are normal subgroups
  of $G$, where $N \leq K$. Define a map: $\theta:G/N \rightarrow G/K$ 
  by  $\theta(aN)=aK$. Show that $\theta$ is well defined.

I know that in order to show a map is well defined, I have to show that $x_1=x_2 \implies \theta(x_1)=\theta(x_2)$. So for this question, is it a case of proving $aN_1 = aN_2 \implies aK_1=aK_2$? I've proved maps are well defined in the past, but I just don't understand the question  here, can someone explain what it is I have to prove please? Thanks in advance.
 A: What you want to prove is that the map is independent of choice of representative. This is what is usually meant by well-defined. Notice that you picked a specific group element $a$ in the definition of $\theta$. Since everything in $aN$ is "equivalent", $\theta(aN)$ should be the same as $\theta(bN)$ if $aN=bN$. That is, if you pick two different representatives $a$ and $b$, you get the same result. That $N$ is a subgroup of $K$ is important here.
A: That's almost it. What you actually want to do is show that $a_1N=a_2N$ implies that $a_1K=a_2K$, since the inputs into $\theta$ are different left cosets of $N$.
A: Since $N$ and $K$ are fixed (i.e. there are no $N_1, N_2$), you rather have to show $$a_1N=a_2N\implies a_1K=a_2K.$$
A: I'm a bit iffy but I think you can say:
aN, bN are in N so (ab^-1)N is in N
aN = bN so (ab^-1) = 1 
θ((ab^-1)N)=(ab^-1)K = aK.(b^-1)K
But as (ab^-1) = 1, θ((ab^-1)N) = θ(1) = 1 (identity element is the same as N is subgroup of K)
So aK.(b^-1)K = θ((ab^-1)N) = 1
so aK = bK
Sorry about my bad formatting, no idea how to do it on this forum!
