# 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.

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.

• Of course :) glad to help Commented Dec 4, 2014 at 19:54

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 $N$.

• Thank you! I'll get to work on the proof. Commented Dec 4, 2014 at 19:53
• You're very welcome. Good luck! Commented Dec 4, 2014 at 19:55

Since $N$ and $K$ are fixed (i.e. there are no $N_1, N_2$), you rather have to show $$a_1N=a_2N\implies a_1K=a_2K.$$

• I knew I was barking up the wrong tree, thank you! Commented Dec 4, 2014 at 19:52

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!

• Add a dollar symbol around any text (MathJax tutorial) you want parsing. Thanks for the proof, but I ended up doing it a slightly different way: $a_1N = a_2N \iff y^{-1}x \in N$ hence $y^{-1}x \in K$ as $N \le K \implies x_1K=x_2K$ (Hope that formats correctly, I don't have a live preview so fingers crossed) Commented Dec 4, 2014 at 21:42
• Thanks =) How's other theriault q's going? I'm really finding this one much harder than the last one! No idea how to show that this is an epimorphism! Commented Dec 4, 2014 at 22:09
• Awful! This coursework has just about killed me and at this rate I'll be pulling an all nighter (answered about 30% so far). I've seen someone else ask for help on the first question so it looks like there's a few of us who are struggling. I'm working on the epimorphism proof now and I'll post it here if I can work it out :) Commented Dec 4, 2014 at 22:49
• I've done all of 1 (though my subgroup lattice is looking rather sparse) I think so if you need any help with that just ask =) Commented Dec 4, 2014 at 23:03
• Thanks so much, I'm working through 2 now but I'll certainly take you up on that one in a bit. Epimorphism: Any coset of M can be written as $xM$ for $x \in G$, and $xM=\theta(xN)$ Commented Dec 4, 2014 at 23:10