# 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.

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. Dec 4 '14 at 19:53
• You're very welcome. Good luck! Dec 4 '14 at 19:55

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 Dec 4 '14 at 19:54

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! Dec 4 '14 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) Dec 4 '14 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! Dec 4 '14 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 :) Dec 4 '14 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 =) Dec 4 '14 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 $xM=\theta(xN)$ Dec 4 '14 at 23:10