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Let $G$ be a group, $H\leq G$, and $N\triangleleft G$ ($H$ is a subgroup, and $N$ is normal subgroup in $G$).
Let us define $H\cap N=K$

Let $\phi:H/K\rightarrow G/N$ be a map between these two quotient groups (I proved that $H\cap N$ is a normal subgroup in H, so the quotient group $H/K$ is indeed a group). $\phi$ defined as:
for every $aK\in H/K$; we have $\phi(aK)=aN$.
I also proved that $\phi$ is well defined, meaning that for every $a,b\in H$, if $aK=bK$, then $\phi(aK)=\phi(bK)$.
Now, there's only one last thing I need to prove: $\phi$ is injective, meaning that for every $aK, bK\in H/K$, if $\phi(aK)=\phi(bK)$, then $aK=bK$.

So far, this is my work;
If $\phi(aK)=\phi(bK)$ then it holds that $aN=bN$, so $a$ and $b$ are in the same coset of $N$, and so $an_0=bn_1$ for some $n_0, n_1\in N$.
And that is it, basically, (not much, I know) but I just couldn't think of anything that will lead me to $aK=bK$.

Any help would be greatly appreciated.

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  • $\begingroup$ Yes, of course. thank you. $\endgroup$ – so.very.tired Dec 30 '13 at 20:29
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Hint: Consider the Homomorphism $\varphi:H \rightarrow G/N$ that is defined by the inclusion $H\rightarrow G$ and the projection $G \rightarrow G/N$. Then what is the kernel of $\varphi$? Use the 1. Isomorphism Theorem.

Note that this also shows that $N\cap H$ is a normal subgroup of $H$.

As a rule of thumb it is often easier to induce homomorphisms of the quotient $H/K$ by homomorphisms of $H$.

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  • $\begingroup$ +1, not only is your rule of thumb spot on, but also the apparently innocuous approach to $\varphi$ as the composition of inclusion and projection is the right way to think about it. At least, that's the way I teach it to my students ;-) $\endgroup$ – Andreas Caranti Dec 30 '13 at 16:10
  • $\begingroup$ Hi benh. thanks for the help. OK, so we didn't learn about inclusion and projection, and this is the first time I hear about these functions, but from reading in Wikipedia, I concluded that the kernel of $\varphi$, as you defined it, is exactly $H/H\cap N$, and using the first Isomorphism Theorem, we'll get that $H/H\cap N\cong Im\varphi$. but what is $Im\varphi$, and how does it relate to my original $\phi$? $\endgroup$ – so.very.tired Dec 30 '13 at 20:58
  • $\begingroup$ @Andreas Caranti: thanks :) I agree. To so.very.tired: So the map $H \mapsto G$ just interprets an element of $H$ as an element of $G$, which is possible because of $H \subset G$. You easily check that it's an injective homomorphism. The projection $\pi: G \mapsto G/N$ maps an element of $G$ to its corresponding coset. That, too, is a homomorphism. We get a homomorphism $\varphi:H \mapsto G/N$ by composing the two. Now the kernel of $\varphi$ must be a normal subgroup of $H$. $\endgroup$ – benh Dec 30 '13 at 21:10
  • $\begingroup$ As the kernel of $\pi$ is $N$, we find that $ker(\varphi)=H \cap N$ ("the elements of the kernel that are already in $H$"). Now the 1. Isomorphism Theorem gives you an injection $H/H\cap N \rightarrow G/N$. The intuition behind this is: if we factor out all elements that get mapped to $0$, we get a map which maps only $0$ to $0$. This means, that $\varphi$ is injective. $\endgroup$ – benh Dec 30 '13 at 21:11
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    $\begingroup$ It's a bijection on the image of $\varphi$, hence an injection. It's a bijection if and only if it was onto before. As a counter-example where $\varphi$ is not onto take $H=0$ and $N\neq G$. By the way, the reason why a bijection on the image is an injection is easy to see if you compose the bijection with the inclusion $im(\varphi) \rightarrow G/N$. We just learned that inclusions are injective, so the map $\varphi$ must be as a composition of a bijection (on the image) with an inclusion ($im(\varphi)$ in $G/N$). $\endgroup$ – benh Dec 30 '13 at 21:34
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Consider the map $i:H\to G/N, h\mapsto hN$. Since each $n\in H\cap N$ is in $N$, we have $H\cap N\subseteq ker(i)$. This implies the well-definedness of $$\tilde i:\dfrac H{H\cap N}\to \frac GN$$Conversely, if $h\in H$ is in $ker(i)$, then $h$ must be in $N$. So we have $ker(i)=N\cap H$. Now recall that a homomorphism with a trivial kernel is injective, so $\tilde i$ is injective.

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  • $\begingroup$ Hi Stefan, thanks for the help. If the kernel of $i$ is trivial, doesn't that mean that $i$ is injective? (rather then $\tilde{i}$ - which is what we want) $\endgroup$ – so.very.tired Dec 30 '13 at 21:05
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    $\begingroup$ @so.very.tired: The kernel of $i$ is $K=H\cap N$, so the kernel of $\tilde i$, which sends $hK$ to $i(h)$, is trivial (it is the singleton $\{K\}$, which is the neutral element in the quotient group $H/K$). So $\tilde i$ is the injective map here. $\endgroup$ – Stefan Hamcke Dec 31 '13 at 1:10

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