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Let $G,H$ be finite groups. Let $N$ be a normal subgroup of $G$. Let $\phi$ be a group homomorphism which maps from $G \rightarrow H$. And let $N \subset \mathrm{ker}(\phi)$

I'm trying to proof that $\exists$ group homomorphism $\tilde{\phi}: G/N \rightarrow H$ s.t. $\tilde{\phi} \circ \pi = \phi$

Here's my proof so far:

Define: $\tilde{\phi}$ by $\tilde{\phi} := \phi(g)$. This is well defined (meaning for each representative of a coset, this gives the same result) because if $gN = g'N \implies \exists n \in \mathbb{N}$ s.t. $g' = gn$ $\implies \phi(g') = \tilde{\phi}(g'N) = \tilde{\phi}(gnN) = \tilde{\phi}(gN) = \phi(g) $ (where we used that for $n \in N$ the function $n \cdot n$ is a bijection of $N$).

So we only need to show that $\tilde{\phi}$ really is a group homomorphism: $\tilde{\phi}((gN)(g'N)) = \tilde{\phi}(gNg'N) = \tilde{\phi}(gg'NN) = \tilde{\phi}(gg'N) = \phi(gg') = \phi(g) \phi(g') = \tilde{\phi}(gN)\tilde{\phi}(g'N)$ Where we have used the normality of $N$ and the property of $n \cdot n$.

I'm just looking for a quick validation that this proof is water proof and I didn't mess up somewhere.

Thanks a lot in advance!

Cheers :)

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For the correctness: You want to prove $gN = n'N \implies \phi(g) = \phi(g')$, that's right. But you cannot use $\tilde{\phi}$ that way because you don't know its definition is correct. That's what you want to prove. You also didn't use the assumption $N ⊆ \ker{\phi}$ (and that's what you need).

For the homomorphism: The property “$n \mapsto n · n$ is bijection” does not hold. Consider $\mathbb{Z}$ which is normal in itself. $(n \mapsto n · n)$ maps $\mathbb{Z}$ to even numbers. $(gN)(g'N) = gg'N$ holds because that's the definition of the operation in $G/N$ (which is correct by the normality of N). The rest is ok.

Also note that $\tilde\phi$ is uniquely determined by the condition $\tilde\phi = π$.

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user87690's excellent answer critiques the proof you gave, but I'd just like to add one way to think about this as follows. If we know that there's a group homomorphism $\phi:G\to H$, then the kernel of $\phi$ consists precisely of those elements of $G$ that we "forget" in $H$ via $\phi$. The quotient group $G/\text{ker}(\phi)$ essentially forgets these elements as well. So, if $N\subseteq \text{ker}(\phi)$, then specifying a group homomorphism $G/N\to H$ just amounts to forgetting fewer elements which we can do. On the other hand, we can't necessarily do this if $N\not\subseteq \text{ker}(\phi)$ because $\text{ker}(\phi)$ consists of precisely the elements that we're forgetting via $\phi$. We can't forget more; nothing else (outside $\text{ker}(\phi)$) is sent to the identity in $H$. (In real life, it's easy to forget more rather than less but in group theory the situation is the opposite!)

I hope this helps!

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  • $\begingroup$ Dear @ronno, thanks very much for your comment. I posted this answer a while ago so I can't remember with certainty my original intention, but I believe it was to give the intuition behind the result as a supplement to the other answer (which was to critique the OP's proof). I agree that the proof posted by the OP does not properly explain the fact that $\widetilde{\phi}$ is well-defined and I have removed the first sentence of my answer ("Yes, it looks great!"); I think the rest of my answer is OK. Would you be willing to upvote user87690's answer since you agree it's good? (I upvoted it.) $\endgroup$ – Amitesh Datta Jun 3 '14 at 15:25
  • $\begingroup$ Also, @ronno, do you agree my answer is fine now? If so, then would you be willing to remove the downvote? If not, then would you be willing to let me know how I can improve it? $\endgroup$ – Amitesh Datta Jun 4 '14 at 3:58

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