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 🙂
Best Answer
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 $\tilde\phi = π$.