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Let $\phi:G\rightarrow K$ be a group homomorphism. Suppose $N$ is a normal subgroup of $G$. Show that if $\psi:G/N\rightarrow K$ defined by $\psi(gN)=\phi(g)$ for all $g\in G$ is well-defined, then $N\subset \ker\phi$.

Attempt: If we take $g_1N=g_2N$, then $g_1^{-1}g_2\in N$ and $\phi(g_1)=\phi(g_2)$. This means $\phi(g_1^{-1}g_2) = \phi(g_1)^{-1}\phi(g_2)=e$. So $g_1^{-1}g_2\in\ker\phi$.

But using this approach, I cannot guarantee $g_1^{-1}g_2$ covers all elements in $N$.

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You're very close!

What do we want to show? We want to show that, given any element $n \in N$, we have $\phi(n) = 1$. But we're told that $\psi$ is well-defined, so $$\phi(n) = \psi(nN) = \psi(N) = \phi(1) = 1$$

This corresponds to taking $g_1 = 1$ and $g_2 \in N$ in your answer.

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In my answer $e$ denotes the identity of $G$.

Based on the fact that $\psi$ is well defined we conclude that $g_{1}N=g_{2}N$ implies $\phi\left(g_{1}\right)=\phi\left(g_{2}\right)$ or equivalently $g_{1}^{-1}g_{2}\in\ker\phi$. This is what you found out yourself.

Now observe that $eN=g_{2}N\iff g_{2}\in N$ so that $g_{2}\in N$ leads to $g_{2}=e^{-1}g_{2}\in\ker\phi$.

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