Define $\tilde{\phi}([g])=\phi(g)$ and show that this is well-defined.

Show that $\tilde{\phi}$ is a homomorphism.

Show that $\tilde{\phi}$ is injective.

Proof.

Define the map $\tilde{\phi}: G/\ker{\phi} \to G’$ by sending $[g]$ to $\phi(g)$. Here $[g]$ is the element of $G/\ker{\phi}$ represented by $g\in G$.

We need to show that this is well-defined.
Namely, we need to show that $\tilde{\phi}$ does not depend on the choice of representative.

So suppose $[g]=[h]$ for $g, h \in G$. Then we have $x:=gh^{-1} \in \ker{\phi}$. Thus we have
\[e’=\phi(x)=\phi(gh^{-1})=\phi(g)\phi(h)^{-1},\]
where $e’\in G’$ is the identity element of $G’$.
Here the third equality follows because $\phi$ is a homomorphism.
Hence we obtain $\phi(g)=\phi(h)$, equivalently $\tilde{\phi}([g])=\tilde{\phi}([h])$. Thus $\tilde{\phi}$ is well-defined.

Now we show that $\tilde{\phi}: G/\ker{\phi} \to G’$ is a homomorphism. Let $[g], [h] \in G/\ker{\phi}$. Then we have
\[\tilde{\phi}([g][h])=\tilde{\phi}([gh])=\phi(gh)=\phi(g)\phi(h)=\tilde{\phi}([g])\tilde{\phi}([h]) \]
and $\tilde{\phi}$ is a homomorphism.

Finally, we prove that $\tilde{\phi}$ is injective. Suppose that $\tilde{\phi}([g])=e’$. Then this means $\phi(g)=e’$, hence $g\in \ker{\phi}$.
Thus $[g]=[e]$, where $e$ is the identity element of $G$.
Hence $\tilde{\phi}$ is injective and the proof is complete.

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