Let $G$ be an arbitrary fixed group.
First, we know that if $f: G \rightarrow G'$ is a group homomorphism, then
$\ker f$ is a normal subgruop of $G$.
Second, we also know that if $H$ is a normal subgroup of $G$, then the canonical map from $G$ into $G/H$ is a homomorphism whose kernel is $H$.
Questions
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If $f_1: G \rightarrow G_1$ and $f_2: G \rightarrow G_2$ are two surjective group homomorphisms whose kernel are same, then is there an isomorphism $h:G_1 \rightarrow G_2$ such that $h \circ f_1 = f_2$?
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If the answer for the above question is true, then can we say that "the cardiality of surjective group homomorphism whose domain is $G$ (up to isomorphism) is equal to the cardinality of the set of normal subgroups of $G$"?
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Is there more relationships between the set of surjective homomorphisms (from $G$) and the set of normal subgroups of $G$?
Thank you for concerning about this question.
Best Answer
The key fact you need is that if $f: G \rightarrow H$ is a surjective group homomorphism, then it induces a group isomorphism $\tilde{f}: G/ker(f) \rightarrow H$. This gives you that $G_1$ and $G_2$ are isomorphic and also gives you a yes for question 2. and a relation for question 3.
Edit: For question 1, I claim that $h=\tilde{f}_1\circ \tilde{f}_2^{-1}$. The equation $h\circ f_1=f_2$ is then equivalent to $\tilde{f}_1^{-1}\circ f_1=\tilde{f}_2^{-1}\circ f_2$ which holds because both sides are the canonical map from $G$ to $G/ker(f_1)$.