# Classification of surjective group homomorphisms

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

1. 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 $$h \circ f_1 = f_2$$?

2. 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 $$G$$"?

3. Is there more relationships between the set of surjective homomorphisms (from $$G$$) and the set of normal subgroups of $$G$$?

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 $$G/ker(f_1)$$.
• I understand that $G_1$ and $G_2$ are isomorphic. However, I'm not sure that my adding condition $h \circ f_1 = f_2$ holds in general. Can we explain it or construct it? Oct 11, 2019 at 14:56
• I think $h=\tilde{f}_1\circ \tilde{f}_2^{-1}$ should be $h=\tilde{f}_2\circ \tilde{f}_1^{-1}$. I understand your explanation. Thank you! Oct 11, 2019 at 15:47
For question 2 we must be careful and work with the isomorphism classes of groups which $$G$$ can surject onto, since otherwise we would be dealing with a proper class. If we look at the set of cardinalities less than or equal to $$|G|$$, and take one set of each cardinality, then the number of possible group structures on each set $$X$$ is a subset of the set of functions $$\circ:X\times X\to X$$, and so we can talk about the set $$I$$ of isomorphism classes of groups that $$G$$ can surject onto.
Let $$A$$ be the set of surjective homomorphisms from $$G$$ onto the elements of $$I$$, modulo the equivalence relation that $$f_1\sim f_2$$ if there exists an automorphism $$h$$ of the target such that $$h\circ f_1=f_2$$. We need to apply the equivalence relation since, for example, the only normal subgroups of $$C_3=\{1,a,a^2\}$$ are $$\{1\}$$ and itself, but since swapping $$a$$ and $$a^2$$ is an automorphism, with the trivial maps $$C_3\to C_3$$ and $$C_3\to\{1\}$$ we would then have $$3$$ surjective homomorphisms but only $$2$$ normal subgroups.
If we let $$B$$ be the set of normal subgroups of $$G$$, and define $$\psi:A\to B$$ via $$\psi:f\mapsto\ker(f)$$, then by question 1 this is a bijection, so $$|A|=|B|$$.