7
$\begingroup$

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 $G$?

Thank you for concerning about this question.

$\endgroup$
5
$\begingroup$

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 $G/ker(f_1)$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ – Doyun Nam Oct 11 '19 at 14:56
  • $\begingroup$ 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! $\endgroup$ – Doyun Nam Oct 11 '19 at 15:47
1
$\begingroup$

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 $|A|=|B|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.