8
$\begingroup$

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.

$\endgroup$

2 Answers 2

6
$\begingroup$

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? $\endgroup$ Commented Oct 11, 2019 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! $\endgroup$ Commented Oct 11, 2019 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 $|A|=|B|$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .