surjective homomorphisms between cyclic groups (wrong question)

Question

This statement does not hold.

Let $C, D$ be cyclic groups and $f_1,f_2:C\rightarrow D$ be surjective homomorphisms. Show that there exists $\xi: C\rightarrow C$ such that $f_2=f_1\circ \xi$.

My method:

Let $C=<a>$, $D=<b>$.

case 1: $ord(b)=\infty$.

Then $ord(a)=\infty$. Hence $f_i(a)= \pm b$ for $i=1,2$. Hence the automorphism $\xi $ is determined.

case 2: $ord(b)=n$.

Then if $f_1(a)=k_1 b$, $f_2(a)=k_2 b$, then $k_1b$ and $k_2b$ are both generators of $D$. Hence $(k_1,n)=(k_2,n)=1$. Hence there exists $x\in \{1,\cdots,n-1\}$ such that $x k_1=k_2$ (mod $n$). Hence $f_1(xa)=xf_1(a)=xk_1b=k_2b=f_2(a)$. Let homomorphism $\xi: C\rightarrow C$, $\xi(a)=xa$. Then $f_1\circ \xi=f_2$.

But I do not know how to prove $\xi$ is automorphism...

Answer 1

For $C$ and $D$ finite we can indeed require $\xi$ to be an automorphism. Let me try to describe what I think is the core of the problem:

Let $C = \Bbb Z /n \Bbb Z$ and $D = \Bbb Z /m \Bbb Z$ and let $\phi:C \rightarrow D$ be a homomorphism onto D. Then $m \mid n$ and thus we can split $\phi$ into the natural projection $\pi:C \rightarrow D$ and an automorphism $\varphi: D \rightarrow D$ such that $\phi = \varphi \circ \pi$.

Now suppose $\phi_1,\phi_2:C \rightarrow D$ are different homomorphisms. We want to show that there is an automorphism $\xi:C \rightarrow C$ such that $\phi_2 = \phi_1 \circ \xi$ are conjugated. If we consider the right-action of $Aut(C)$ on $Hom(C,D)$, then this statement is equivalent to the transitivity of the action. We may show transitivity by showing that every element of $Hom(C,D)$ is conjugated to $\pi \in Hom(C,D)$.

By splitting homomorphism in $Hom(C,D)$ into projection and automorphism $\varphi$ of $D$, the latter statement is equivalent to the following Lemma:

For any $\varphi \in Aut(\Bbb Z /m \Bbb Z)$ there is a lift $\xi \in Aut(\Bbb Z /n \Bbb Z)$ such that the following diagram commutes:

$$\require{AMScd}\begin{CD} \Bbb Z /n \Bbb Z @>\displaystyle \exists\xi>> \Bbb Z /n \Bbb Z\\ @VV\displaystyle\pi V @VV \displaystyle\pi V\\ \Bbb Z /m \Bbb Z @>>\displaystyle\varphi> \Bbb Z /m \Bbb Z \end{CD}$$

Proof of the Lemma: $\pi$ induces a surjective homomorphism $$\pi:\left(\Bbb Z / n \Bbb Z \right)^\times \rightarrow \left(\Bbb Z / m \Bbb Z \right)^\times$$ As $\varphi$ is onto, $\varphi(1)$ is a generator of $\Bbb Z /m \Bbb Z$. Hence, we can choose $g \in \left(\Bbb Z /n \Bbb Z\right)^\times$ with $\pi(g) = \varphi(1)$. Define $\xi$ by $1 \mapsto g$. Then $\xi $ is an automorphism of $\Bbb Z /\Bbb nZ$ because $g$ is a generator, and $\varphi \circ \pi(1) = \pi \circ \xi(1)$ coincide on a generator, hence the diagram commutes.

Answer 2

I hope the case involving infinite groups is clear to you.

For $C$, $D$ cyclic, let $D=Z/m$. Then the surjectivity of $f_1$, $f_2$ implies that $C=Z/um$ for some integer $u$. Suppose that $f_{1}(1)=a$, $f_{2}(1)=b$, then by surjectivity $a,b$ are relatively prime to $m$. Let $k=a^{-1}b\mod m$ be an integer($b^{-1}$ just means the residue class which is the inverse of the class containing $b$), then $1 \rightarrow k$ defines an automorphism ξ on $C$ satisfying the requirement.