Does every group have a 'cyclization'?

Question

Here's the question:

Does every group have a 'cyclization'? That is, let $G$ be a group. Does there necessarily exist a cyclic group $C$ and a surjective homomorphism $\varphi:G\rightarrow C$ such that for every other cyclic group $H$ such that there is a surjective homomorphism $\psi:G\rightarrow H$ then there is a unique homomorphism $\phi:C\rightarrow H$ such that $\psi=\phi\circ \varphi$?

This question is inspired by the existence of the abelianization of a group. Here's what I have done:

Let $G$ be a group and $C$ a cyclic group such that $\varphi:G\rightarrow C$ is a surjective homomorphism. Since $C$ is cyclic, $C=\langle c\rangle$ for some $c\in C$. Let $c^m$ and $c^n$ be two other elements of $C$. Since $\varphi$ is surjective, there are elements $g$ and $h\in G$ such that $\varphi(g)=c^m$ and $\varphi(h)=c^n$ (assume $m$ and $n$ are non-zero). Thus

$$\varphi(h)^{\text{lcm}(m,n)/n}=c^{\text{lcm}(m,n)}=\varphi(g)^{\text{lcm}(m,n)/m}$$

This implies that $$\varphi\left(g^{\text{lcm}(m,n)/m}h^{-\text{lcm}(m,n)/n}\right)=1_C$$

We can conclude then that $$N=\langle\;g^{\text{lcm}(m,n)/m}h^{-\text{lcm}(m,n)/n}\;|\;g,h\in G\text{ and }\;m,n\in\Bbb Z-\{0\}\rangle\subseteq\ker\varphi$$

In addition, since $$xg^{\text{lcm}(m,n)/m}h^{-\text{lcm}(m,n)/n} x^{-1}=(xgx^{-1})^{\text{lcm}(m,n)/m}(xhx^{-1})^{-\text{lcm}(m,n)/n}$$ for any $x\in G$, we get that $N$ is a normal subgroup of $G$. Further, we see that $[G,G]\subseteq N$ since $$xyx^{-1}y^{-1}=(xy)^{\text{lcm}(-1,1)/1}(x^{-1}y^{-1})^{-\text{lcm}(-1,1)/-1}\in N$$

So, I seem to be on the right track. Perhaps $G/N$ is itself cyclic. However, even though I know that $G/N$ is abelian, I have no idea on how to show it is cyclic (if it necessarily is).

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Stepping back from the problem, it would suffice to show that every abelian group has a cyclization since every map onto a cyclic quotient would have to factor through the abelianization of a group. This seems to be obvious. If $G$ is an abelian group which is finitely generated then the Structure Theorem for Finitely-generated Abelian Groups tells us that $G$ is isomorphic to a group that looks like $$C_{d_1}\times C_{d_2}\times\cdots\times C_{d_n}\times FA_m$$

where $d_1\mid d_2\mid \cdots\mid d_n$, each $C_{d_i}$ represents a cyclic group of order $d_i$, and $FA_m$ represents a free abelian group on $m$ generators.

In the absence of a $FA_m$ term, the cyclization would be $C_{d_n}$. In the presence of a $FA_m$ term, the cyclization would be $FA_1$.

For an abelian group that is not finitely generated, I believe the cyclization would be the trivial group. But I'm having difficulty with this claim. Maybe there's a counter-example here.

Answer 1

Cyclizations need not exist in general. Consider the group $G=\oplus_{i=1}^\infty \mathbb Z / p_i \mathbb Z$, where $p_i$ is the $i$-th prime. $G$ surjects onto $\mathbb Z / q \mathbb Z$ for any prime $q$, so $\mathbb Z/ n \mathbb Z$ cannot be a cyclization for any $n$. On the other hand, all elements are of finite order, so it's clear that $\mathbb Z$ is not a cyclization.

Edit. After some more thought, I've realized that we can generalize the above: cyclizations cannot exist for any direct sum of cyclic groups that is not itself cyclic. Roughly, the universal property that you've specified forces the cyclizaton map to be injective, which forces the original group to be cyclic. More precisely, suppose

$$G \simeq \oplus_{i\in \mathbb N} G_i$$ where each $G_i$ is cyclic. If $\varphi: G \rightarrow C$ is a cyclization, then for each $i$ we must have $f_i: C \rightarrow G_i$ so that

$$\require{AMScd} \begin{CD} G @>\varphi>> C \\ @VV\pi_i V @V f_i VV \\ {G_i} @>1>> G_i \end{CD}$$ commutes, where $\pi_i$ is the $i$-th projection.

For any $n$-tuple $x \in G$ with a non-identity component, we can choose $\pi_i$ so that $\pi_i(x) \neq e$. Since $\pi_i=f_i\circ \varphi$, this forces $\varphi(x) \neq e$. Therefore $\varphi$ must be an injection, which implies $G \simeq \varphi(G) \subset C$. Since subgroups of cyclic groups are cyclic, this is impossible unless $G$ itself is cyclic.

Answer 2

The answer is negative. For a counterexample, one can consider $G=C_p\times C_p$. Then the only cyclic image of $G$ is $C_p$. If there would be a "cyclization" $C$ of $G$, then it would be isomorphic to $C_p$. Moreover, there is an exact sequence $$1\to U\to G\stackrel{\varphi}\to C\to 1,$$ where $U$ is some copy of $C_p$ in $G$. But one then can construct a surjection $G\to U$ which obviously does not factor through $\varphi$.

Answer 3

Let $G=\mathbb{Q}$. If $C$ is cyclic, hence isomorphic to $\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$, then $\mathrm{Hom}(\mathbb{Q}, C)=0$, and no surjective homorphism is possible.

Answer 4

Let $G = \mathbf Z_p$ be the additive group of $p$-adic integers. It has arbitrarily large finite cyclic quotients (that's part of what being an inverse limit of arbitrarily large finite cyclic groups is all about), so if $G$ has a cyclization $C$ then $C$ can't be finite, and thus $C \cong \mathbf Z$. Thus there would have to be a surjective group homomorphism $\mathbf Z_p \rightarrow \mathbf Z$. We want to show that is impossible. [Edit: See Alex's comment for a simpler approach than what I write below.]

Homomorphisms of abelian groups are the same thing as $\mathbf Z$-module homomorphisms, and free modules are projective modules, so if there were a surjective group homomorphism $\mathbf Z_p \rightarrow \mathbf Z$ then there would be an isomorphism of abelian groups $\mathbf Z_p \cong \mathbf Z \oplus H$ for some abelian group $H$. Then, since $\mathbf Z$ has every finite cyclic group as a quotient, we can realize every finite cyclic group as a quotient of $\mathbf Z_p$, but the only finite quotient groups of $\mathbf Z_p$ are of $p$-power order, so we have a contradiction.

Here is why all finite quotient groups of $\mathbf Z_p$ have $p$-power order. Suppose there is a surjective group homomorphism $\mathbf Z_p \rightarrow A$ where $A$ is a nontrivial finite group, necessarily abelian. Let $q$ be a prime other than $p$. From the decomposition of $A$ as a direct product of its Sylow subgroups, there's a surjective homomorphism $A \rightarrow A_q$, where $A_q$ is the $q$-Sylow subgroup of $A$. Therefore we'd have a surjective group homomorphism $f \colon \mathbf Z_p \rightarrow A_q$. Set $|A_q| = q^m$. Then $q^m\mathbf Z_p \subset \ker f$, but $q^m\mathbf Z_p = \mathbf Z_p$ since $q$ and $p$ are distinct primes, so $\mathbf Z_p \subset \ker f$. Thus $q^m = 1$, so the $q$-Sylow subgroup of $A$ is trivial for every prime $q$ other than $p$, and therefore $A$ is a group of $p$-power order.