Is $f(Z(G)))\subseteq Z(H)$ always true for a group homomorphism $f:G\to H$?

Question

The statement I am trying to prove or disprove is: does $$f(Z(G)))\subseteq Z(H)$$ always hold for a group homomorphism $f:G\to H$?

if $f$ is surjective, then this statement is easily seen to be true. However, $f$ is not necessarily surjective in general, and this is where I am stuck. Intuitively, I feel like the statement should not always be true, but I am struggling to construct a counterexample.

Could someone help me prove or disprove this statement? If it's false, could you provide a counterexample? Any insights or guidance would be greatly appreciated!

Answer 1

No. Take the injective homomorphism $f:A_3\hookrightarrow S_3$. Then $Z(A_3)=A_3$, since $A_3$ is commutative, and $Z(S_3)=1$. So, with $G=A_3,H=S_3$ we have $$ f(Z(G))=f(G)=G \not\subseteq 1=Z(H). $$

Answer 2

TL;DR. Take any non-abelian group $G$, choose $g \in G \setminus Z(G)$. Then the inclusion map $\langle g \rangle \hookrightarrow G$ doesn't preserve centers.

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Surprisingly, it seems this precise question hasn't been asked before on the site, but there are many highly related questions:

• Let $f : G → G_1$ be a surjective homomorphism from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$. is about showing $f(Z(G)) \subseteq Z(H)$ when $f : G \to H$ is an epimorphism. Same for The center of a group and a homomorphism
• Question about group isomorphism is about showing $f(Z(G))=Z(H)$ when $f : G \to H$ is an isomorphism
• Let $G$ be a group, and $H \le G$. For what pairs $(G,H)$ is $Z(H)=Z(G)\cap H$ true? asks for conditions when $Z(H) = Z(G) \cap H$ holds for subgroups $H \subseteq G$, which is actually equivalent to the question being asked here. A counterexample is mentioned as well. So technically this is a duplicate, but the example there is much too complicated.
• Why is there no functor $\mathsf{Group}\to\mathsf{AbGroup}$ sending groups to their centers? is in particular about the impossibility to have $f(Z(G)) \subseteq Z(H)$ in a categorical setup (actually, something stronger is proven: there might not be any homomorphism $Z(G) \to Z(H)$ at all), hence not really a duplicate since this question assumes more knowledge.
• Riehl's Category Theory in Context - Exercise 1.3.ix. falls into the same, well, category.

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Now to add some content here. I will explain how to systematically develop a counterexample, or rather, "all counterexamples". I want to emphasize that there is no need to "come up" with some special groups.

Every homomorphism of groups $f : G \to H$ factors as $$G \xrightarrow{p} f(G) \xrightarrow{i} H,$$ where $p$ is a surjective homomorphism and $i$ is an injective homomorphism, in fact the inclusion of a subgroup. We know that the claim is true for $p$, and the homomorphisms for which the claim holds are stable under composition. Thefore, the whole problem reduces to the case of the inclusion of a subgroup:

If $H \subseteq G$ is a subgroup, do we have $Z(H) \subseteq Z(G)$? Well, the center is most easy to describe when the group is abelian. When $G$ is abelian, $H$ is also abelian, and the answer is trivially yes. But what if $H$ is abelian? Then we ask ourselves if $H \subseteq Z(G)$ holds for every abelian subgroup $H$ of $G$. But then this holds for every cyclic subgroup in particular (and the converse also holds, since trivially every abelian group is the sum of its cyclic subgroups). So we get $\langle g \rangle \subseteq Z(G)$, which means $g \in Z(G)$.

This means every non-abelian group $G$ provides a counterexample. Take any $g \in G \setminus Z(G)$, and define $H := \langle g \rangle$. Then $Z(H) = H$ is not contained in $Z(G)$. And thus the inclusion homomorphism $i : H \to G$ does not preserve centers.