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I am working on a school assignment and have been stuck on this question for some time.

Let $f : G \rightarrow G_1$ be a surjective homomorphism (also called epimorphism) from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$

I know we start by assuming we have an element in $f(Z(G))$ then showing it is also in $Z(G_1)$. Also, $Z(G)$ is the center of $G$, so $Z(G)$ is all the elements of $G$ that commute with all elements in $G$. But how do we know that if we map the elements of $Z(G)$ to $G_1$, then their image will also commute with all the elements in $G_1$? Does it have something to do with using commutativity with the homomorphism? i.e. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.

Martin Brandenburg
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Iden Craven
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  • Urgh... don’t use “epimorphism” synonimously with “surjective homomorphism.” They don’t mean the same (though in the context of the category of all groups, the two concepts agree, in other contexts, such as monoids, or even rings with unity, they do not agree). – Arturo Magidin Feb 08 '19 at 05:30

2 Answers2

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Really all you need to do is follow the definitions and the result will fall into your lap.

Hint. Let $h\in G_1$ and $y\in f(Z(G))$. We can write $h=f(g)$ because $f$ is surjective, and $y=f(x)$ where $x$ is...

Then we have $$hy=f(g)f(x)=\cdots=yh\ .$$ I'll leave you to fill in the dots and supply reasons.

David
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This is really straight forward. If $g\in G_1$, then $\exists h_1\in G$ such that $f(h_1)=g$ (since $f$ is an epimorphism). Let $z\in f(Z(G))$. Then $z=f(h_2)$ with $h_2\in Z(G)$. Then look at $gz=f(h_1)f(h_2)=f(h_1h_2)=f(h_2h_1)=f(h_2)f(h_1)=zg\implies z\in Z(G_1)\square$.