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

Question

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. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.

Answer 1

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.

Answer 2

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 $gz=f(h_1)f(h_2)=f(h_1h_2)=f(h_2h_1)=f(h_2)f(h_1)=zg\implies z\in Z(G_1)\square$.