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If $G_1,G_2$ are groups, $\varphi:G_1 \rightarrow G_2$ is a homomorphism and $a \in Z(G_1)$, is it true that $\varphi(a) \in Z(\operatorname{im}\varphi)$?

My attempt so far: I define $\varphi:G_1 \rightarrow G_2, \varphi(a)= \frac1a, a\in\mathbb Z$. which is not an integer. Also, $\varphi$ is a homomorphism since $a,b\in \mathbb Z :\varphi(ab)=\frac1a\frac{1}{b}$.

Is this a valid counterexample?

Shaun
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Jut3
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    What are $G_1$ and $G_2$ in your example? For abelian groups the claim is obviously true. – Desperado Mar 22 '21 at 17:59
  • We do not know whether or not they are abelian groups. – Jut3 Mar 22 '21 at 18:01
  • Your use of $\frac{1}{a}$ is strange. Do you mean $a^{-1}$? Because, in $\Bbb Z$, the operation is addition, so the inverse of $a$ in $\Bbb Z$ is $-a$. – Shaun Mar 22 '21 at 18:06
  • What does $\varphi(a)$ even mean, if $a\in\mathbb{Z}$? $G_1$ is a group, so on what grounds do you even think that you can plug in an integer into $\varphi$? – Arturo Magidin Mar 22 '21 at 18:07
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    You understand that the notation $Z(G)$ means "the center of $G$", i.e., $Z(G)={g\in G\mid gx=xg\text{ for all }x\in G}$ ? It has nothing to do with integers. – Arturo Magidin Mar 22 '21 at 18:08

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