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Is the map $f: G \rightarrow Z(G)$ where $f$ maps elements not in the center to the identity and is the identity map restricted to the center.

I think it is by my computation, but I just want to make sure. Could anyone confirm it for me?

Thanks!

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Have you thought that the product of two non-central elements can be central? – Joanpemo Mar 23 at 14:45

Have you thought that the product of two non-central elements can be central? For example, take the dihedral group of order $\;8\;$ :

$$D_{2\cdot4}=\left\{\,1,s,t,t^2,t^3,st, st^2,st^3\,\right\}\implies Z(D_8)=\{\,1,t^2\,\}\;$$

So $\;t\notin Z(G_8)\;$, so by your definition

$$f(t)=1\;,\;\;\text{but}\;\;\; t^2=f(t^2)=f(t)^2=1^2=1$$

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If this conjecture is true, then the set containing the identity and everything not in the center and nothing else, would be the kernel of the homomorphism. And so that set would be a subgroup. As a subgroup, it would have a number of members that divides the order of the whole group.

What would all that imply?

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take $g\in G-Z(G)$ and $c\in Z(G)-\{e\}$. Then $gc^{-1}$ is not in the center either (if it were, $g$ would be). Then $f(g)=e$ by definition but $e=f(g)=f(gc^{-1}c)=f(gc^{-1})f(c)=ec=c$, a contradiction.

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Consider the group $G$ of symmetries of a square. Let us pay particular attention to two reflections $F_V$ and $F_H$ (vertical and horizontal) whose mirror planes meet at a right angle:

enter image description here

Their composition is a $180^\circ$ rotation. Your map $f$ would send $F_V$ and $F_H$ to the identity of $G$, but $f(F_V F_H)$ would be the $180^\circ$ rotation.

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By your definition of $f$ $$|\ker f|=|Z(G)|+1$$

So every group $G\quad \text{s.t}\quad |Z(G)|+1\nmid |G|$ is a counterexample .

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