# Show that $Z\left(G/Z(G)\right)=\{1\}$

This problem was given and I didn't give and point any ideas about it in the class in time. It says: if the group $G$ is perfect, then $Z\left(G/Z(G)\right)=\{1\}$, where $Z(G)$ is the centre of $G$. Can someone help me to tackle it well. Thank you!

Edit. A group $G$ is called perfect if $G=[G,G]$.

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Start with the definition of the centre (or center) of a group, i.e. those elements which commute with every other element of the group. In particular, why does $G/Z(G)$ make sense?? –  hardmath Nov 28 '12 at 10:38
This result is known as Grun's Lemma. Try to google it. –  user26857 Nov 28 '12 at 10:42
Note first that $gZ(G) \in Z(G/Z(G))$ if and only if $[g,x] \in Z(G)$ for all $x \in G$. Using this fact you can show that when $gZ(G) \in Z(G/Z(G))$, the map $x \mapsto [g,x]$ is a homomorphism $G \rightarrow Z(G)$. Since $Z(G)$ is Abelian and $G$ is perfect, the kernel of this map is all of $G$. Thus $g \in Z(G)$, which proves the claim.