central extension implies $\frac{G}{Z(G)}\cong \frac{K}{\pi(Z(G))}.$

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Let $K$ be a group then a short exact sequence
$$1 \rightarrow N\rightarrow G \stackrel{\pi}{\rightarrow} {K}\rightarrow 1$$ is said to be central extension of $K$ if $\operatorname{ker}(\pi)=N\subseteq Z(G)$. Now I want to prove that if above represents a central extension then $\frac{G}{Z(G)}\cong \frac{K}{\pi(Z(G))}.$

Please give some hints that how can we proceed.