# short exact sequence and center

Let us prove:

Let $1\to A\stackrel{f}\to B\stackrel{g}\to B/A\to 1$ be a short exact sequence, if the center $Z(B/A)=1$  then $Z(B)

Proof:  When $x\in Z(B)$ then $g(x)\in Z(B/A)$, so $g(x)=1$.

Therefore $x\in\ker (g)={\rm im}(f)=A$

$\Box$