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)<A

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

 

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Filed under 3-manifold, algebra, group theory, word algebra

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