Heisenberg group is nilpotent

Question

$F$ is a field and $H(F)$ is the Heisenberg group over $F$. Is it nilpotent? Is it solvable?

I did all the math and I found that the commutator subgroup is in the center $Z(H(F))$, so $H(F)/Z(H(F))$ is abelian. The only thing I have to do is to show that $Z(H(F)/Z(H(F)))=Z_{2}(H(F))=H(F)$ so $H(F)$ nilpotent. Is it obvious that $Z(H(F)/Z(H(F)))=Z_{2}(H(F))=H(F)$? What should I say?

Answer 1

Since you have already proved that the commutator subgroup is in the center, you can conclude that the group is nilpotent of class 2 since it is not abelian.