I have just learned the definition of a nilpotent group. My book seems to claim that $Z_2 \times S_3$ is not nilpotent, because they say, for the upper central series, $Z(G) = Z_1(G) = Z_2(G) = Z_n(G)$ "has order $2$ for all $n$. But here is my argument that $G$ is nilpotent:
Let $G = Z_2 \times S_3$, and let $Z_2 = \langle x \rangle$. We construct the upper central series. $Z_0(G) = 1 $, and $Z_1(G) = Z(G) = \{(1, 1), (x, 1) \}$. Therefore $G / Z_1(G)$ has order $3$, so it is abelian, and therefore $Z(G/Z_1(G)) = G/Z_1(G)$, so $Z_2(G) = G ,$ and $G$ is nilpotent.
Could you please let me know if my argument is correct, and $G$ is indeed nilpotent? Thank you very much.
Every subgroup $H$ of a nilpotent group $G$ is again nilpotent. So suppose that $G=\Bbb Z_2\times S_3$ is nilpotent. Then the subgroup $H=S_3$ is nilpotent. Since nilpotent groups have non-trivial center and $Z(S_3)=1$ this is a contradiction. Hence $G$ is not nilpotent.
Reference: $S_3$ is soluable but not nilpotent