# Why isn't $Z_2 \times S_3$ nilpotent?

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.

• The order of $G$ is $12$, so $G/Z_1(G)$ has order $12/2=6$, not $3$. Commented Nov 28, 2019 at 17:18
• @AnalysisStudent0414 Oh lol thanks
– Ovi
Commented Nov 28, 2019 at 17:18
• @DietrichBurde Thanks for your answer! Haha in some cases I made the mistake of forgetting, but in most cases I think the answer just doesn’t address the question fully. In this case I wanted to accept it, but you can’t accept in the first 15 minutes of a question, so I just walked away from the computer with the intent of returning later and accepting. I almost always upvote though.
– Ovi
Commented Nov 28, 2019 at 19:51
• Ovi, never mind! Thank you for accepting!. I hope everything is clear with the question. Your question has an upvote, too. Commented Nov 28, 2019 at 19:51
• @DietrichBurde Haha thanks! (It is Thanksgiving day in the US after all :) )
– Ovi
Commented Nov 28, 2019 at 20:51

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