# A nilpotent group has a normal abelian non-central subgroup

I am working on a proof which involves nilpotent groups. In a proof that I have read about it, it says that, if we have a non-abelian nilpotent group $G$, then there is a normal abelian non-central subgroup $A \trianglelefteq G$. However, I do not immediatly see why this is true?

I have tried the following:

Suppose $G$ has the central series $$G = G_0 \trianglerighteq G_1 \trianglerighteq ... \trianglerighteq G_n = 1$$ such that $G_i/G_{i+1} \subset Z(G/G_{i+1})$. Now my guess was that the group $A$ that we are looking for is $G_{n-1}$, because we then get that $G_{n-1} \subset Z(G)$. However I do not see why $A \neq Z(G)$?

Assuming that $G$ is nonabelian, let $g \in Z_2(G) \setminus Z(G)$, and put $A=\langle Z(G),g \rangle$.
• I have used a different, but equivalent definition of nilpotent, therefore something is not really clear to me yet. $Z_2(G) = \{ g \in G \ | \ \forall y \in G [g,y] \in Z(G) \}$ correct? I do believe that $Z_2(G) - Z(G)$ is non empty, however I do not see that immediatly when I use my definition. – Krijn May 6 '15 at 19:46
• If you are seriously studying nilpotent groups, then you shoud familiarize yourself with the upper central series $1 \le Z(G) \le Z_2(G) \le Z_3(G) \cdots$, where $Z_{i+1}(G)$ is defined by $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$. So $Z_2(G)/Z(G) = Z(G/Z(G))$, and the centre of a nontrivial nilpotent group is nontrivial, and hence $Z_2(G) \setminus Z(G)$ is nonempty. – Derek Holt May 6 '15 at 22:13