If $G$ is nilpotent of class $n$, then $G/Z(G)$ is nilpotent of class $n−1$.

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I saw this here: If $G$ is nilpotent of class $c$, then $G/Z(G)$ is nilpotent of class $c-1$.

However, the definition of Nilpotent I have to work with is equivalent but different.

A group is Nilpotent if it has a normal series

$1=G_n \unlhd G_{n-1} \unlhd \dots G_1 \unlhd G_0= G$ such that $G_{i-1} / G_i \leq Z(G/G_i)$

If $G$ is Nilpotent and $n$ is the minimum such that the condition of nilpotency is true, we say that $G$ is nilpotent of class $n$.

In particular, $G_{n-1}/G_n \leq Z(G/G_n)$ simple means that $G_{n-1} \le Z(G)$

So if taking the quotient we have $G_{n-1} / Z(G) \leq Z(G)/ Z(G) =\{ Z(G)\}$.

Therefore, $\{ Z(G)\}=G_{n-1}/ Z(G) \unlhd \dots \unlhd G_1 / Z(G) \unlhd G / Z(G)$ works, so $ G/Z(G) $ is nilpotent of class $n−1$.

Is this correct for this definition?