Suppose $G$ is a nilpotent group, i.e. it has an upper central series:
\begin{align} Z_{0}(G) \leq Z_{1}(G) \leq Z_{2}(G) \leq ... \end{align} where $Z_{i+1}(G) = \{x \in G : [x,y] \in Z_i(G) \, \forall y \in G \}$
which terminates at $G$, say $Z_{c}(G) = G$ for some $c \in \mathbb{N}$.
I am trying to show each element of $G$ is nilpotent, in the sense that the adjoint action ${\rm ad}_{g}:G \rightarrow G, \, x \mapsto [g,x]$ is a nilpotent function for each $g \in G$. Specifically, I am trying to show for any $g \in G , \, {\rm ad}_{g}^c(x) =x , \, \forall x \in G$.
First, $g \in G = Z_c(G)$ by definition means ${\rm ad}_g(x) =[g,x] \in Z_{c-1}(G) \; \forall x \in G$.
Hence $[[g,x],g] \in Z_{c-2}(G)$. And as $Z_{c-2}(G)$ is a group: $[[g,x],g]^{-1} = [g, [g,x]] = {\rm ad}_g^{2}(x)\in Z_{c-2}(G)$. Continuing in this way, we eventually obtain ${\rm ad}_g^c(x) \in Z_0(G) = {1_G}$.
But this conclusion is not correct, since I need ${\rm ad}_g^c(x) =x.$ The approach feels like it is on the right track, but clearly there is an error somewhere.
The working definition of an adjoint element was wrong. The map ${\rm ad}_{g}^c$ should be trivial, not the identity. So the above proof turns out to be correct.