About strict inequality in Groups Locally Nilpotent Let $G$ be a locally nilpotent group and $x \in G$. How can I prove that $[G,x] \neq G$?
Note that if $G$ is a nilpotent group then this statement is true, because $G'=[G,G]<G$ and $[G,x]\leq G'$.
However, $G$ can be a locally nilpotent group such that $G=G'$ that this statement is still true.
How can I prove this?

If $G$ is nilpotent and has central series $1 <G_1 < \cdots < G_n=G$ and $x \in G_{i+1} \setminus G_i$ for some $i$, then $[G,x] \le G_i$, and so $x \not\in [G,x]$.
Now suppose that $G$ is locally nilpotent and $x \in [G,x]$. Then $x$ is a finite product of elements of the form $[g_i,x]^{\pm 1}$. But the finitely many $g_i$ in this product together with $x$ generate a nilpotent subgroup $H$ of $G$, and $x \in [H,x]$, contradicting what we proved above.
So $x \not\in [G,x]$ and $[G,x] \ne G$.
