Minimum size of generating set of associated Lie ring equals minimum size of generating set of quotient of group by nilpotent residual

Statement
Suppose $$G$$ is a group and $$L(G)$$ is its associated Lie ring. Then, the minimum size of generating set of $$L(G)$$ equals the minimum size of generating set of $$G/\gamma_\infty(G)$$, the quotient group of $$G$$ by its nilpotent residual $$\gamma_\infty(G)$$. Both these numbers are equal to the minimum size of generating set for the abelianization of $$G$$.

In particular, if $$G$$ is a nilpotent group, then $$G/\gamma_\infty(G) = G$$, so the minimum size of generating set for $$L(G)$$ equals the minimum size of generating set for $$G$$.

Similar facts

 * Explicit description of lower central series of associated Lie ring of a group
 * Nilpotency class of associated Lie ring equals nilpotency class of quotient of group by nilpotent residual
 * Order of associated Lie ring equals order of group for nilpotent group
 * Exponent of associated Lie ring divides exponent of group