The commutators of a generating set with a set generating one lower central factor generate the next factor

Let $G$ be a group with lower central series $G = \Gamma_0 \supseteq \Gamma_1 \supseteq \cdots$, where $\Gamma_{k+1}$ is the commutator subgroup of $\Gamma_k$ with $G$. Let $X$ generate $G$, and let $Y$ be a subset of $\Gamma_k$ whose image generates $\Gamma_k/\Gamma_{k+1}$ -- precisely, $\Gamma_k$ is contained in the subgroup generated by $Y$ together with $\Gamma_{k+1}$. Then $\Gamma_{k+1}$ is contained in the subgroup generated by the commutators $xyx^{-1}y^{-1}$, for $x \in X$ and $y \in Y$, together with $\Gamma_{k+2}$; that is, those commutators generate $\Gamma_{k+1}/\Gamma_{k+2}$.

The mechanism is that $\Gamma_{k+1}/\Gamma_{k+2}$ is central in $G/\Gamma_{k+2}$, so the commutator is bilinear modulo $\Gamma_{k+2}$ and it suffices to check the generators. The commutator is written out as a product rather than with bracket notation, and is $xyx^{-1}y^{-1}$ in that order.

Three points of precision. The indexing starts at zero, $\Gamma_0 = G$ and $\Gamma_1 = [G,G]$; authors who index from one call this subgroup $\gamma_{k+2}$. The family of commutators is asymmetric -- only $xyx^{-1}y^{-1}$ with $x \in X$ and $y \in Y$, not the other order and not commutators within $X$ or within $Y$. And the conclusion is stated as one containment, that $\Gamma_{k+1}$ lies in the join of the generated subgroup with $\Gamma_{k+2}$; the reverse holds automatically, since each such commutator already lies in $\Gamma_{k+1}$, so the two sides are in fact equal and the commutators really do generate the factor.

Nothing here is vacuous -- taking $X = G$ and $Y = \Gamma_k$ satisfies the hypotheses for every $G$ and $k$ -- but two degenerate cases are permitted. At $k = 0$ the hypothesis on $Y$ is automatic, and $Y$ may be empty, in which case the hypotheses force $\Gamma_k = \Gamma_{k+1}$ and the conclusion reduces to the true statement $\Gamma_{k+1} = \Gamma_{k+2}$.