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Let $G$ be a group. Suppose that $F^{\bullet}G$ is a filtration on $G$. If $q: G \to K$ is a quotient map, then we get an induced filtration of $K$ given by $F^aK:= q(F^aG) = (F^aG)/(F^aG \cap \ker q)$.

Suppose that $F^{\bullet}G$ and $T^{\bullet}G$ are two filtrations on $G$. Define $$\mathsf{gr}_F^i(G) = {F^iG}/{F^{i+1}G}.$$ As mentioned, there is an induced filtration $T^{\bullet}\mathsf{gr}_F^i(G)$. Similarly, there is an induced filtration $F^{\bullet}\mathsf{gr}_T^j(G)$. Then we get graded pieces $\mathsf{gr}_T^j\mathsf{gr}_F^iG$ and $\mathsf{gr}_F^i \mathsf{gr}_T^j G$. This produces two bigraded groups $\mathsf{gr}_F\mathsf{gr}_TG$ and $\mathsf{gr}_T\mathsf{gr}_FG$.

Recall the Zassenhaus lemma.

Suppose that $A \unlhd \tilde{A} \leq G \geq \tilde{B} \unrhd B$. Then we have a group isomorphism $${A\cdot (\tilde{A} \cap \tilde{B})}/{A \cdot (\tilde{A} \cap B)} \cong {B \cdot (\tilde{A} \cap \tilde{B})}/{B\cdot (A \cap \tilde{B})}.$$

I have read that a corollary of this is that $$\mathsf{gr}_T^j\mathsf{gr}_F^iG \cong \mathsf{gr}_F^i \mathsf{gr}_T^j G.$$

In fact, sometimes this seems to be called the Zassenhaus lemma. But I can't find a proof of this statement online.

Exactly how does one derive this corollary?

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