# Using Zassenhaus to obtain an isomorphism of bigraded groups

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?