The relationship between $G_i H / G_{i+1} H$ and $G_i /G_{i+1}$ Suppose that $G$ is a group, $G_{i+1} \triangleleft G_i \triangleleft G$, $H \triangleleft G$. I need to investigate the relationship between the groups $G_i H / G_{i+1} H$ and $G_i /G_{i+1}$, but so far I can't see how to do it. Any hints?
 A: I want to preface this answer with the fact that I'm including an extra assumption, as I'm not sure how to answer the problem without it. The assumption is that $G_i \cap H = 1$, and I'll explain why I assume that when I reach that section of the proof.
First, we use the Zassenhaus or Butterfly Lemma (page 20 of Lang's Algebra provides a proof), which states that, given $A,C \leq G$, $B \triangleleft A$, and $D \triangleleft C$, then $$ \frac{(A \cap C)B}{(A \cap D)B} \cong \frac{(A \cap C)D}{(B \cap C)D}$$
Choosing $A=G_i$, $B=G_{i+1}$, $C=G$, and $D=H$, we have $$ \frac{(G_i \cap G)G_{i+1}}{(G_i \cap H)G_{i+1}} \cong \frac{(G_i \cap G)H}{(G_{i+1} \cap G)H}$$
Since $G_i, G_{i+1} \leq G$ then $G_i \cap G = G_i$ and $G_{i+1} \cap G = G_{i+1}$. We also have $G_i G_{i+1} = G_i$ since $G_{i+1} \leq G_i$. Thus, after all of these simplifications, we have $$ \frac{G_i}{(G_i \cap H)G_{i+1}} \cong \frac{G_i H}{G_{i+1} H}$$
Now, this holds regardless of any conditions on $G_i \cap H$. With the extra assumption that $G_i \cap H = 1$, we have the far prettier $$ \frac{G_i}{G_{i+1}} \cong \frac{G_i H}{G_{i+1} H}$$
This, of course, fails dramatically in the case that $G_i \cap H \ne 1$. However, based on the fact that you are specifically looking for a relationship between $G_i / G_{i+1}$ and $G_i H / G_{i+1} H$ and the similarities between these two congruences, I'd hazard a guess that the problem was supposed to stipulate that their intersection is the identity. Feel free to let me know what you think Alexei.
