# Subnormal series and indices

I'm trying to solve this problem:

Let be $H, K$ subgroups of a finite group $G$. Suppose that exists one serie of subgroups such that

$G=G_{0}\triangleright G_{1} \triangleright \ldots \triangleright G_{r}=H$. $(*)$

Show that $(H:H\cap K)\Big|(G:K)$ and $|H.K|\Big | |G|$.

I tried do this way:

Define $G_{iK}=(G_{i}\cap K).G_{i+1}$ for each $i\in \{0,\ldots, r-1\}$. This way we have the refinement of the serie $(*)$

$G=G_{0}\triangleright \underbrace{(G_{0}\cap K).G_{1}}_{G_{0K}} \triangleright G_{1}\triangleright \underbrace{(G_{1}\cap K).G_{2}}_{G_{1K}} \triangleright G_{2} \triangleright \ldots \triangleright G_{r-1}\triangleright \underbrace{(G_{r-1}\cap K).G_{r}}_{G_{r-1 K}} \triangleright G_{r}=H$

then,

$|G|=|G_{0}|=\displaystyle\Big|\frac{G_{0}}{G_{0K}}\Big|.\Big|\frac{G_{0K}}{G_{1}}\Big|.\Big|\frac{G_{1}}{G_{1K}}\Big|. \ldots . \Big|\frac{G_{r-1K}}{G_{r}}\Big|=\frac{|G_{0K}|}{|G_{r}|}$

Since $K\leq G=G_{0}$, we have that $G_{0K}=(G_{0}\cap K).G_{1}=KG_{1}$. Furthermore, as $G_{r}=H$ follow that

$|G|=\displaystyle\frac{|KG_{1}|}{|H|}=\displaystyle\frac{\displaystyle\frac{|K|.|G_{1}|}{|K\cap G_{1}|}}{|H|}=\frac{|K|.|G_{1}|}{|H\cap G_{1}||H|}=\frac{|H|.|K|.|G_{1}|.|H\cap K|}{|H\cap G_{1}|.|H\cap K|.|H|.|H|}$

That is

$|G|=\displaystyle\frac{|H|.|K|}{|H\cap K|}.\frac{|G_{1}|.|H\cap K|}{|H\cap G_{1}|.|H|.|H|} (**)$

Note that as $K\leq G$ then $\displaystyle\frac{|G|}{|K|}$ is an integer and, futhermore, we can write

$\displaystyle\frac{|G|}{|K|}=\displaystyle\frac{|H|}{|H\cap K|}.\frac{|G_{1}|.|H\cap K|}{|H\cap G_{1}|.|H|.|H|}$

Equivalently,

$(G:K)=(H:H\cap K).q$, where $q=\displaystyle\frac{|G_{1}|.|H\cap K|}{|H\cap G_{1}|.|H|.|H|}$.

Then,

$(H:H\cap K)\Big|(G:K)$

The expression $(**)$ we obtain immediately that

$|H.K|\Big | |G|$.

But this argumentation have at least one problem: Is necessary to show that:

$G_{i+1} \triangleleft (G_{i}\cap K).G_{i+1} \triangleleft G_{i}$ and I no have ideas how do this. Some sugestion? And this raciocinio is correct?

Let $N=G_1$. By induction on $r$ we can assume that $|H(K \cap N)|$ divides $|N|$.
Now $|HK| = |H||K|/|H \cap K|$ and |$H(K \cap N)| = |H||K \cap N|/|H \cap K \cap N| = |H||K \cap N|/|H \cap K|$, so
$|HK|/|H(K \cap N)| = |K|/|K \cap N| = |KN/N|$, which divides $|G/N|$, and we are done.