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 $(*)$

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


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

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

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


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


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

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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.