# $U,H\leq G$ Subgroups. H of finite Index. Does $|U:H\cap U|\mid |G:H|$ hold?

Let $G$ be a group and $U,H\leq G$ subgroups, such that $H$ is of finite Index in $G$ (not necessary U, too). May $n=|G:H|$. One can easily show with an Injection between the two appropriate cosets that $|U:H\cap U|\leq n$, but does $|U:H\cap U|\mid n$ hold? So does the Index of the Intersection $H\cap U$ in $U$ divides the Index of $H$ in $G$?

I don't think that the claim holds, so I'm searching for an example to disprove it. Is there a difference between the two cases, where $G$ is infinite or finite? Thanks for help.

• This holds if $K$ and $H$ are normal, by one of the isomorphism theorems (the $HK/H\cong K/(H\cap K)$ one). You could try mimicking the proof using cosets. It should be a good exercise, if nothing else... – user1729 Sep 10 '13 at 19:04
• What you mean, holds if $H$ "or" $U$ is normal in $G$, beacause then $\langle U,H \rangle=UH=HU$. Then the Indices coincides. That's not what I was asking. – Tomas Sep 10 '13 at 19:38
• Try two subgroups of order $2$ in $S_3$. – Derek Holt Sep 10 '13 at 19:43
• @Tomas I am confused by what you mean by "indices". What indices? – user1729 Sep 10 '13 at 19:46
• It's the plural form of index. – Tomas Sep 10 '13 at 19:47

This is not true in general. Check the nonabelian group of order $6$ for a counterexample.
For a more general example, let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Suppose that $G$ does not have normal Sylow $p$-subgroup. If $H$ and $U$ are distinct Sylow $p$-subgroups of $G$, then $p$ divides $[U : U \cap H]$ but $p$ does not divide $[G:H]$.