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I know that if we got two finite index subgroups $U,H$ of a group $G$. Then the index of the intersection $H\cap U$ of both is smaller than the product of the indices of $H,U$. We proof this with an injective map from the cosets of $H\cap U$ in H to the cosets of $U$ in $G$. But is there a chance to show that this map is surjective in some cases? Especially in the case:

$U\neq H< G$ infinite groups, $p$ prime number. $|G:U|=p=|G:H|$. Then $|G:H\cap U|= p^2$?

Or when the indices are coprime?

Or can we show that the index of the intersection $H\cap U$ in $G$ divides the product of the indices of $H$ and $U$ in $G$.

If we can't do this for infinite groups. Do anyone have an example where $|G:H\cap U|$ does not divide $|G:U|\cdot|G:H| $?

Thanks for help.

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  • $\begingroup$ Since $H$ and $U$ are assumed to have finite index, $H\cap U$ has finite index, and so contains a normal subgroup of $G$ of finite index. Moding out by this subgroup you can reduce to the case of $G$ finite. $\endgroup$ May 28, 2012 at 13:04

2 Answers 2

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$S_4$ has two (in fact, three) subgroups of index 3, but their intersection is not of index 9. Yes, I know, $S_4$ is a finite group, but just do $S_4\times G$ for some infinite group, $G$.

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  • $\begingroup$ Okay. But the index of the interesection divides the product of the indices of the subgroups, so it divides 9. right? $\endgroup$
    – Peter
    May 28, 2012 at 13:20
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    $\begingroup$ Does it? The only numbers dividing 9 are 1, 3, and 9. The index of the intersection can't be 9 cuz 9 doesn't divide the order of the group $S_4$. And it can't be 3 or less since the intersection is smaller than either subgroup. But you know it's not that hard to find those two subgroups of $S_4$ and see for yourself what the intersection is and what its index is. Try it! $\endgroup$ May 28, 2012 at 13:24
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Finiteness or infiniteness is irrelevant; a finite example yields an infinite one by taking a direct product with an infinite group. And every infinite example can be turned into a finite example by moding out by a normal subgroup of $G$ of finite index contained in $H\cap U$.

In general, if $G$ is a group, $H$ and $K$ are subgroups, then $$[H\colon H\cap K] \leq [G:K]$$ in the sense of cardinalities; but you may or may not get equality. In particular, if $[G:H]=[G:K]=p$, then $[G:H\cap K]\leq p^2$, and is a multiple of $p$, but you may or may not have equality to $p^2$.

Proposition. Let $G$ be a group, $H$ and $K$ subgroups. Then $$[H:H\cap K] \leq [G:K]$$ in the sense of cardinalities, with equality if and only if $HK=G$.

Proof. We define a map from the left cosets of $H\cap K$ in $H$ to the left cosets of $K$ in $G$ by mapping $h(H\cap K)$ to $hK$. I claim the map is well-defined and one-to-one: $$\begin{align*} h(H\cap K) = h'(H\cap K) &\iff h'^{-1}h\in H\cap K\\ &\iff h'^{-1}h\in K\quad\text{(it is always in }H\text{)}\\ &\iff hK = h'K. \end{align*}$$ The map is onto if and only if for every $g\in G$ there exists $h\in H$ such that $gK=hK$, if and only if for every $g\in G$ there exists $h\in H$, $k\in K$ such that $g=hk$, if and only if $G\subseteq HK$, if and only if $G=HK$. $\Box$

In particular, for you situation, if $HU$ is a subgroup of $G$ (it need not be!) then $HU=G$, so you will have $[G:H\cap U] = [G:H][H:H\cap U] = [G:H][G:U] = p^2$. But if $HU$ is not a subgroup, then the best you have is that $p$ divides $[G:H\cap U]$, and $p\lt [G:H\cap U] \leq p^2$; so it could be any of $2p$, $3p,\ldots,(p-1)p,p^2$.

In the example Gerry Myerson gives, for instance, the intersection has index $6=2p$ ($p=3$).

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  • $\begingroup$ Thank you very much. For my field of interest, I wanted to have that the index is a power of $p$ not a multiple of $p$. But I thoughted that it won't work, too. But didn't know how to proof it. $\endgroup$
    – Peter
    May 28, 2012 at 14:28

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