# $[H\vee K:H]\ge [K:H\cap K]$

I'm trying to prove that If $H$ and $K$ are subgroups of a group $G$, then $[H\vee K:H]\ge [K:H\cap K]$

I know that $K\subset HK\subset H\vee K$, this easily implies that $[HK:H]\le [H\vee K:H]$.

Now, I need to prove $[K:H\cap K]\le [HK:H]$ in order to solve the question.

Definition

$H\vee K$ is the subgroup generated by $H\cup K$

I really need help

Thanks

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@amWhy is it ok now? :) – user42912 Apr 27 '13 at 19:04
@amWhy thank you for helping me to improve my question. – user42912 Apr 27 '13 at 19:06
You're welcome: we all appreciate your efforts! – amWhy Apr 27 '13 at 19:07

## 1 Answer

$[K:H\cap K]$ counts the cosets of $H\cap K$ in $K$. (I'll use left cosets but the same proof works with right cosets.) So what I'd like to do is to map these cosets, $a\cdot(H\cap K)$ (where $a\in K$) in a one-to-one manner to cosets of $H$ in $H\lor K$. The first (indeed the only reasonably natural) map that comes to mind is to send $a\cdot(H\cap K)$ to $a\cdot H$. Before doing anything else, I'd better check that this is really a function, i.e., that if I can express the same coset of $H\cap K$ in two different ways, say $a\cdot(H\cap K)=b\cdot(H\cap K)$, then both expressions produce the same result, $a\cdot H=b\cdot H$. Fortunately, that's obvious, because the first of these equations means $a^{-1}b\in H\cap K$, the second means $a^{-1}b\in H$, and the implication is clear since $H\cap K\subseteq H$. Now that I have a map, what's left is to check that it's one-to-one, i.e., that if $a\cdot H=b\cdot H$ (with $a,b\in K$ as above), then $a\cdot(H\cap K)=b\cdot(H\cap K)$. So that means I need to prove the converse of what I did earlier; I need to show that $a^{-1}b\in H$ implies $a^{-1}b\in H\cap K$. Fortunately, that's also easy, because $K$ contains $a$ and $b$, and therefore also $a^{-1}b$.

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