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Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.

Can somebody help me out?

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Step 1: Show that $H \cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H \cap K][H\cap K : H]$. –  Clive Newstead Apr 5 '12 at 22:41
    
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[G\colon(H\cap K)]\leq [G\colon H]\cdot[G\colon K].$ You can try to find an injection from a certain set to another. –  user23211 Apr 5 '12 at 23:02
    
A related question. –  user23211 Apr 5 '12 at 23:06
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@CliveNewstead The formula $[G:H] = [G:H \cap K][H\cap K : H]$ seems incorrect to me. What is $[H\cap K : H]?$ Did you mean $[G:H\cap K]=[G:H][H:H\cap K]?$ –  user23211 Apr 5 '12 at 23:23
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@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment. –  Clive Newstead Apr 7 '12 at 11:45
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3 Answers

Proof $1$:

$$[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K].$$

We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. It is however a disjoint union of left cosets of $K$, so the index makes sense.

Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/H\times G/K$. The latter is finite so the orbit of $H\times K$ is finite, and the stabilizer of it is simply $H\cap K$. Invoke orbit-stabilizer.

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There's probably a thread in which your second equality is proved, too. I'll go digging. –  Dylan Moreland Jul 14 '12 at 23:10
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Found one: math.stackexchange.com/questions/168942/… –  Dylan Moreland Jul 14 '12 at 23:15
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Let $\{H_1,\dots,H_m\}$ be the left cosets of $H$, and let $\{K_1,\dots,K_n\}$ be the left cosets of $K$. For each $x\in G$ there are unique $h(x)\in\{1,\dots,m\}$ and $k(x)\in\{1,\dots,n\}$ such that $x\in H_{h(x)}$ and $x\in K_{k(x)}$. Let $p(x)=\langle h(x),k(x)\rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:

Proposition: If $x$ and $y$ are in different left cosets of $H\cap K$, then $p(x)\ne p(y)$.

It follows immediately that $H\cap K$ can have at most $mn$ left cosets.

It may be easier to consider the contrapositive of the proposition:

If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $H\cap K$.

You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}y\in H$.

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Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 \subset H$ and $N_2 \subset K$. Then $G/N_1 \times G/N_2$ is a finite group; do you see why this implies that $N_1 \cap N_2$ has finite index in $G$?

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I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal. –  anon Jul 14 '12 at 22:59
    
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind. –  Dylan Moreland Jul 14 '12 at 23:03
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