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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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    Proof of Eklavya in https://math.stackexchange.com/questions/1605121/prove-that-h-cap-k-have-finite-index-in-g: Awra Dip uses $g(H\cap K) = gH \cap gK$, but proof of Eklavya uses only $g(H\cap K) \subseteq gH \cap gK$. Since there are only finitely many choices for $gH$ and for $gK$, there are only finitely many choices for $g(H \cap K)$.2018-10-14

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Proof $1$: $\quad[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K].$

We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:H\cap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $H\cap K$.

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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    @AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $h\in H$.2018-09-24
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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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    @Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.2017-09-05
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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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    @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.2012-07-14
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$H, K$ be subgroups of $G$. Any $A \in \frac{G}{H \cap K}$ can be written as $A = B \cap C$, where $B \in G/H$ and $C \in G/K$ as follows. For any $g \in G$ $g(H\cap K) = gH \cap gK$.

Hence, $[G:H \cap K] \leq [G:H] [G:K]$.

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Let $C$ be the set of left cosets of $S = H \cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C \to C_1 \times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| \leq |C_1| \cdot |C_2|$, which proves that $[G: S]$ is finite.

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Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l \in G$ and $G$ be partitioned as $G=h_1H \cup ... \cup h_lH = k_1K \cup ... \cup k_mK$

I will find $a_1,...a_n \in G$ s.t. $G$ is partitioned as $G=a_1(H \cap K) \cup ... \cup a_n(H \cap K)$ which means $n=[G:H \cap K] < \infty$.

Let $b_1 \in G$. Then $\exists i_1 \in \{1,...,m\}, j_i \in \{1,...,l\}$ s.t. $b_1 \in h_{i_1}H \cap k_{j_1}K=b_1H \cap b_1K = b_1(H \cap K)$. Next, let $b_2 \in G \ \setminus \ b_1(H \cap K)$. Then $b_2 \in b_2(H \cap K)$ where $b_2(H \cap K) \cap b_1(H \cap K) = \emptyset$ because $b_1H \cap b_2H = h_{i_1}H \cap h_{i_2}H = \emptyset = k_{j_1}K \cap k_{j_2}K = b_1K \cap b_2K$

where $h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 \in \{1,...,m\} \ \setminus \ \{i_1\}, j_2 \in \{1,...,l\} \ \setminus \ \{j_1\}$ This process continues at most $lm$ times for $a_p=b_p, p \in \{1,2,...,n\}$ Thus, $n \le lm < \infty.$