# Does the intersection of two finite index subgroups have finite index?

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?

• 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
• @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
• @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

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.

• 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
• – Dylan Moreland Jul 14 '12 at 23:15
• The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of $\{gK \mid g \in HK\}$. But then $g=hk$ and so any element of $\{gK \mid g \in HK\}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense? – Al Jebr Sep 23 '18 at 21:08
• @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 $h\in H$. – anon Sep 24 '18 at 0:29

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 $x^{-1}y\in H$.

• Please, professor, could you clarify what does the $p(x)$ mean? – Danilo Gregorin Afonso Sep 4 '17 at 1:20
• @Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values. – Shahab Sep 5 '17 at 15:32
• @Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else. – Danilo Gregorin Afonso Sep 5 '17 at 19:52

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 $G$?

• 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

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

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.

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, $$n \le lm < \infty.$$

I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.

Since both $$H$$ and $$K$$ are finite index subgroups of $$G$$, we can consider the right cosets $$G/H$$ and this is finite by hypothesis. In particular $$G$$ acts (transitively) on $$G/H$$ by right multiplication, i.e. $$G/H \times G \to G, \qquad (Hx, g) \mapsto Hxg.$$ If we now restrict the action to $$K$$, and consider the $$K$$-orbit of $$H$$, call it $$\mathcal{O}_K(H)$$, then it must be finite since it's contained in $$G/H$$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality $$|\mathcal{O}_K(H)| = [K : \text{Stab}_K(H)],$$ where $$\text{Stab}_K(H)$$ is the stabiliser of $$H$$ under the $$K$$-action. Notice that $$\text{Stab}_K(H) = K \cap \text{Stab}_G(H)$$, and what's the stabiliser of $$H$$ for the $$G$$-action? But of course $$H$$ itself, so $$\text{Stab}_K(H) = K \cap H$$, and finally we conclude that $$[K : K \cap H]$$ is finite. Hence $$[G : H \cap K] = [G : K][K : K \cap H]$$ must be finite as well.