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et $G$ be an infinite group and $H,K \leq G$ with $[G:H] = m$ and $[G:K] = n$. I want to show that $[G:H \cap K] < \infty$.

My idea:

Let $\{h_1,\cdots,h_m\}$ be representants of $\{gH \mid g \in G\}$ and similar $\{k_1,\cdots,k_n\}$. Then each $g \in G$ is in $h_iH \cap k_jK$ for some $1 \leq i \leq m$ and $1 \leq j \leq n$. Then I want to show that if $g \in h_iH \cap k_jK$ then $g( H \cap K) = h_iH \cap k_jK$.

Does this show that $[G:H \cap K] <\infty$ ? I think that this shows that $[G: H \cap K] \leq mn$.

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Show that $\,\{h_ik_j\}\,$ is a complete set representative of the cosets of $\,H\cap K\,$ in $\,G\,$ (though, most probably, not a minimal one!) – DonAntonio Jan 5 '13 at 14:32
You mean $\{h_i,k_j \mid 1 \leq i \leq n , 1 \leq j \leq m \}$ ? – Andre Jan 5 '13 at 14:35
No @Andre, I meant what I wrote. Read my answer below. – DonAntonio Jan 5 '13 at 16:47

4 Answers 4

up vote 3 down vote accepted

Let $H$ act on the set of left cosets of $K$ in $G$ by left multiplication. (This action is not necessarily transitive) The stabilizer of the coset $K$ is $H \cap K$. By the orbit-stabilizer theorem, $[H: H \cap K]$ is finite since $[G:K]$ is finite. Since $[G:H]$ is also finite the result follows.

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Why is the action transitive ? – Andre Jan 5 '13 at 14:56
Straightforward approach. I favorited it. +1 – Babak S. Jan 5 '13 at 15:02
Just to clarify it for myself: Let $A:= \{gK \mid g \in G\}$. Then $|A| = [G:K]< \infty$ and $[H : H \cap K] = [H: Stab(K)] = |Orb(K)| < \infty$ because $Orb(K) \subseteq A$. How to proceed then ? – Andre Jan 5 '13 at 15:35
Sure. Thanks ! Nice approach . – Andre Jan 5 '13 at 15:38

Let $A$ be the set of all right cosets of $H\cap K$ in $H$ and $B$ be the set of all right cosets of $K$ in $G$. Then the map $\varphi:A\to B$ given by $(H\cap K)h\mapsto Kh$ for $h\in H$ is well defined since $(H\cap K)h'=(H\cap K)h$ implies $h'h^{-1}\in H\cap K\subseteq K$, hence $Kh'=Kh$. To see that $\varphi$ is injective, suppose $Kh_1=Kh_2$. Then $h_1h_2^{-1}\in H\cap K$, hence $h_1\in (H\cap K)h_2$ and we have $(H\cap K)h_1=(H\cap K)h_2$. Then $[H:H\cap K]=|A|\leq |B|=[G:K]$. Then notice that $H\cap K$ is a subgroup of $H$, and in turn $H$ is a subgroup of $G$, thus we have $$ [G:H\cap K]=[G:H][H:H\cap K]\leq [G:H][G:K], $$ therefore, $H\cap K$ has finite index in $G$.

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Thanks. Why is $[G: H \cap K] = [G:H][H:H \cap K]$? Is this a general resut if $H,K \leq G$ ? – Andre Jan 5 '13 at 14:59
It follows from the third isomorphism theorem. Since $(G/H\cap K) \big/ (H/H\cap K)\cong G/H$, we multiply the index over and obtain the equality. Hope this helps! – Clayton Jan 5 '13 at 15:02
Yes this helps :) Bute $K,N$ are maybe not normal in $G$ ? – Andre Jan 5 '13 at 15:04
Yes, since we know $H\cap K\leq H$ and $H\leq G$. That is all the idea is, really. – Clayton Jan 5 '13 at 15:09
I have chosen the answer of Serkan because its faster. But yours is nice,too because its very precise. – Andre Jan 5 '13 at 15:39

Let $(H\cap K)a$ be the rigth coset of subgroup $H\cap K$ in your group $G$. It can be shown that $(H\cap K)a=Ha\cap Ka$. In fact as you noted, every right(or left) coset of $H\cap K$ is an intersection of a right coset of $H$ and a right coset of $K$. It means as you also noted the right cosets of $H\cap K$ resemble $Hx\cap Ky$. If we enumerate the distinct right cosets of $H\cap K$, we will find the number of them less than all above compositions, $Hx\cap Ky$, in $G$. So the index of $H\cap K$ in the group is finite and less than $mn$. Moreover if $(m,n)=1$ then $[G:H\cap K]=mn$.

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+1 I think I saw this question asked again this month! Maybe you came to the rescue for that newer question, too. – amWhy Feb 23 '13 at 3:26

Addressing Andre's question:

$$\forall\,x\in G\;\;\exists\, !\,\,h_i\in\{h_1,...,h_m\}\,\,s.t.\,\,x=h_i\alpha\,\,,\,\,\alpha\in H$$

But for this $$\alpha\,\,\,\exists\,!\,k_j\in\{k_1,...,k_n\}\,\,s.t.\,\,\alpha=k_j\beta\Longrightarrow \left\{h_ik_j\right\}_{1\leq i\leq m\,,\,1\leq j\leq n}$$

is a complete, though perhaps not minimal, set of representatives of the cosets of $\,H\cap K\,$ in $\,G\,$

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This is what you commented Andre correctly. – Babak S. Jan 5 '13 at 17:20

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