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Let $G$ be a group and $H$ and $K$ be subgroups of $G$ of finite indices. Show that $H\cap K$ is of finite index.

I was trying to solve this problem by using the formula $o(HK)=\frac{o(H)o(K)}{o(H\cap K}.$ But the problem is, that nowhere it's given $H$ and $K$ are finite, so we cant use this formula. Next, I tried using the fact: $H$ and $K$ are two subgroups of $G.$ Let $a,b$ in $G$ then $Ha\cap Kb=\phi$ or there exists a $c\in G$ such that $(H\cap K)c=Ha\cap Kb.$ I proved this result stated above but I am not attaching it here because I found out that this is a rather popular one. (However, if someone feels that attaching the proof of this would be helpful incase, please, feel free to let me know.) Finally, after searching a while on the internet and looking at some strategies to solve these problems, I wrote out a solution which goes like this:

Given $H,K\leq G.$ Let the set of right cosets of $H,K$ be $H'=\{Hg_{i_1},...Hg_{i_k}\},K'=\{Kg_{j_1},...,Kg_{j_k}\}$ respectively. Now, say, $(H\cap K)g_{k_1}$ is a right coset of $H\cap K.$ Then, $Hg_{k_1}\in H'$ and $Kg_{k_1}\in K'$ as $H',K'$ are finite. Since $g_{k_1}$ is arbitary so we see $\forall (H\cap K)g_{k_1}$ $\exists i_p,j_p$ such that $Hg_{k_1}=H{i_p}\in H'$ and $Kg_{k_1}=K{j_p}\in K'$, thus, we conclude the set of cosets of $H\cap K$ finite since $H',K'$ is finite and we see that we can map every cose of $H\cap K$ to a coset of $H$ and $K$.

Is the above solution, correct? If not, where is it going wrong?

This link Does the intersection of two finite index subgroups have finite index? does not answer my question as there the user asks for a solution but I am asking this to validate my solution, i.e whether this solution is correct or not? I posted a solution and edited the OP to make it a solution verification post.

Arthur
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  • Okay, my answer before was dead wrong, I am sorry. I hadn't worked with algebra for a while and jumped a bit to fast with my conclusions. Maybe have a look at: https://math.stackexchange.com/questions/128538/does-the-intersection-of-two-finite-index-subgroups-have-finite-index?rq=1 – Symplectic Witch Feb 05 '23 at 18:38
  • @Nuke_Gunray I have edite my question a bit? Mind looking at it, to ensure whether it's valid, I mean the solution, I posted?What do you think?.... – Arthur Feb 06 '23 at 04:54
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    Looks correct to me, but so did my proof ^^ – Symplectic Witch Feb 06 '23 at 21:38
  • The notation makes no sense. What is $i_p$ in terms of the elements? I think you mean $H_{g_{i_p}}$, etc. Even so, the argument is at least incompete (and thus incorrect). You would need to show that distinct cosets correspond to distinct pairs of cosets. Otherwise, all you have is a function from cosets of $H\cap K$ to pairs of cosets $(Ha,Kb)$. If this association is not one-to-one, then the finiteness of the target set does not imply finiteness of the domain. – Arturo Magidin Feb 07 '23 at 04:59

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With $B$ a subgroup, $A$ a union of cosets of $B$, we denote $\frac{|A|}{|B|}$ the number of cosets of $B$ in $A$.

We have $\frac{|H|}{|H\cap K|}=\frac{|HK|}{|K|}$.

Note that $H=\{h(H\cap K)\mid h\in H\}$ and $HK=\{hK\mid h\in H\}$.

For $h_1,h_2\in H$, we have $$h_1 K=h_2 K\iff h_1^{-1} h_2\in K \iff h_1^{-1} h_2\in H\cap K\iff h_1 (H\cap K)= h_2 (H\cap K),$$ that is, we have a bijection between $h(H\cap K)$ and $hK$.

The above proof works for infinite groups.

We have $$\frac{|H |}{|H\cap K|} =\frac{|HK|}{|K|}\leq \frac{|G|}{|K|}<\infty,$$ $$\frac{|G|}{|H\cap K|}=\frac{|G|}{|H|}\cdot \frac{|H |}{|H\cap K|}<\infty.$$

stlinex
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