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As a graduate student I will present my first P.S (problem solution) lecture for undergraduates. One of my question is proving the following theorem:

Let H and K be finite subgroups of a group G. Then $$|HK|=\frac{|H||K|}{|H \cap K|}$$

We have nice proofs here, but I will introduce you mine:

(1) We said that H and K are subgroups of G. Let $HK=\{hk| h \in H, k \in K\}$.Then, we can say that $HK= \bigcup_{h \in H}K$ (union of the left cosets of K). By using the property of cosets, it is known that $|hK|=|K|$ for each $h \in H$. Now, if we can find the number of distinct left cosets of K, we can find $HK$.

(2)Assume that X is the set of left cosets of K, then K is the element of X,because $eK=K$, so K is left coset (realize that the identity element for H and K are same because of $K \leq G, H \leq G$). If we find the orbit of element $eK$ of X under action H, we can find the elements of distinct left cosets of K.

(3) We know that orbit of a group element $x \in X $ can be defined as $orb_G(x)=\{y\in X | y=gx, g \in G \}$. Then, $orb_H(eK)=\{y \in X | y=h(eK)\}$, but we said that $eK=K$,so $orb_H(eK)=\{y \in X | y=hK\}$.

As a result when we find the orbit of $eK$, we find the distinct left cosets of K, because $orb_H(eK)$ denotes a set, so the elements its inside must be distinct.

(4) Now, the question is how many distinct left cosets of K there is,i.e, what is the cardinality of $orb_H(eK)$. By Orbit-Stabilizer Theorem, $$stab_H(eK)=\{h \in H | h(eK)=eK=K\}= \{h \in H | h \in K\}$$ (because if $hK=K$, then the elements of H for h must also be the element of K because of the closure property of subgroup K)

(5) So, $orb_H(eK)=\frac{|H|}{|H \cap K |}$ and because of the coset property in (1), $$|HK|=\frac{|H||K|}{|H \cap K |}$$

My question: Can you please review my proof to detect any missing or wrong part ? I want my proof open as possible as. Moreover, I am hesitating about using symbol $eK$ instead of $K$. What do you think about it ?

Are $orb_H(eK)$ and $stab_H(eK)$ fine instead of $orb_H(K)$ and $stab_H(K)$ ?

Not a Salmon Fish
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  • The last sentence of (1) is misleading. You do not want to find the number of distinct (left) cosets of $K$. You are trying to find the number of distinct left cosets of $K$ that are represented by (or that contain) elements of $H$. Just those. And you don't want to find $HK$, you want to find $|HK|$. – Arturo Magidin Jun 05 '24 at 15:25
  • @ArturoMagidin very good catch, i noted it :) thanks – Not a Salmon Fish Jun 05 '24 at 15:26
  • @ArturoMagidin do you have any comment for using notation $eK$ ? It hesitated me as I mentioned – Not a Salmon Fish Jun 05 '24 at 15:33
  • Seems right, if a bit wordy. You can also consider $H\times K$ acting on $HK$ by $(h,k)x:=hxk^{-1}$, so ${\rm Orb}(e)=HK$ and ${\rm Stab}(e)$ is the "diagonal" copy ${(g,g)\mid g\in H\cap K}$ of $H\cap K$ within $H\times K$. – coiso Jun 05 '24 at 15:34
  • @coiso thanks for your comment, but I hesitate about whether it is unnecessary or not,it is already true – Not a Salmon Fish Jun 05 '24 at 15:41
  • I have commented on all I wish to comment on. Please don't poke me to ask me to do more; it's annoying. Also, this is not really a "here is my proof, please let me know if it is right" site. Read the description of the solution-verification tag. – Arturo Magidin Jun 05 '24 at 15:42

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