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 $stab_H(K)$ ?
solution-verificationtag. – Arturo Magidin Jun 05 '24 at 15:42