Let $H$ and $K$ be finite subgroups of a group $G$. Then $|HK|=\frac{|H||K|}{|H\cap K|}$.
Sketch of proof: $C=H\cap K$ is a subgroup of $K$ of index $n=|K|/ |H\cap K|$ and $K$ is the disjoint union of right cosets $Ck_1\cup Ck_2 \cup … \cup Ck_n$ for some $k_i\in K$. Since $HC= H$, this implies that $HK$ is the disjoint union $Hk_1 \cup Hk_2 \cup … \cup Hk_n$. Therefore, $|HK|=|H|\cdot n=|H||K|/|H\cap K|$.
Filling details: $(1)$ $[K:H\cap K]=|\{(H\cap K)k\mid k\in K\}|\leq |K|$. Since $K$ is finite, $[K:H\cap K]=n\in \Bbb{N}$.
$(2)$ $HC=H(H\cap K)=H$. Let $x\in H(H\cap K)$. Then $x=ab$, where $a\in H$ and $b\in H\cap K$. Since $H\cap K\subseteq H$, we have $b\in H$. So $x=ab\in H$. Thus $H(H\cap K)\subseteq H$. Conversely, let $x\in H$. Since $H, K$ are subgroup of $G$, we have $H\cap K$ is subgroup of $G$. So $e\in H\cap K$ and $x=xe\in H(H\cap K)$. Thus $H(H\cap K)\supseteq H$. Hence $H(H\cap K)= H$.
$(3)$ $H(\bigcup_{i\in J_n}Ck_i)=\bigcup_{i\in J_n}HCk_i$. For seek of simplicity, we generalize $P(\bigcup_{i\in J_n}Q_i)=\bigcup_{i\in J_n}PQ_i$. Let $x\in P(\bigcup_{i\in J_n}Q_i)$. Then $x=pq$, where $p\in P$ and $q\in \bigcup_{i\in J_n}Q_i$. So $q\in Q_j$, for some $j\in J_n$. Thus $x=pq\in PQ_j\subseteq \bigcup_{i\in J_n}PQ_i$. Conversely, let $x\in \bigcup_{i\in J_n}PQ_i$. Then $x\in PQ_j$, for some $j\in J_n$. So $x=pq$, where $p\in P$ and $q\in Q_j\subseteq \bigcup_{i\in J_n}Q_i$. Thus $x=pq\in P(\bigcup_{i\in J_n}Q_i)$. Hence $P(\bigcup_{i\in J_n}Q_i)=\bigcup_{i\in J_n}PQ_i$.
$(4)$ $\{Hk_1,…,Hk_n\}$ are pairwise disjoint, that is $Hk_i\cap Hk_j=\emptyset$ if $i\neq j$. Assume towards contradiction, $Hk_i\cap Hk_j\neq \emptyset$. Since two equivalence class are either disjoint or equal, we have $Hk_i= Hk_j$. It’s easy to check, $(H\cap K)b=Hb\cap Kb$ and $Kb=K$, for all $b\in K$. So $$\begin{align} Ck_i\cap Ck_j &= [(H\cap K)k_i]\cap [(H\cap K)k_j]\\ &= (Hk_i\cap Kk_i)\cap (Hk_j\cap Kk_j)\\ &= (Hk_i\cap K)\cap (Hk_j\cap K)\\ &= Hk_i\cap K\\ &=(H\cap K)k_i\\ &=\emptyset .\end{align}$$ But $k_i=ek_i\in (H\cap K)k_i$. Thus we reach contradiction. Hence $Hk_i\cap Hk_j=\emptyset$ if $i\neq j$. Is my filling of details correct?
Do you have any motivation to prove this kind of theorem? I tired to solve this problem by my own for weeks, but made no progress.