----- > [!proposition] Proposition. ([[cardinality of frobenius product of subgroups]]) > Let $H,K$ be finite [[subgroup]]s of a [[group]] $G$. We have that $|HK|=\frac{|H|\cdot |K|}{|H \cap K|}.$ > ![[CleanShot 2023-09-10 at 17.14.06.jpg]] **Lemma.** ![[k-to-1 correspondence#^76a2c6]] ![[k-to-1 correspondence#^0f256d]] ![[k-to-1 correspondence#^a12726]] > [!proof]- Proof. ([[cardinality of frobenius product of subgroups]]) > Define $f: H \times K \to HK$ to be given by $f(h,k)=hk$. We will show that $f$ is an $n$-to-$1$ [[k-to-1 correspondence|correspondence]], where $n=|H \cap K|$. Then since $|H \times K|=|H||K|$, the result will follow from [[k-to-1 correspondence|the n-to-1 rule]]. \ Let $hk \in HK$. How many elements in $HK$ are equal to $hk$? Because $H,K$ are closed under the [[group]] operation, we know that an arbitrary member of $H \times K$ has the form $g =(h\alpha, \beta k)$ for $\alpha \in H$ and $k \in H$. Computing $f(g)=f\big( (h \alpha, \beta k) \big)=h \alpha \beta k,$ we see that $f(g)=f((h \alpha, \beta k))=hk \iff \alpha ^{}=\beta ^{-1} \iff \alpha, \beta \in H \text{ \textbf{ and } }\alpha, \beta \in K,$ ($H$ and $K$ are closed under inverses). Therefore, $f(g)=hk$ for precisely as many values of $g$ as there are elements in $H \cap K$. Since $hk$ was arbitrary we conclude $f$ is $n$-to-$1$, where $n=|H \cap K|$. Via the **lemma** this concludes the proof. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```