The Order of HK of Two Subgroups H and K of a Group G

Recall from The Product HK of Two Subgroups H and K of a Group G page that if $G$ is a group and $H$, $K$ are subgroups of $G$ then we defined the product of $H$ and $K$ to be:

(1)

\begin{align} \quad HK = \{ hk : h \in H, k \in K \} \end{align}

We noted that in general, $HK$ need not be a subgroup of $G$, and proved that $HK$ is a subgroup of $G$ if and only if $HK = KH$.

The following proposition tells us how large the $|HK|$ is.

Proposition 1: Let $G$ be a group and let $H$ and $K$ be subgroups of $G$. Then $\displaystyle{|HK| = \frac{|H| |K|}{|H \cap K|}}$.

Note that since $H \cap K$ is a subgroup (by the proposition on the The Intersection and Union of Two Subgroups page) we have that $|H \cap K| \geq 1$. So in the equation above - we are never dividing by $0$.

• Proof: Since $H$ and $K$ are subgroups of $G$ we have that $H \cap K$ is a subgroup of $G$. Thus, if $z \in H \cap K$ then $z^{-1} \in H \cap K$.

• Furthermore, since $H \cap K$ is a subgroup of $H$ and a subgroup of $K$, we see that if $h \in H$, $k \in K$, and $z \in H \cap K$ then $hz \in H$ and $z^{-1}k \in K$.

• Thus, if $hk \in HK$ then $hk = (hz)(z^{-1}k)$. So every $hk \in HK$ can be written in at least $|H \cap K|$ different ways as an element from $H$ times an element from $K$.

• But also, note that if $hk = h'k'$ then $h^{-1}h' = kk'^{-1}$. So $h^{-1}h' = kk'^{-1} \in H \cap K$ and so there exists a $z \in H \cap K$ $h^{-1}h' = z$ and $kk'^{-1} = z$. so $h' = hz$ and $k' = z^{-1}k$, i.e., $h'k' = (hz)(z^{-1}k)$ for some $z \in H \cap K$.

• Therefore, every $hk \in HK$ can be written in exactly $|H \cap K|$ different ways as an element from $H$ times an element from $K$. Therefore:

(2)

\begin{align} \quad |HK| = \frac{|H| |K|}{|H \cap K|} \quad \blacksquare \end{align}