If $H$ and $K$ are finite subgroups of a group $G$, then we have
\[|HK|=\frac{|H| |K|}{|H \cap K|}.\]

Proof.

Let $p^{\alpha}$ is the highest power of $p$ that divides $|G|$.
That is, we have
\[|G|=p^{\alpha}n,\]
where $p$ does not divide the integer $n$.

Then the orders of the Sylow $p$-subgroups $H, K$ are $p^{\alpha}$.

Since the intersection $H\cap K$ is a subgroup of $H$, the order of $H \cap K$ is $p^{\beta}$ for some integer $\beta \leq \alpha$ by Lagrange’s theorem.
As $H$ and $K$ are distinct subgroups, we must have $\beta < \alpha$.

Then the number of elements of the product $HK$ is
\begin{align*}
|HK|&=\frac{|H| |K|}{|H \cap K|}\\[6pt]
&=\frac{p^{\alpha} p^{\alpha}}{p^{\beta}}=p^{2\alpha-\beta}.
\end{align*}
Since $\beta < \alpha$, we have $2\alpha-\beta > \alpha$.

It follows that the product $HK$ cannot be a subgroup of $G$ since otherwise the order $|HK|=p^{2\alpha-\beta}$ divides $|G|$ by Lagrange’s theorem but $p^{\alpha}$ is the highest power of $p$ that divides $G$.

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