In The Theory of Finite Groups: An Introduction of Hans Kurzweil, Bernd Stellmacher(http://gen.lib.rus.ec/book/index.php?md5=C1F95CF5F64359D6DFD3C6E5BD7EB12F), on page 9, let $A,B$ be subgroups of the finite group $G,$ then $[G:A\cap B]\leq[G:A][G:B].$ Morever, on page 7, let $A$ and $B$ be finite subgroups of a group $G,$ then $|AB|=\dfrac{|A|\cdot|B|}{|A\cap B|}.$
Thus, $|AB|=1.$ Is it correct?
For any subgroups $A$ and $B$ of any group $G$. $\left\vert AB\right\vert=1$ iff $\left\vert A\right\vert=\left\vert B\right\vert=1$.
Clearly, if $\left\vert A\right\vert=\left\vert B\right\vert=1$, then $\left\vert AB\right\vert=1$.
Conversely, suppose $\left\vert AB\right\vert=1$. Then since $A\subseteq AB$ and $B\subseteq AB$, we also have that $\left\vert A\right\vert=\left\vert B\right\vert=1$.