On the permutability of two subgroups of coprime indexes Let $G$ be a finite group and let $H$ and $K$ two subgroups of coprime indexes. Is it true that $G=HK$.
Obviously $G=\langle H,\, K\rangle$ but why the permute each other? Any ideas?
 A: $\newcommand{\index}[1]{\lvert #1 \rvert}$Yes, here's the standard argument, as worked out for instance in B. Huppert, Endliche Gruppen I, Hilfssatz 2.13.
In general (without the assumption that indices are coprime), we have $$\index{G:H\cap K} \le \index{G:H} \cdot \index{G:K}.$$ This is because we have
$$
\index{H K} = \frac{\index{H} \cdot \index{K}}{\index{H \cap K}}
$$
(we are not claiming that $HK$ is a subgroup), and thus
$$
\index{H : H \cap K} = \index{HK: K} \le \index{G:K}.
$$
Thus
$$
\index{G : H \cap K} = \index{G:H} \cdot \index{H:H \cap K} \le \index{G:H} \cdot \index{G:K}.
$$
Now, since $\index{G:H}$ and $\index{G:K}$ divide $\index{G:H\cap K}$, and $\gcd(\index{G:H}, \index{G:K}) = 1$, we have that $\index{G:H} \cdot \index{G:K}$ divides $\index{G:H\cap K}$. 
So in your situation $\index{G:H\cap K} = \index{G:H} \cdot \index{G:K}$.
Now we have
$$
\index{H K} = \frac{\index{H} \cdot \index{K}}{\index{H \cap K}} =
\index{G} \cdot \frac{\index{G:H\cap K}}{\index{G:H} \cdot \index{G:K}}
= \index{G},
$$
so $G = H K$.
