Two Subnormal subgroups with index of one and order of other relatively prime Let $H,K$ two subgroups of a finite group $G$. Suppose that $\gcd(|G:H|,|K|)=1$
Prove that if $K\triangleleft\triangleleft\; G$ then $K\subseteq H$.
My idea: Consider before the case $K\triangleleft G$. Then $HK\leq G$.
Remember that $|HK|=\dfrac{|H||K|}{|H\cap K|}$. Using this formula it's easy to show that $|G:HK|=\dfrac{|G:H|}{|K:K\cap H|}$. This must be an integer but numerator and denominator are coprime so the only possibility (except for the trivial case $G=H$) is $|K:K\cap H|=1$ so $K\subseteq H$.
In the general case this argument doesn't work because $HK$ is not a subgroup.
So i was trying to use induction on subnormal depth.
$K=K_0\triangleleft K_1\triangleleft...\triangleleft K_t=G$. But from here i don't know how to proceed.

Let $\pi$ be the set of primes dividing $|K|$, and let $O_\pi(G)$ be the largest normal subgroup of $G$ whose order is divisible only by primes in $\pi$. Then $O_\pi(G)$ is characteristic and hence normal in $G$, so by the case you have solved, $O_\pi(G) \le H$.
So now we have to prove that $K \le O_\pi(G)$ You can do that by induction on $t$. It is true when $t=1$. By induction $K \le O_\pi(K_{t-1})$. But since $O_\pi$ is a characteristic subgroup, $O_\pi(K_{t-1})$ is normal in $G$ and hence is contained in $O_\pi(G)$ and we are done.
