A finite group with three nilpotent subgroups

Question

Suppose $G$ is a finite group with subgroups $H_i\leq G$, for $i\in\{1,2,3\}$, such that

• $H_i$ is nilpotent for all $i\in\{1,2,3\}$,
• if $i,j\in\{1,2,3\}$ are distinct then $\text{gcd}([G:H_i],[G:H_j])=1$.

Prove that $G$ is nilpotent.

Answer 1

We show each Sylow $p$-subgroup is normal.

Let $p$ be a prime dividing $|G|$. Since the $[G:H_i]$ are relatively prime, $p$ can only divide one of the $[G:H_i]$. Let $\{i,j,k\} =\{1,2,3\}$ and let $P_j$ be a Sylow $p$-subgroup of $H_j$. Since $p$ does not divide $[G:H_j]$, $P_j$ is also a Sylow $p$-subgroup of $G$. Similarly, let $P_k$ be a Sylow $p$-subgroup of $H_k$ and conclude it is a Sylow $p$-subgroup of $G$. By Sylow's theorem, some conjugate of $H_k$ (WLOG $H_k$ itself) contains $P_j$, so (by the same WLOG) we can take $P=P_j=P_k$. Since both $H_j$ and (that conjugate of) $H_k$ are nilpotent, $P$ is normal in both $H_j$ and (that conjugate of) $H_k$. Let $q$ be a prime dividing $|G|$. Since $\gcd([G:H_j],[G:H_k])=1$, $q$ divides at most one of $[G:H_j],[G:H_k]$, so one of $H_j$, (that conjugate of) $H_k$ contains a Sylow $q$-subgroup $Q$ of $G$. Since both $H_j$ and $H_k$ are nilpotent, we get that $Q$ normalizes $P$. Hence the normalizer $N=N_G(P)$ of $P$ in $G$ contains a Sylow $q$-subgroup of $G$ for every prime divisor $q$ of $G$, so $[G:N_G(P)]$ is not divisible by any prime $q$, so $N_G(P)=G$ and $P$ is normal in $G$.