A finite group G is called a 2-Frobenius group if it has a normal series $1\unlhd H\unlhd K\unlhd G$, where $K$ and $G/H$
are Frobenius groups with kernels $H$ and $K/H$, respectively.


We consider the relationship between $A=\{2\text{-Frobenius groups}\}$ and $B=\{\text{Frobenius groups}\}$:



> 
> **QUESTION.** Is it $A\cap B=\varnothing$?
> 
> 
> 



Is there a finite group that is both 2-Frobenius and Frobenius?
Attempt at correction ( or more precisely, justification that original attempt was basically correct): let $G$ be a Frobenius group with kernel $L$. Then $L \leq F(G)$ by Thompson's theorem, so $Z(F(G)) \leq C\_{G}(L) \leq L,$ as $L$ is a Frobenius kernel. Then $F(G) \leq C\_{G}(Z(F(G)) \leq L,$ so in fact $F(G) = L.$


Now suppose that $ 1 \lhd H \lhd K \lhd G$ with $K$ a Frobenius group with kernel $H$, and $G/H$ a Frobenius group with kernel $K/H$ (ie, that $G$ is also a $2$-Frobenius group). Then $K/H = F(G/H)$ by the argument above, so $F(G) \leq K,$ then $F(G) \leq F(K) \leq F(G)$ and $F(G) = F(K)$. However, $H = F(K),$ again by the argument above, since $K$ is a Frobenius group with kernel $H$. 
Thus $H = F(G)$. 


Since $G$ is a Frobenius group with kernel $L$, then we have $L = F(G) = H$.
Hence $G/H$ is isomorphic to a Frobenius complement as well as being a Frobenius group in its own right ( the Frobenius kernel of $G/H$ is $K/H$ and the Frobenius complement is isomorphic to $G/K$).


If $G/H$ has even order, then (since it is a Frobenius complement), it contains a unique involution $tH$ which is central in $G/H,$ so lies in $F(G/H) = K/H$ . This is a contradiction to the fact that $G/H$ is a Frobenius group with kernel $K/H.$ Hence $G/H$ has odd order, and all its Sylow subgroups are cyclic, as it is a Frobenius complement. By elementary transfer, $G/H$ is solvable. It follows that $F(G/H) = K/H$ is a cyclic Hall subgroup of $G/H.$


However, W. Burnside already observed that if $p,q$ are distinct prime divisors of the order of of a Frobenius complement, any subgroup of order $pq$ of the complement is cyclic. If we choose a prime divisor $p$ of $[K:H]$ and a prime divisor $q$ of $[G:K]$, we obtain a contradiction, since $G/H$ can't contain an element of order $pq$ as it's a Frobenius group with kernel $K/H.$ On the other hand, an element of order $q$ in $G/H$ normalizes $\Omega\_{1}(O\_{p}(K/H)),$ which is cyclic of order $p,$ so $G/H$ does contain a subgroup of order $pq.$ 


