
> 
> **Theorem $1$(Burnside):** A simple nonabelian finite group can not have a conjugacy classes with prime power elements.
> 
> 
> **Theorem $2$:** A group of order $p^nq^m$ is solvable.
> 
> 
> 


Theorem $1$ depends on character theory. Theorem $2$ is a direct consequences of Theorem $1$. Burnside proved Theorem $2$ using character theory, but later a character-free proof was found.



> 
> **Theorem 3:**
> Let $G$ be a finite group with nilpotent Hall $p'$ subgroup $H$. Then $G$ is solvable.
> 
> 
> 


Now we can directly prove Theorem $3$ by Theorem $1$. I wonder that can we prove Theorem $3$ by Theorem $2$ and induction as it is generalization of Theorem $2$, so that Theorem $3$ can be established without character theory as well.


Thanks.


**Note:** I have asked this question [there](https://math.stackexchange.com/questions/1549481/the-groups-with-nilpotent-hall-p-subgroup).



The groups with nilpotent Hall $p'$ subgroup
It may be that Wielandt's proof that if a finite group $H$ has nilpotent Hall subgroups $A$ and $B$ with $H = AB$ is solvable assumes Burnside's result about
non-Abelian simple groups having no non-identity conjugacy class of prime power
order as a starting point ( so it can be assumed that both $|A|$ and $|B|$ have at least two different prime divisors- and after this no further character theory is needed). I think this can be circumvented by methods from the proof of the odd order theorem of Feit and Thompson, and revised by Bender and Glauberman. Thompson commented when the proof of the odd order theorem appeared that the methods of the paper could be adapted to give a direct group-theoretic proof of Burnside's $p^{a}q^{b}$-theorem when $p$ and $q$ are both odd. D. Goldschmidt and H. Bender (independently) later carried out such a proof, and a proof by similar methods was later devised to deal with the prime $2$.


Here I outline a group-theoretic proof that if $G$ is a minimal non-Abelian simple group of odd order, then $G$ does not have a nilpotent Hall $p^{\prime}$-group
for any prime $p$. These assumptions are stronger than required to fully answer the question, but they are illustrative of the methodology of the odd order paper and its relevance to the question. The proof could probably be further streamlined in this special situation, but I haven't attempted to optimize it.


So suppose that our minimal non-Abelian simple group $G$ of odd order has the form $G = PH$ where $P$ is a Sylow $p$-subgroup of $G$ and $H$ is a nilpotent (Hall) $p^{\prime}$-subgroup.


We first note that $H$ normalizes no non-identity $p$-subgroup of $G$ (and similarly $P$ normalizes no non-identity $p^{\prime}$-subgroup). For suppose that $H$ normalizes a non-trivial $p$-group $Q$. Then $HQ = H(P \cap HQ)$ by Dedekind's modular law, and $Q = (P \cap HQ)$ is the unique Sylow $p$-subgroup of $HQ$. Thus $Q \leq P$. But $G = HP^{g}$ for any $g \in G$, so we have $Q \leq P^{g}$ for every $g \in G$. Hence $\cap\_{ g \in G}P^{g}$ is a non-identity normal $p$-subgroup of $G$ ( for it contains $Q$), a contradiction. The proof that $P$ normalizes no non-identity $p^{\prime}$-subgroup is similar.


We next prove that $P$ is contained in a unique maximal subgroup of $G$, and that $H$ is contained in a unique maximal subgroup of $G$.


For let $M \geq P$ be a maximal subgroup of $G$. Then $M = P(M \cap H)$ and $O\_{p^{\prime}}(M) = 1$. Thus $ZJ(P) \lhd M$, so that $M = N\_{G}(ZJ(P))$ is uniquely determined by $M$. Similarly, if $L$ is a maximal subgroup of $G$ containing $H$, then $L = N\_{G}(ZJ(H))$, where $J(H)$ is the subgroup of $H$ generated by its Abelian subgroups of maximal order. We retain the notation $M$ and $L$ for these maximal subgroups


Now we have to go more deeply into the Feit-Thompson and Bender-Glauberman (and Goldschmidt) methodology. The first aim is to prove that if $X$ is any maximal subgroup of $G$, then either $F(X)$ is a $p$-group, or $F(X)$ is a $p^{\prime}$-group. Once this is established, it follows that $X = N\_{G}(ZJ(Y))$, where $Y$ is a Sylow $p$-subgroup of $X$ (in the former case) or $Y$ is a (nilpotent) Hall $p^{\prime}$-subgroup of $X$ (in the second case). It then follows that $X$ is conjugate to one $L$ or $M$. This (when $O\_{p}(X) = 1$) does require a theorem of Wielandt that all $p^{\prime}$-subgroups of $G$ are conjugate to subgroups of $H$. Once we know that $X$ is conjugate to one of $L$ or $M$, we have that $P \cap P^{g} = 1$ for all $g \in G \backslash M$ and $H \cap H^{g} = 1$ for all $g \in G \backslash L$. Hence $|G| \geq |P|^{2}$ and $|G| \geq |H|^{2}$, contrary to the fact that $|G| = |P||H|$.


It remains then to prove that we either have $F(X)$ a $p$-group or $F(X)$ a $p^{\prime}$-group.


Suppose first that $X$ contains no elementary Abelian subgroup of order $r^{3}$ for any prime $r$. Then Feit-Thompson proved that $X$ has a normal $s$-complement where $s$ is the smallest prime divisor of $|X|$, and then that $X$ has a normal Sylow $q$-subgroup where $q$ is the largest prime divisor of $|X|$.
Then $X = N\_{G}(Q)$ by maximality of $X$, and $Q$ is a Sylow $q$-subgroup of $G$. If $q = p$, then $X$ is conjugate to $M$. If $q \neq p$, then we may replace $X$ by a conjugate if necessary and assume that $Q \leq H$. Then $H \leq N\_{G}(Q) = X$ and $X = L$ as $L$ is the unique maximal subgroup of $G$ containing $H$.


Suppose then that $X$ contains an elementary Abelian subgroup $A$ of order $r^{3}$ for some prime $r$. Replacing $X$ by a conjugate if necessary, we may suppose that $A \leq P$ if $r =p$ and $A \leq H$ if $r \neq p$.


By the Feit-Thompson uniqueness Theorem (later shortened somewhat by Bender and Glauberman), $A$ is contained in a unique maximal subgroup of $G$. If $r = p$, this must be $M$, the unique maximal subgroup of $G$ containing $P$, while if $r \neq p$, this must be $L$, the unique maximal subgroup of $G$ containing $H$.


Hence we have indeed established that $X$ is either a conjugate of $M$ or a conjugate of $L$ in all cases.


