It's known all groups of order $p^{m}q^{n}$ and all groups of odd order are
solvable (By Burnside theorem and Feit-Thompson theorem).


Let $G$ be a group of order $2^{m}\cdot p^{n}\cdot q^{t}$ where $p\neq 3$
and $q\neq 3$ are prime.


Does anyone knows why $G$ is solvable?



Why a group of order $2^{m}\cdot p^{n}\cdot q^{t}$ is solvable?
Let $G$ be an unsolvable group. Since $G$ is a finite group, then it
has a chief series. Since $G$ is unsolvable it is easy to show that $G$ has series: $1 \unlhd N \lhd H \unlhd G$ such that $H/N$ is a non-abelian simple group or $H/N$ is a direct product
of isomorphic non-abelian simple groups. Since $G$ has three prime divisors,
then $H/N$ has three prime divisors too. Since $3\notin \pi (G)$, by the
following Lemma, we can get a contradiction. Therefore $G$ is solvable group.


Lemma: If $G$ is a simple with three prime divisors, then $G$ is
isomorphic to one of the following groups: 
$A\_{5}$, $A\_{6}$, $PSL(2,7)$, $PSL(2,8)$, $PSL(2,17)$, $PSL(3,3)$, $PSU(3,3)$ or $PSU(4,2)$.(Ref: Herzog, M, On finite simple groups of order divisible by three primes only, J. Algebra, 120 (10), (1968), 383-388.)


Similar to the above discussion we can prove that all finite groups of order $
2^{m}\cdot 3^{n}\cdot p^{k}$, where $p$ is prime but not in {5,7,13,17}
are solvable groups.


