I have to prove that every group of order $2p^n$, p is prime, is solvable.
When $n=1$, the group either is cyclic or dihedral, in any case the group is solvable.
When $n=2$, if P is p-sylow subgroup, then [G:P]=2, and then P is normal in G. Then we got solvable sequence ${e}<P<G$, and G is solvable.
In general case, according Burnside’s Theorem, any group with order $p^aq^b$ must be solvable. But how can I proof this case, without Burnside?
Thank you.
If $p=2$, $G$ is a $2$-group. Since every $p$-group is solvable (in fact, nilpotent), for any prime $p$, that case is covered.
If $p\ne 2$, then by Sylow's theorem there is a $p$-Sylow subgroup $P$ of order $p^n$ and thus of index $2$ and therefore normal. But $P$ is solvable and so has a composition series with cyclic factors. Since $P$ is normal in $G$, just add $G$ to the end to get a composition series for $G$, and that last factor is also cyclic, since it has order $2$. Thus, $G$ has a composition series with cyclic factors and so is solvable. Alternatively, note $G/P$ is abelian since it's order $2$, so $G^{\prime}\le P$, so $G$ is solvable since $P$ is.