group of order $p^nq^m$ is solvable

Question

I know Burnside theorem says this is true

I wonder can i go easy way if there is an additional condition:

$p<q$ and the order of $[p] \in \mathbb{Z}_q^{\times}$ is larger than $n$

Answer 1

i think i solved it

By First Sylow theorem, there are subgroups $H_1, \dotsi, H_m$ of order $q, q^2, \dotsi, q^m$ where each $H_j$ is normal in $H_{j+1}$ for $j=1,\dotsi, m-1$. Since $[p] \in {\mathbb{Z}}/{q \mathbb{Z}}^{\times} $ is larger than $n$, we have $p^i \not\equiv 1\; (\text{mod}\; p) \; \text{ for } i=1,\dotsi,n$ By Third Sylow theorem, the number of Sylow $q$-subgroup which is 1 (mod $q$) divides the order of the group $G$, that is, $ qk+1 \mid p^nq^m $ If $k>0$, then $qk+1$ must divide $p^n$ and so $qk+1=p^j$ for some $1\le j \le n$ which is a contradiction. Hence $k$ must be zero. Since there is one and only one Sylow $q$-subgroup $H_m$ of $G$, $H_m$ is normal in $G$.

Now consider the quotient group ${G}/{H_m}$ of order $p^n$. By applying First Sylow theorem to ${G}/{H_m}$, there are subgroups ${K_{1}}/{H_m}, \dotsi, {K_{n}}/{H_n}$ of order $p, p^2, \dotsi, p^n$ where each ${K_j}/{H_m}$ is normal in ${K_{j+1}}/{H_m}$ for $j=1,\dotsi, n-1$. For each $j=1,\dotsi,n$, $p^j = \Big\vert {K_j}/{H_m} \Big\vert =\frac{\vert K_j\vert}{q^m} \; \text{ and thus } \vert K_j \vert = p^j q^m $ In particular, $\vert K_m \vert = p^n q^m = \vert G \vert $ and thus $K_m=G$. For each $j=1,\dotsi,n-1$, $ {K_j}/{H_m} \triangleleft {K_{j+1}}/{H_m} \; \Leftrightarrow \; K_j \triangleleft K_{j+1} $ Moreover, since $H_m$ is normal in $G$, so is in $K_1$. Then we have $H_1 \triangleleft \dotsi \triangleleft H_m \triangleleft K_1 \triangleleft \dotsi \triangleleft K_n=G $ Since the order of each quotient group is a prime, it must be cyclic. Put $K_i=H_{i+m}$ for each $i=1, \dotsi, n$.