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Czhang271828 发表于 2023-3-16 06:23
basic calculations. Since the number of Sylow $7$-group is $7k+1$, which is also a divisor of $25$, such group has a normal subgroup with order $7$ (aka. unique maximal Sylow $7$-group)
方法与Theorem 2一样
Theorem 2. If $|G|=p^k q^s$ where $p$ and $q$ are prime numbers $k, s \in \mathbb{N}$ and $1 \bmod p \neq$ $q^t$ for $t=1,2, . .$, s then $G$ is solvable.
Proof. Since $N_p$ divides $|G|$ and it is equal to $1 \bmod p$ and since $q^t \neq 1 \bmod p$ for $t=1,2, . ., s$ we get that $N_p=1$. Let $P$ be the only p-Sylow subgroup of $G$. $P$ is a normal subgroup of $G$. Since the order of $P$ is $p^k$ where $p$ is prime we have that $P$ is a p-group and by our basic example-theorem $P$ is solvable. Also $[G: P]=q^s$ thus $\frac{G}{P}$ is a q-group and again by our basic example-theorem $\frac{G}{P}$ is solvable. By the tool theorem, now $G$ is solvable.
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Czhang271828
发表于 2023-3-16 13:23
