Prime divisor greater than Sylow index is Sylow-unique

Statement

Suppose $p$ is a prime number and $m$ is a natural number less than $p$. Suppose:

$n=p^{k}m$

for some nonnegative integer $k$. Then, $p$ is a Sylow-unique prime divisor (?) of $n$.

Facts used

1. congruence condition on Sylow numbers
2. divisibility condition on Sylow numbers

Related facts

• Order is product of Mersenne prime and one more implies normal Sylow subgroup

Proof

Given: A prime $p$, a natural number $m<p$, a group $G$ of order $n=p^{k}m$ for $k\geq 0$.

To prove: $n_{p}=1$ in $G$.

Proof: By fact (1), $n_{p}\equiv 1\mod p$. By fact (2), we also have $n_{p}|m$, Thus, $n_{p}\leq m$.

Combining the conditions $n_{p}\equiv 1\mod p$ and $n_{p}\leq m$ yields that $n_{p}=1$.