Abstract Algebra: Theory and Applications · Problem 15.4.22

The key is to use Sylow's theorems to find a subgroup whose size forces it to be normal. Since $|G| = p^n q$ with $p > q$, a Sylow $p$-subgroup has order $p^n$, and the number of such subgroups divides $q$ and is congruent to $1 \pmod{p}$. Because $q < p$, the only divisor of $q$ that satisfies the congruence is $1$, meaning the Sylow $p$-subgroup is unique—and any unique Sylow subgroup must be normal.

The key is to use Sylow's theorems to find a subgroup whose size forces it to be normal. Since $|G| = p^n q$ with $p > q$, a Sylow $p$-subgroup has order $p^n$, and the number of such subgroups divides $q$ and is congruent to $1 \pmod{p}$. Because $q < p$, the only divisor of $q$ that satisfies the congruence is $1$, meaning the Sylow $p$-subgroup is unique—and any unique Sylow subgroup must be normal.