Sylow p-subgroup

Let $G$ be a finite group and let $p$ be a prime. Write $|G|=p^n m$ with $n\ge 0$ and $p\nmid m$ (so $p^n$ is the largest power of $p$ dividing $|G|$). A Sylow $p$-subgroup of $G$ is a subgroup $P\le G$ such that $|P|=p^n$. In particular, every Sylow $p$-subgroup is a $p$-group .

The first Sylow theorem guarantees that Sylow $p$-subgroups exist whenever $p$ divides $|G|$. The second Sylow theorem states that any two Sylow $p$-subgroups are conjugate (in the sense of conjugation of subgroups).

Examples:

• In $S_3$ (order $6=2\cdot 3$), Sylow $2$-subgroups have order $2$ (each is generated by a transposition), and the unique Sylow $3$-subgroup is $A_3$.
• In $A_4$ (order $12=2^2\cdot 3$), a Sylow $3$-subgroup has order $3$ and is generated by a $3$-cycle.
• If $|G|=p^n$, then $G$ itself is the unique Sylow $p$-subgroup.