Sylow p-Subgroups of a Group

Definition: Let $G$ be a finite group of order $n$ and let $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ be the prime power decomposition of $n$. A Sylow $p_i$-Subgroup ($1 \leq i \leq k$) of $G$ is a subgroup $H$ of $G$ with order $p_i^{e_i}$.

In other words, a subgroup $H$ of $G$ is said to be a Sylow $p$-subgroup of $G$ if the order of $H$ is the highest power of $p$ that divides $|G|$.

Here is an alternate way to define a Sylow subgroup: Let $G$ be a group of order $n$ and let $p$ be a prime such that $p \mid n$ so that $p^k | n$ for some $k \geq 1$. If $n = p^km$ where $p \nmid m$, then a Sylow $p$-subgroup of $G$ is a subgroup $H$ of $G$ for which $|H| = p^k$.

Definition: Let $G$ be a finite group of order $n$ and let $p \mid n$. The Set of All Sylow $p$-Subgroups of $G$ is denoted by $\mathrm{Syl}_p(G)$. The Number of All Sylow $p$ Subgroups of $G$ is denoted by $n_p = |\mathrm{Syl}_p(G)|$.

Example 1

For example, consider the group $(\mathbb{Z}/12\mathbb{Z}, +)$. We have that $|\mathbb{Z}/12\mathbb{Z}| = 12$, and $12 = 2^23^1$. So a subgroup $H$ of $\mathbb{Z}/12\mathbb{Z}$ is a Sylow $2$-subgroup of $\mathbb{Z}/12\mathbb{Z}$ if $|H| = 4$. Similarly, a subgroup $H$ of $\mathbb{Z}/12\mathbb{Z}$ is a Sylow $3$-subgroup of $\mathbb{Z}/12\mathbb{Z}$ if $|H| = 3$.

Now since $\mathbb{Z}/12\mathbb{Z}$ is a cyclic group we have that any subgroup of $\mathbb{Z}/12\mathbb{Z}$ is cyclic. The only cyclic subgroup of order $4$ is the subgroup $\{0, 3, 6, 9 \}$, and the only cyclic subgroup of order $3$ is the subgroup $\{ 0, 4, 8 \}$.

Therefore $\{0, 3, 6, 9 \}$ is the only Sylow $2$-subgroup of $\mathbb{Z}/12\mathbb{Z}$, and $\{0, 4, 8 \}$ is the only Sylow $3$-subgroup of $\mathbb{Z}/12\mathbb{Z}$.