$3$-Sylow in $\mathbb Z/ 6\mathbb Z$ In $\mathbb Z/6\mathbb Z$, by the definiton of Sylow $p$-subgroups, we can talk about  Sylow $2$-subgroups or Sylow $3$-subgroups. The textbook asserts that $\langle 2\rangle$ is a 3-Sylow and the only one. 
I understand that $2^3\equiv1 \pmod 6$ (and $3$ is the smallest natural number such that $2^n\equiv1$), and we have the following subgroup $\{0,2,4\}\subseteq \mathbb Z/ 6\mathbb Z$ in which the order of each element is a power of 3. 
However, if we take $\langle 4\rangle$, we have $4^3\equiv12\equiv1\pmod6$ and we get the same subgroup of $\mathbb Z/ 6\mathbb Z$. 
Can you help me understand which part am I getting wrong?

$\langle 2\rangle$ and $\langle 4\rangle$ are the same subgroup, which (as a set), is the classes of $\{0,2,4\}$ in $\mathbb Z/6\mathbb Z$.
