# $3$-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?

• Where did you get $2^3\equiv 1$ from? Sep 12, 2019 at 16:11
• That would be $3\cdot 2$, and that's still not $\equiv 1$, that's $\equiv 0$. Do you mean to say that in the group $\langle 2\rangle$, the element $2$ has order $3$? Sep 12, 2019 at 16:13
• @Lowkey it's confusing to use multiplicative notation when you really mean addition. Sep 12, 2019 at 16:14
• Yes, but I wrote it with multiplicative notations and got confused. Sep 12, 2019 at 16:15
• $\langle 2\rangle = \langle 4\rangle$ is the only Sylow $3$-subgroup. What's the problem? The book doesn't say the generator has to be unique, does it? Sep 12, 2019 at 16:26

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