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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?

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    $\begingroup$ Where did you get $2^3\equiv 1$ from? $\endgroup$
    – Arthur
    Sep 12, 2019 at 16:11
  • $\begingroup$ 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 $3$? $\endgroup$
    – Arthur
    Sep 12, 2019 at 16:13
  • $\begingroup$ @Lowkey it's confusing to use multiplicative notation when you really mean addition. $\endgroup$ Sep 12, 2019 at 16:14
  • $\begingroup$ Yes, but I wrote it with multiplicative notations and got confused. $\endgroup$
    – Lowkey
    Sep 12, 2019 at 16:15
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    $\begingroup$ $\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? $\endgroup$
    – AlvinL
    Sep 12, 2019 at 16:26

1 Answer 1

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

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