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I want to find all Sylow 2-subgroups of $D_6 = \{ e, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5\}$. $r$ is the rotation element, and $s$ is the reflection element.

$|D_6| = 12$, and the order of a Sylow 2-subgroup of $D_6$ must be the highest order of $2$ that divides $12$ which is $4$. Also, the order of each element of the Sylow 2-subgroup must be a power of $2$.

Using this information I was able to come up with only this Sylow 2-subgroup $\{ e, s, r^3, sr^3 \}$. However, I do not know if there are more Sylow 2-subgroups of $D_6$.

Lastly, I figured out, by Sylow's Third Theorem, that the number of Sylow 2-subgroups of $D_6$ must divide the order of $D_6$, and also be congruent to 1 mod 2. The first part of this statement suggests that there can only be either $1,2,3,4,6$, or $12$ Sylow 2-subgorups of $D_6$, but I am unsure of what congruent to 1 mod 2 means. I know that 1 mod 2 $= 1$, but what does congruent mean?

PiccolMan
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    "congruent to 1 mod 2" is simply a fancy way of saying "an odd number". –  Dec 02 '16 at 18:28
  • So generally, congruent to 1 mod p means that the remainder must be 1 if divided by p? – PiccolMan Dec 02 '16 at 18:32
  • Yes, that's right. –  Dec 02 '16 at 18:33
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    This suggests that the number of Sylow 2-subgroups of $D_6$ are either 1 or 3. Is there a quick way to find the other 2, or prove that there is only 1? – PiccolMan Dec 02 '16 at 18:35
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    There are in fact 3 Sylow 2-subgroups. You have found one of them. Have you tried conjugating it by various elements of $D_6$? –  Dec 02 '16 at 18:37
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    Focus on the elements of the form $sr^k$. Note that $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = e$, so each of these elements has order $2$. And every element of order $2$ must live inside some Sylow $2$-subgroup. –  Dec 02 '16 at 18:44
  • Oh right, the second Sylow theorem says that each Sylow 2 subgroup is a conjugate of the other. Also, why must every element of order 2 live inside a sylow 2 subgroup? Can I say generally that every element of order p must live in a Sylow p - subgroup? – PiccolMan Dec 02 '16 at 18:55
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    Yes, every element of order $p$ must live in a Sylow $p$-subgroup. More generally, any subgroup of order $p^k$ for some power of $k$ must live inside a Sylow $p$-subgroup. Some authors include this with the Sylow theorems, e.g. Martin Isaacs calls it the "Sylow development theorem" - every $p$-subgroup can be "developed" into a Sylow $p$-subgroup. –  Dec 02 '16 at 18:57

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