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Consider the symmetric group of$S_{20}$ and it's subgroup $A_{20}$ consisting of all even permutations. Let $H$ be a $7$-Sylow subgroup of$A_{20}$. Is $H$ cyclic? And is correct the statement which says that any $7$-Sylow subgroup of $S_{20}$ is subset of $A_{20}$?

I know that order of H is 49 and H is not normal subgroup of $A_{20}$. But I don't understand whether it is cyclic or not.

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  • $\begingroup$ If $H$ were cyclic, then there would be an element of order $49$ in $H \subset A_{20}$. Is that possible? $\endgroup$
    – user348338
    Jun 27 '16 at 9:57
  • $\begingroup$ yes there is a element of order 49 in $A_{20}$ $\endgroup$ Jun 27 '16 at 10:01
  • $\begingroup$ I know $S_3,S_4,S_5$ $\endgroup$ Jun 27 '16 at 10:10
  • $\begingroup$ In $S_3$ orders of elements are 1,2,3 $\endgroup$ Jun 27 '16 at 10:12
  • $\begingroup$ But we can write 20 as 7+7+1+1+1+1 +1+1 $\endgroup$ Jun 27 '16 at 10:13
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In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$.

Group of order $7^2$ is either cyclic or isomorphic to $Z_7\times Z_7$.

If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is possible? You will reach to your answer.

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  • $\begingroup$ yes it has a element of order 49 but what about the statement any 7-Sylow subgroup of $S_{20}$ is subset of$ A_{20}$ $\endgroup$ Jun 27 '16 at 10:07
  • $\begingroup$ No! It has no element of order $49$! Consider small order groups $S_4$ and $S_5$. Can you write down orders of elements here? What standard results you are familiar with about permutations and their orders? For example, what is order of $(12)$? of $(123)$? of $(12)(345)$? $\endgroup$
    – p Groups
    Jun 27 '16 at 10:10
  • $\begingroup$ Hmm, can you say what the element of order 49 you are thinking of ? $\endgroup$
    – user348338
    Jun 27 '16 at 10:10
  • $\begingroup$ (1234567)(891011121314) $\endgroup$ Jun 27 '16 at 10:21
  • $\begingroup$ The element $(1234567)(891011121314)$ doesn't have order 49... Maybe thinking smaller might help, what is the order of $(12)(34)$ ? $\endgroup$
    – user348338
    Jun 27 '16 at 10:25
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For the last option Sn has two types of subgroups: a subgroup which contains half even and half odd permutations or it contains only even permutations.

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