Number of sylow 5 subgroups in $S_{7}$ is? $|S_{7}| =5040= 2^4 \times 3^2 \times 5\times 7$. number of sylow 5 subgroups is by sylow 3rd theorem n5=1+5k=possibilities are$\left\{1,6,16,21,36,56,126,336,1008\right\}$.we know that sylow subgroup exist of order 5 is cyclic . O(Sylow 5 subgroup)=5.Suppose we have 6 sylow subgroup then each sylow subgroup order 5 elements is 4 totally we have 24 elements of order 5 is it possible?
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This has been answered on many forums, e.g., here. Every such Sylow subgroup is cyclic, so it suffices to count the elements of order $5$. – Dietrich Burde Sep 14 '23 at 09:12
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In a p group number of p order elements is (p-1). So number of 5 order elements in cyclic group of order 5 is 4. – Rajagopal S Sep 14 '23 at 09:19
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Yes, but your list of possibilities is not correct, because $s_5\mid 2^4\cdot 3^2\cdot 7=1008$, so $s_5$ can be bigger than $56$. How many Sylow subgroups can you form from the $504$ elements of order $5$? – Dietrich Burde Sep 14 '23 at 09:27
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1008 =3(mod 5) but we need 1(mod 5) so 1008 not in the list.1+5k does not divide 1008 – Rajagopal S Sep 14 '23 at 09:32
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I didn't say $s_5=1008$. It is only a divisor. Your list ends with $56$. Why? – Dietrich Burde Sep 14 '23 at 09:36
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Yes 336 also in the list. Thanks – Rajagopal S Sep 14 '23 at 09:47
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Thanks i need one clarification. But sylow 2 subgroups exist of order 16 . but 16 order group need not be cyclic. Then what technique need to count number of sylow 2subgroup inS7 – Rajagopal S Sep 14 '23 at 10:00
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One step after the other. You still forgot some numbers in your list. For Sylow-$2$ subgroups, have a look at the methods, e.g., here. – Dietrich Burde Sep 14 '23 at 10:11
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So number of sylow 5 subgroups in S7=126 is it correct? – Rajagopal S Sep 14 '23 at 10:52
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$No.of Sylow-p~ subgroup~ in~ S_{n}=\frac{number ~ of p~ order ~element in~ S_{n}}{\phi(p)}$ – Rajagopal S Jan 26 '25 at 03:50