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Let $G=S_{15}$ (symmetric group on $15$ symbols). We say that any two subgroups $H$ and $K$ of $G$ are conjugate if $gHg^{-1}=K$ for some $g\in G$. This is an equivalence relation and the equivalence classes are known as conjugate classes. Find the number of conjugacy classes of cyclic subgroup of order $15$ in $S_{15}$.

I know that any two elements in $S_n$ are conjugate if they have the same cyclic structure. Is there any such result to determine the conjugacy classes for subgroups? Please help.

PAMG
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    No. I am looking for subgroups, not for elements. – PAMG Oct 07 '22 at 10:12
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    A cyclic group of order 15 is generated by an element of order 15, which is necessarily the product of a 3-cycle and a disjoint 5-cycle, or a 15-cycle. The two cases are necessarily non-conjugate (think about fixed points). Count the number of elements of each type, and then be careful about distinct elements yielding the same sungroup. – Arturo Magidin Oct 07 '22 at 18:46
  • @ArturoMagidin The number of $15$-cycles are $14!$ and $(3, 5)$-cycles are $\frac{15!}{5!\times 3!}$. How to think forward? – PAMG Oct 09 '22 at 04:28
  • How many 15-cycles generate the same cyclic group? how many 3-5 cycles? – Arturo Magidin Oct 09 '22 at 04:44
  • @ArturoMagidin This is given by Euler's function that is $\phi(15)=8$. – PAMG Oct 09 '22 at 05:03
  • So then you can count the number of subgroups. – Arturo Magidin Oct 09 '22 at 05:25
  • @ArturoMagidin yes. The number of subgroups because of $15$ cycles are $\frac{14!}{8}$ and that of $(3,5)$ cycles are $\frac{15!}{8!\times 3!\times 8}$. – PAMG Oct 09 '22 at 05:30
  • I gave you the hints. I don't need a blow-by-blow report of your progress. – Arturo Magidin Oct 09 '22 at 05:31
  • @ArturoMagidin I thought with my blow-by-blow report, you are taking me to the answer. Anyway, I thank you very much. – PAMG Oct 09 '22 at 05:37
  • You have the answer. And you aren't asking any questions. You are just pinging me to inform me of your progress. To what end? None. – Arturo Magidin Oct 09 '22 at 05:39
  • @ArturoMagidin I am sorry for pinging you many times. This is one last time. Can you please tell me how to proceed forward because I really don't understand how to get only $5$ conjugacy classes out of these many subgroups. – PAMG Oct 09 '22 at 06:03
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    Any two 15 cycles are conjugate, so the subgroups they generate are conjugate. Any rwo 3-5 cycles are conjugate, so the subgroups they generate are conjugate. Don't know where you get 5 conjugacy classes... – Arturo Magidin Oct 09 '22 at 06:06
  • @ArturoMagidin Is it that all the cycles $(15), (3, 5), (3, 3, 5), (3, 3, 3, 5), (3, 5, 5)$ (total $5$) generate cyclic subgroups of order $15$? Am I right? – PAMG Oct 09 '22 at 06:21

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