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Does $S_{10}$ have a subgroup that is isomorphic to $\Bbb{Z}/30\Bbb{Z}$?

I tried to use the fact that if such subgroup $H$ exists, then $|H|=|\Bbb{Z}/30\Bbb{Z}|=30$, however I don't see why such subgroup can't exist.

Beyond that I really have no idea how to proceed. Can anyone give a hint?

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    $\begingroup$ Does $S_{10}$ have an element of order 30? $\endgroup$ – Edward Evans Jun 13 '19 at 21:10
  • $\begingroup$ I noticed it doesn't, but I don't know if $S_{10}$ is cyclic (is it?), so it doesn't cover all subgroups. $\endgroup$ – איתן לוי Jun 13 '19 at 21:16
  • $\begingroup$ $S_{10}$ a cyclic group? What is the element of order $10!=3628800$??? A cyclic group is abelian, but $S_n$ for $n\ge 3$ is never abelian. $\endgroup$ – Dietrich Burde Jun 13 '19 at 21:43
  • $\begingroup$ It is not a cyclic group. For some reason I thought for a second that $10!=1000$ and then it didn't seem that far-fetched. $\endgroup$ – איתן לוי Jun 13 '19 at 21:45
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Hint: What would the cycle type of a generator of such a subgroup be?

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  • $\begingroup$ Alright I think I understand, $\phi(1)$ would have to be the generator of H, correct? $\endgroup$ – איתן לוי Jun 13 '19 at 21:18
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    $\begingroup$ What do you mean by $\phi$ and $1$ and $\phi(1)$? $\endgroup$ – Servaes Jun 13 '19 at 21:20
  • $\begingroup$ If I define $\phi: \mathbb{Z}_{30} \rightarrow H$, then because $<1>=\mathbb{Z}_{30}$, $<\phi(1)>=H$ is necessary as well. $\endgroup$ – איתן לוי Jun 13 '19 at 21:23
  • $\begingroup$ In that case yes, if $\phi$ is an isomorphism. $\endgroup$ – Servaes Jun 13 '19 at 21:24
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    $\begingroup$ That connection didn't seem that simple to me when I asked the question, I didn't notice it. I prove this using the fact that $o(\sigma_1\sigma_2...\sigma_n)=lcm(k_1,...,k_n)$ when $o(\sigma_1)=k_1,o(\sigma_2)=k_2,...,o(\sigma_n)=k_n$ if those cycles are co-prime cycles (not sure this the correct terminology in English), and we can't find such cycles in $S_{10}$ $\endgroup$ – איתן לוי Jun 13 '19 at 21:43
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The question is equivalent to

Does $S_{10}$ have an element of order $30$ ?

Now, $30=2 \cdot 3 \cdot 5$ and $2+3+5 \le 10$, and so the answer is yes: just take a permutation with cycle structure $2-3-5$. The simplest one is $$ (1,2)(3,4,5)(6,7,8,9,10) $$ This permutation has order $30=\operatorname{lcm}(2,3,5)$.

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