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

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?

• Does $S_{10}$ have an element of order 30? – Edward Evans Jun 13 '19 at 21:10
• I noticed it doesn't, but I don't know if $S_{10}$ is cyclic (is it?), so it doesn't cover all subgroups. – איתן לוי Jun 13 '19 at 21:16
• $S_{10}$ 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. – Dietrich Burde Jun 13 '19 at 21:43
• 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. – איתן לוי Jun 13 '19 at 21:45

Hint: What would the cycle type of a generator of such a subgroup be?

• Alright I think I understand, $\phi(1)$ would have to be the generator of H, correct? – איתן לוי Jun 13 '19 at 21:18
• What do you mean by $\phi$ and $1$ and $\phi(1)$? – Servaes Jun 13 '19 at 21:20
• If I define $\phi: \mathbb{Z}_{30} \rightarrow H$, then because $<1>=\mathbb{Z}_{30}$, $<\phi(1)>=H$ is necessary as well. – איתן לוי Jun 13 '19 at 21:23
• In that case yes, if $\phi$ is an isomorphism. – Servaes Jun 13 '19 at 21:24
• 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 $S_{10}$ – איתן לוי Jun 13 '19 at 21:43

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 $$30=\operatorname{lcm}(2,3,5)$$.