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In permutation group $S_n (n \geq 5)$, if $H$ is the smallest subgroup containing all $3$-cycles, then which one of the following is true?

(i) $H=S_n$,
(ii) $H=A_n$,
(iii) $H$ is Abelian,
(iv) $|H|=2$

In the above problem, if $H=S_n$, then $H$ is definitely not smallest(, Although I don't have a clear inductive proof for the same). If $H=A_n$ then I think, this might be an option, but I don't have any solid evidence to conclude it. $H$ need not be Abelian, that's a conclusion we can say for sure by trying out various examples. Any hint(s) will be highly appreciated.

I found this post the closest related to this topic Show that if $K$ is a normal subgroup of $H$ such that $H/K$ is abelian, then K contains all of the 3-cycles., but it's quite different as I want to prove that "In permutation group $S_n (n \geq 5)$, if $H$ is the smallest subgroup containing all $3\text{-cycles}$, $\color{green}{\text{a possible conclusion holds}}$, but this question provides two informations and then asks to prove the subgroup contains all $3\text{-cycles}$. I think there might be numerous post associated with the same topic, but unfortunately I couldn't find any...

Arthur
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Xander Henderson Feb 02 '23 at 01:21

1 Answers1

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A sequence of hints, which you will hopefully be able to flesh into a full argument:

  1. Every product of 3-cycles is an even permutation. Can you find an element of $S_n$ which is not a product of 3-cycles?
  2. Any group containing all 3-cycles contains the alternating group. Which subgroups of $S_n$ contain $A_n$?
  3. Write down an example of two 3-cycles which don't commute.
  4. We said earlier that $H$ contains $A_n$. How big is $A_n$?
preferred_anon
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  • How can we find such an element in $S_n$ which is not a product of $3$ cycles as no information is given about $n$ apart from $n\geq 5$ ? How can we search in $S_n$ when we don't know what $n$ is? – Arthur Jan 30 '23 at 11:55
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    @Franklin Since every product of 3-cycles is an even permutation, what sort of permutations are not products of 3-cycles? – preferred_anon Jan 30 '23 at 12:02
  • The odd permutations are not product of $3$ cycles. As for your second hint, I dont understand which subgroups of $S_n$ contains $A_n$? I have no idea about it... – Arthur Jan 31 '23 at 12:44
  • @Franklin So if you can find an odd permutation in $S_n$, then that won't be in $H$. $A_n$ is exactly half of $S_n$ (since there are an equal number of odd and even permutations). Since $A_n \le H \le S_n$, $|A_n| \le |H| \le |S_n| = 2|A_n|$, and there are no divisors of $x$ between $x$ and $x/2$. Hence $H$ is either $A_n$ or $S_n$. – preferred_anon Jan 31 '23 at 14:04