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I've been battling at this for a bit, $S_4$ isn't cyclic so that makes things a bit more difficult. I know by Lagrange a subgroup G with order 6 exists. I actually found some examples. Then found by research that each order 6 group is precisely the $S_3$. I'm failing to see why this is important, hence failing to prove the actual headline of this question. Any help is much appreciated.

Florian Suess
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    "I know by Lagrange a subgroup $G$ with order $6$ exists." - Lagrange doesn't allow you to conclude this. Instead, deduce the existence of a subgroup isomorphic to $S_3$ by considering the set of elements of $S_4$ that fix a particular point. –  Aug 23 '18 at 00:52
  • You're right, I just worked through the notes and I found that I was very accustomed to using the fundamental theorem of cyclic subgroups in conjunction with Lagrange. Thanks so much for pointing that out. – Florian Suess Aug 23 '18 at 01:25

1 Answers1

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There are two groups of order 6, up to isomorphism: $S_3$ and $\mathbb{Z}_6$.

Is it possible for a subgroup of $S_4$ to be isomorphic to $\mathbb{Z}_6$?

Hint: Think about the orders of elements of $S_4$.

Daniel Mroz
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    So if I pick the generated group <{$(23),(123)$}> for example, I built something of order 6 (lcm of cycle lengths) right? And it's cyclic and finite, hence isomorphic to ${\mathbb Z}_6$. But to generalise, if I choose any two elements in $S_4$ such that the cycle lengths of each element is 2 and 3 respectively, this is also isomorphic to ${\mathbb Z}_6$. Am I on the right track or not really? – Florian Suess Aug 23 '18 at 01:04
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    @Florian Suess Careful! The order of the element $(2 3)(1 2 3)$ is not 6. The order of an element of a symmetric group is the LCM of the cycle lengths of its disjoint cycle decomposition. Can you see how in fact no element of $S_4$ is of order 6? Therefore a subgroup of $S_4$ of order 6 can not be cyclic, and so it must be isomorphic to $S_3$. – Daniel Mroz Aug 23 '18 at 01:13
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    Oh... yes that's right, cyclic by definition is a group generated by a single element and lcm of the disjoint cycle decomposition determines the order. Indeed, there is no way for a subgroup of $S_4$ to be cyclic if it has order 6. There simply aren't big enough permutations. Now I'm puzzled on, how can we jump to the conclusion, given a subgroup of G such that it's order is 6, it must be isomorphic to $S_3$, is it the only non-cyclic group of order 6? – Florian Suess Aug 23 '18 at 01:23
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    Indeed! And yes, that’s it - a non cyclic group of order 6 is isomorphic to $S_3$. You can find a nice proof of that result here. – Daniel Mroz Aug 23 '18 at 01:28
  • You have been incredibly helpful, you're answer has been perfect and I'm sure if people see this post they'll be able to follow how to solve this question. I wish you the best. – Florian Suess Aug 23 '18 at 01:30