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We have the group $(S_3, \circ)$, where $S_3=\{(1)(2)(3),(1)(2,3),(1,2)(3),(1,2,3),(1,3,2),(1,3)(2)\}$. How can I find all cyclic subgroups from this?

I know that I could go through each element and exponentiate it to see if I get all other terms. However, I would guess there is a more efficient way of solving this problem.

Shaun
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Ron Snow
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    Welcome to Mathematics Stack Exchange. There are only $6$ elements, so it's not much work to go through them. Do you know how to tell the order of an element of $S_3$? Do you know that the order of an element must divide the order of the group (which is $6$ in this case, not leaving many possibilities for the order of an element)? – J. W. Tanner Sep 24 '19 at 03:26
  • I do not, and thank you for the welcome! – Ron Snow Sep 24 '19 at 03:28
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    The order of a cycle is equal to its length – J. W. Tanner Sep 24 '19 at 03:30
  • So I just completed all of the calculations. I got 5 cyclic subgroups (including the trivial cyclic subgroup). 3 of these are of order 2. One is of order 3. So, I see what you are saying in that the order of one of the cyclic subgroups must divide 6. However, are there any further patterns I can draw from this? – Ron Snow Sep 24 '19 at 03:38
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    Good job. You are correct that $S_3$ has five cyclic subgroups: one of order $1$, three of order $2$, and one of order $3$. (The only other subgroup is the entire group, which is not cyclic.) – J. W. Tanner Sep 24 '19 at 03:41
  • Would there be an efficient way to show that the entire group is not cyclic, rather than searching for a generator? – Ron Snow Sep 24 '19 at 03:54
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    For one thing, $S_3$ is not Abelian, and cyclic groups are Abelian – J. W. Tanner Sep 24 '19 at 03:56
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    Cf. this question, which asks about all subgroups of $S_3,$ not just cyclic ones – J. W. Tanner Sep 24 '19 at 04:03
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    Note: $S_3$ is not Abelian because $(1,2)(1,3)\ne(1,3)(1,2)$ – J. W. Tanner Sep 24 '19 at 04:04
  • Thank you for your extensive help, J.W. Tanner. – Ron Snow Sep 24 '19 at 04:29

1 Answers1

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The order of an element must divide the order of the group. The order of $S_3$ is $6$, and $S_3$ is not cyclic; that leaves $1, 2, $ and $3$ as possible orders for elements of $S_3$. The cyclic group of order $1$ has just the identity element, which you designated $(1)(2)(3)$. The order of an element in a symmetric group is the least common multiple of the lengths of the cycles in its cycle decomposition. Therefore, the orders of $(1)(2,3), (1,2)(3), $ and $(1,3)(2)$ are $2$, and the orders of $(1,2,3)$ and $(1,3,2)$ are $3$. [The elements of order $2$ are self-inverses. $(1,2,3)$ and $(1,3,2)$ are inverses of each other.] $S_3$ is a good example group to study, because it is the smallest non-Abelian group.

J. W. Tanner
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