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I would like to know if my proof below is correct. I do not have issues proving that $S_{\Omega}$ is a group; what I am not sure is whether my proof that $\vert S_{\Omega} \vert = \infty$ is correct.

Problem Prove that if $\Omega = \{1,2,3,\ldots\}$, then $S_{\Omega}$ is an infinite group.

Solution: First it is easy to show that it is a group by checking the definition. The only slightly non-trivial part is $\vert S_{\Omega} \vert = \infty$. Given any $n$, we have the element $(1,2,\ldots,n)$ is of order $n$, i.e., the group generated by $(1,2,\ldots,n)$ has $n$ distinct elements. We also have that the order of the group is greater than or equal to the order of the group generated by any element, i.e., all these elements are also elements of $S_{\Omega}$. Hence, we have $\vert S_{\Omega} \vert \geq n$ for all $n \in \mathbb{Z}^+$. Hence, $\vert S_{\Omega} \vert = \infty$.

Thanks

John Smith
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    You have no need to use the fact that a group is at least larger than the order of the subgroup generated by any of its elements: you have exhibited infinitely many distinct elements so the group is infinite! – Mariano Suárez-Álvarez Dec 26 '13 at 18:24
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    For extra fun, you could try figuring out exactly what cardinal your group has. – Mariano Suárez-Álvarez Dec 26 '13 at 18:24
  • "We also have that the order of the group is greater than the order of the group generated by any element" - to be precise it's better to write greater or equal to instead of greater. – sdcvvc Dec 26 '13 at 18:25
  • So Marijano you are saying that for any $n$, we have $(1),(1,2),(1,2,3),\ldots,(1,2,\ldots,n)$ belong to $S_{\Omega}$ and hence $\vert S_{\Omega} \vert \geq n$? – John Smith Dec 26 '13 at 18:27
  • No, I am saying that the elements $(1,2,3,\dots,n)$ for positive integers $n$ are all different and infinitely many, so thet your group has to be infinite. – Mariano Suárez-Álvarez Dec 27 '13 at 05:32
  • This question was listed among related questions: http://math.stackexchange.com/questions/73566/if-omega-1-2-3-cdots-then-s-omega-is-an-infinite-group (And you can also find a few questions about the cardinality of $S_\Omega$.) – Martin Sleziak Dec 27 '13 at 07:27
  • @MarianoSuárez-Alvarez I am saying the same thing. Also, the cardinality of $S_n$ is $n!$, so heuristically the cardinality of $S_{\Omega}$ should be same as the cardinality of reals. Am I right? – John Smith Dec 27 '13 at 16:16
  • You are not saying the same thing, really. – Mariano Suárez-Álvarez Dec 27 '13 at 17:33
  • As for your "heuristic", I can't make sense of it. – Mariano Suárez-Álvarez Dec 27 '13 at 17:34
  • @MarianoSuárez-Alvarez Why am I not saying the same thing? If I understand you right, you are saying $(1,2) \in S_{\Omega}$, $(1,2,3) \in S_{\Omega}$ and so on? Also, is the cardinality same as reals? If so, then my heuristic is right. – John Smith Dec 27 '13 at 17:38
  • You can reach a correct conclusion by a faulty argument... – Mariano Suárez-Álvarez Dec 27 '13 at 17:42
  • @MarianoSuárez-Alvarez You are putting words in my mouth. What I said was not a proof or a rigorous argument, that's why I called it a heuristic. In mathematics, you need to have some sort of heuristic/guess first before going ahead and proving anything. Also, what about my first question? – John Smith Dec 27 '13 at 17:45

3 Answers3

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Yes, your argument looks fine.

For a different, perhaps easier way to proceed, just think about

$$(12), (13), (14), ...$$

as elements of the group. How many such elements are there?

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A fine argument. Another, more explicit one is noting that the transpositions $(2i-1\ 2i)$ are all in your group.

Igor Rivin
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Why not take the permutation that increases every odd number by $2$, decreases every even number by $2$, except that it sends $2$ to $1$? Obviously of infinite order.

Lubin
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