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The Question: Prove that (1 2) cannot be written as the product of 3 disjoint cycles.

The Attempt: Suppose (1 2) has a cycle decomposition into 3 disjoint cycles $m_1, m_2$, and $m_3$. Then (1 2) = $m_1 m_2 m_3$ and the order of (1 2) should be the least common multiple of $m_1, m_2,$ and $ m_3$.

Where do I go from here?

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  • $\begingroup$ Do you mean the permutation $(1\;2)$ as an element of $S_2$? $\endgroup$ – J.R. Oct 17 '12 at 18:18
  • $\begingroup$ @IHaveAStupidQuestion - if it is of $S_2$ this becomes a trivial question. I believe the intention is that it is an element of $S_n$ $\endgroup$ – Belgi Oct 17 '12 at 18:21
  • $\begingroup$ @IHaveAStupidQuestion It's for S_n. $\endgroup$ – user39794 Oct 17 '12 at 18:25
  • $\begingroup$ @Belgi thats why I asked :-D. $\endgroup$ – J.R. Oct 17 '12 at 18:33
  • $\begingroup$ For large $n$ consider what the cycle form is. It means things inside each cycle get moved... $\endgroup$ – coffeemath Oct 17 '12 at 18:36
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If $m_1, m_2, m_3$ are disjoint cycles of length $l_1, l_2, l_3$ respectively, then $m_am_2m_3(x)\ne x$ for all $l_1+l_2+l_3$ elements occuring in any of the cycles. From $l_i\ge2$, we conclude that $m_1m_2m_3(x)\ne x$ for at least $6$ elements, whereas $(1\, 2)$ permutes only $2$ elements.

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