Here, $\sigma \in S_n$. The backwards (<=) direction is fairly straightforward. Assume $\sigma$ is a product of disjoint 2-cycles. Then the order of $\sigma$ is simply lcm$(2,2,2,...) = 2$. Conversely, if we assume $\sigma$ has order 2, I'm not sure how to show that it necessarily follows that $\sigma$ is a product of disjoint 2-cycles. However, I do think that it does have something to do with the fact that 2 is prime.
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$\begingroup$ Possible duplicate of A permutation is a product of disjoint transpositions iff its order is $1$ or $2$ $\endgroup$ – user198044 Oct 10 '18 at 3:41
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Hint: Every permutation can be written as a product of disjoint cycles. What happens if one of these cycles has length $>2$?
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$\begingroup$ Do we once again use the fact that the lcm of the lengths of the disjoint cycles is the order of $\sigma$? If any cycle has length $> 2$, then the lcm of the cycle lengths will always be $> 2$. Is this a contradiction? If we extend this to order 3, we would have to show the same for cycles of length $> 3$ and $2$. But since 3 is prime, the lcm of the lengths of non-3-cycles will never be 3. I think this makes sense, but I could be missing something. $\endgroup$ – playitright Nov 1 '16 at 8:59
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Suppose $\sigma$ has order 2. Then $\sigma^2$ is the identity permutation. Suppose to the contrary in the disjoint cycle representation there exists a cycle of length $k>2$, say $(a_1, a_2, \dots, a_k)$. Then $\sigma^2$ maps $a_1$ to $a_3 \neq a_1$, so $\sigma^2 \neq e$, where $e$ represents the identity permutation. Contradiction.
