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Let $H_n = \{\alpha \in S_n:|\alpha| \text{ is odd}\}$. For which values of $n \ge 2$ is $H_n$ a subgroup of $S_n$?

Ok so I figure that since the order is odd, then $\alpha$ can be written as a product of disjoint cycles of odd length and therefore and be represented as an even number of $2$-cycles, and if $H_n$ is to be a subgroup, the inverse and product has to be represented as a product of two cycles as well. I'm having trouble making the connection. Any help would be appreciated.


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up vote 4 down vote accepted

A subgroup has to be closed. Can you think of two $3$-cycles which when multiplied together yield a permutation with even order?


Now think about why this works. Is there any smaller possible example?

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