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I want to show that the set of odd permutations is not a subgroup of Sn. Let the set $H= \{\text{odd permutations}\}$.

Is it enough to say that the Identity permutation which sends all of its elements on themselves is even therefore, $e \notin H$ therefore $H$ is not a subgroup of $S_n$.

Bernard
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Sylvester Stallone
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1 Answers1

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Yes, that's enough. You can also say that the product of any two odd permutations is even, and therefore not an element of $H$. This shows that $H$ isn't closed (as long as $H$ has at least one element in it). Either argument works just fine as a proof.

Arthur
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