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So I am aware a subgroup $H \subset S_N$ can consist of only even permutations (i.e. taking the set of all even permutations will produce a normal subgroup), but can it consist of only odd permutations?

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Is the identity (neutral element of $S_N$) an odd permutation? – Hans Giebenrath Dec 16 '12 at 7:46
up vote 2 down vote accepted


The identity is an even permutation.

In particular, any subgroup must contain the identity, and thus has at least one even permutation.

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Suppose $H \leq S_n$ is a subgroup. Then either every element of $H$ is even or half of the elements of $H$ are odd.

Let $\sigma_1, \ldots, \sigma_k$ be the even elements in $H$. If this list contains every element of $H$, then every element of $H$ is even. Suppose then that $H$ contains an odd element $\sigma$. I claim that then $\sigma \sigma_1, \ldots, \sigma \sigma_k$ are all the odd elements of $H$. For if $\tau \in H$ is odd, then $\sigma^{-1} \tau \in H$ is even. Thus $\sigma^{-1} \tau = \sigma_i$ for some $i$ and $\tau = \sigma \sigma_i$. Furthermore, the list $\sigma \sigma_1, \ldots, \sigma \sigma_k$ contains no repetitions since $\sigma \sigma_s = \sigma \sigma_t$ implies $\sigma_s = \sigma_t$.

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Put another way: sign is a homomorphism from $S_n$ to the two element group $\{-1,1\}$. For any group homomorphism $\phi: G \to H$ and any $y \in \phi(G)$, multiplication by $x \in G$ takes $\phi^{-1}(e)$ to $\phi^{-1}(\phi(x))$, so the inverse images of all members of $\phi(G)$ have equal cardinalities. – Robert Israel Dec 16 '12 at 8:28
Or $[H : H \cap A_n] = 2$ when $H$ contains an odd element, which is basically what I prove in my answer (by showing that $H - H \cap A_n$ is a coset). – Mikko Korhonen Dec 16 '12 at 9:55

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