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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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No.

The identity is an even permutation.

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

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 $\sigma_s = \sigma_t$.

The identity element can be written as composition of two 2-cycles of the form (ab)(ab)=1 (where 1 is the identity permutation) proving that it is an even permutation. We know that every subgroup must contain the identity element so if H is consisting of all odd permutations then it will miss the identity permutation,i.e., 1. Hence, H could not be a subgroup.

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