I've been given the following question:
"Let $H=\{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4), (2 \ 3) \}$.
Show that $H$ is a subgroup of $S_{4}$"
Every element in this set is contained in $S_{4}$, so it is a subset. Now, we must show that $H$ is closed under the operation, that the elements of $H$ all have inverses contained in $S_{4}$, and that the operation is associative.
If I start by trying to show that $\forall x,y \in H, \quad xy \in H$, we find that $(1 \ 2)(3 \ 4) \circ (1 \ 3)(2 \ 4)=(1 \ 4)(2 \ 3)$. However, $(1 \ 4)(2 \ 3)$ is not in $H$, so it is not closed and therefore not a subgroup.
Is the error in the question, or in my workings?
Lagrange's theorem states that the order of the subgroup must divide the order of the group. We know that $|S_n|=n!$, so in our case does the order of the subgroup divide $4!$?