TITLE: positions of a methane molecule with carbon atom at the origin
QUESTION [10 upvotes]: Let $\text{CH}_4$ be the molecule of Methane:

The four hydrogen atoms form vertices of a regular tetrahedron with the carbon atom in the center of the regular tetrahedron.
Here we regard all atoms to be points in $\mathbb{R}^3$ with volume zero (or we can consider the centers of these atoms, which are points with volume zero). We want to embed the $\text{CH}_4$ molecule into $\mathbb{R}^3$ such that the $\text{C}$-atom is at the origin $0$.

The collection of all such embeddings of $\text{CH}_4$ molecule into $\mathbb{R}^3$ form a manifold $G$. In fact, 
$$
G=O(3)/{\frak S}_4=SO(3)/{\frak A}_4,
$$
where ${\frak S}_4$ is the symmetric group and ${\frak A}_4$ the alternating group. Here we regard $O(3)$ as the isometric embedding Lie group of the $\text{CH}_4$ molecule with the four hydrogen atoms distinct, labelled by $1,2,3,4$. Then we quotient the permutation action of $S_4$ on the four hydrogen atoms, regarding all the four hydrogen atoms the same and do not distinguish them.
Question: as a manifold, what is the cohomology ring (with cup product)
$$
H^*(G;\mathbb{Z}_2)
$$
and the Steenrod square $Sq$'s acting on the cohomology ring?

A contradiction with the following answer by W. Schlieper:
We consider a vector bundle
$$
\xi: \mathbb{R}^4\to O(3)\times_{S_4}\mathbb{R}^4\to O(3)/S_4
$$
where $S_4$ permutes the order of coordinates in $\mathbb{R}^4$. The action of $S_4$ on $\mathbb{R}^4$ may have determinant $-1$, hence it is not contained in $SO(4)$, but only in $O(4)$. Therefore, the first Stiefel-Whitney class $w_1(\xi)\neq 0$. However, $w_1(\xi)\in H^1(O(3)/S_4;\mathbb{Z}/2)$. This contradicts your result
$$
H^1(O(3)/S_4,\mathbb{Z}/3) \simeq H^2(O(3)/S_4,\mathbb{Z}/3) \simeq \mathbb{Z}/3.
$$
Why we have such a contradiction?

REPLY [8 votes]: Over $\mathbb{Z}/2$ it turns out to be pretty boring.  Your space $G$ (which you really shouldn't call that, as it's merely a manifold, not a group), which you define as a quotient of $O(3)$ or equivalently as one of $SO(3)$, can be thought of as the quotient of this group's double cover $SU(2) \cong S^3$ by the binary tetrahedral group $2T$, which turns out to be your fundamental group as $S^3$ is simply connected.
The first homology group over $\mathbb{Z}$ will be $2T/[2T,2T]$, but $[2T,2T]$ is the quaternion group $Q_8$, and $2T/Q_8 \simeq \mathbb{Z}/3$, so $H_1(G,\mathbb{Z}/2)=H^1(G,\mathbb{Z}/2)=0$ and by Poincaré duality, $H^2(G,\mathbb{Z}/2)=0$ will have the same values as that of the $3$-sphere.
Over $\mathbb{Z}/3$ you'll get $H^1(G,\mathbb{Z}/3) \simeq H^2(G,\mathbb{Z}/3) \simeq \mathbb{Z}/3$ and a nontrivial Bockstein, though.