What is $S^3/\Gamma$?

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Let G is a group and H is a subgroup of G. I know $G/H$ is the quotient space but I have no idea about what $S^3/\Gamma$ is, where $S^3$ is the sphere and $\Gamma$ is a finite subgroup of $SO(4)$. In this case $S^3$ has not structure of a group and $\Gamma$ is not subgroup of the sphere. So, what is $S^3/\Gamma$?

Thanks for your help.

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Groups have applications in many areas of math because of the notion of a group action. For instance, a group $G$ can act on a set $X$ through a homomorphism:

$$\rho:G\to \text{Bij}(X)$$

Where $\text{Bij}(X)$ is the group of all set-map bijections of $X$, under composition. If we give these sets more structure, we can require that the group operation respect the structure, so we can have groups acting on other groups, rings, topological spaces, smooth manifolds, etc.

If $M$ is a manifold, and $G$ is a group, we can define a group action of $G$ on $M$ by defining a group homomorphism $\rho:G\to \text{Diff}(M)$, where $\text{Diff}(M)$ is the group of all diffeomorphisms of $M$ under compositions. In a sense, every group element represents a diffeomorphism of $M$ in a way that is consistent with the group law (the diffeomorphism represented by $gh$ is the composition of the diffeomorphisms represented by $g$ and $h$, and so on). For notational purposes, we often write $g\cdot x$ for $\rho(g)(x)$, where $x\in M$.

In this case, we can speak of the quotient space $M/G$, which is the set of equivalence classes under the relation $x\sim y$ if and only if there is some $g\in G$ with $g\cdot x=y$. In other words, the equivalence classes are the orbits of the group action. Now, if the action happens to be sufficiently nice, we can actually place a manifold structure on the set of equivalence classes $M/G$, giving a new manifold, called the quotient manifold of $M$ by $G$.

A brief example. The group of $n$th roots of unity, $\mu_n$ acts on $S^1$ via multiplication, if we view $S^1$ as a subset of the complex plane. In that case, each orbit has a unique representative whose argument lies between $0$ and $2\pi/n$. It turns out that the quotient manifold structure is diffeomorphic to the original $S^1$, under the map $\overline{z}\mapsto z^n$.

In your case, we have $S^3$ being acted on by a subgroup $\Gamma$ of $SO(4)$, and we are considering the quotient manifold $S^3/\Gamma$.

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$\mathrm{SO}(4)$ acts on $S^3$ in the obvious way if we think of $S^3$ as the set of unit vectors in $\Bbb R^4$. Then any subgroup $\Gamma \leq G$ acts on $S^3$ as well. $S^3/\Gamma$ is the the orbit space of this $\Gamma$-action, i.e. $$S^3/\Gamma = S^3/\!\sim$$ where $$x \sim y \quad \iff \quad x = g \cdot y \text{ for some } g \in \Gamma.$$