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We see $S_3$ as the quotient of the free group on two elements and the normal subgroup $R$ generated by $\langle\sigma^3,\tau^2,\sigma\tau\sigma\tau\rangle$ where $\sigma$ and $\tau$ are the generators of the free group. The covering space corresponding to $R$ of the bouquet of 2 circle sould be the following:


Now $S_3$ acts on this covering space, and the action should have two orbits. Could you explain me how is this action? (I mean, what are the images of the single edges?)

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Let us write $t=(12)$ and $s=(123)$, two elements of the symmetric group $S_3$ of degree $3$.

Construct a directed graph $\Gamma$ as follows:

  • the vertices are the elements of $S_3$,

  • if $g\in S_3$ is a vertex, there are two edges coming out of $g$ in $\Gamma$: one going from $g$ to $gt$ and the other going from $g$ to $gs$.

In other words, the set of edges is $$E=\{(g,gt)\in S_3\times S_3:g\in S_3\}\cup \{(g,gs)\in S_3\times S_3:g\in S_3\}.$$

We can draw a picture:

enter image description here

There is an action of $S_3$ on $\Gamma$ as follows: if $h\in S_3$, then

  • the action of $h$ on the vertices of $\Gamma$ is by left multiplication by $h$: that is, a vertex $g\in S_3$ is mapped to $hg$;

  • on the other hand, the action of $h$ on the edges is the induced one: if $(g_1,g_2)$ is one of the edges, then $h\cdot(g_1,g_2)=(hg_1,hg_2)$. It is easy to see that this latter element is, indeed, an edge of $\Gamma$.

It is very easy to see that the action of $S_3$ on the vertices of $\Gamma$ is simply transitive, so that the quotient graph $\Gamma/S_3$ has exactly one vertex, and that the action of $S_3$ on the edges of $\Gamma$ has exactly two orbits. It thus follows that $\Gamma/S_3$ is a two-leaved rose.

$$♦ ♦ ♦$$

Can you see how to go from this action of $S_3$ on $\Gamma$ to an action of $S_3$ on a CW-complex of dimension $1$, which is what you want?

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It might help to remember that $S_3$ is the same thing as $D_3$. So let's look for a "triangle-ish" shape for $D_3$ to act on. As you've drawn your picture, you can almost see it: the upper and lower three-cycles form a pair of triangles; in this way the whole graph is best visualized as a sort of triangular prism. Now $\sigma$ will act as a rotation, and $\tau$ will swap the upper and lower triangles. From this way of looking at things, it's perhaps easier to see that $S_3$ acts transitively on triangle edges (the non-vertical edges in your picture) and also acts transitively on the vertical edges. So these are your two orbits.

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