I was recently reading an expository paper by Milnor and came across the following claim (in the footnote on the 2nd page):
If $I$ is the subgroup of $SO3$ consisting of rotational symmetries of the icosahedron, then $\pi_1(SO3/I)$ is a perfect group of order 120.
I'm having trouble seeing why the order of the group should be 120. My calculation below seems to suggest it's 60. Where have I gone wrong?
My calculation:
The group $I$ of rotational symmetries of the icosahedron is isomorphic to $A_5$.
Proposition 1.40 in Hatcher states:
If a group $G$ acts on a space $Y$ and satisfies (*), then $G \cong \pi_1(Y/G)/ p_*(\pi_1(Y))$ if $Y$ is path-connected and locally path connected. ($p:Y \rightarrow Y/G$ is the quotient map.)
Here (*) is
Every $y \in Y$ has a neighborhood $U_y$ for which $g_1(U_y) \cap g_2(U_y) \ne \emptyset$ implies that $g_1 \ne g_2$.
Since the action of $I$ on $SO3$ satisfies (*), we have
$T \cong \pi_1(SO3/I) / p_*(\pi_1(SO3))$.
It is well-known that $SO3$ is diffeomorphic to $RP^3$, so $\pi_1(SO3) \cong Z/2$. Hence to calculate $p_*(\pi_1(SO3))$ we only need to consider where a generator of $\pi_1(SO3)$ goes.
Consider a path $r_{\theta}$ in $SO3$ consisting of rotations around a fixed axis, beginning with the identity and increasing to a rotation of $\theta$. It is a short exercise to check that $r_{2\pi}$ is a generator of $\pi_1(SO3)$.
As such, when this generator is projected to the quotient $SO3/I$, the resulting path $r_{2\pi}'$ must have order dividing 2. It in fact has order 1, as can be seen by thinking about $r_{2\pi/5}'$. Since $r_{2\pi/5}$ is an element of $I \cong A_5$, it has order at most 6. If $r_{2\pi}$ had order 2, then $r_{2\pi/5}$ would have order 10, which cannot happen. Hence $r_{2\pi} = 1$, and so $p_*(\pi_1(SO3)) = \{ 1 \}$, meaning $I \cong \pi_1(SO3/I)/\{1\} \cong \pi_1(SO3/I)$.
But this implies that $|\pi_1(SO3/I)| = |I| = 60$.
Where have I gone wrong?
$r_{2\pi/5}$ is the path starting from identity and increasing to a rotation of angle $2\pi/5$. So it is a path in $SO3$, not an element of $SO3$ itself.
In fact, since $p$ is a covering projection, $p_*$ is injective and therefore the cardinality of $p_*(\pi_1 (Y))$ is the same as the cardinality of $\pi_1 (Y)$. The generator of $\pi_1 (SO3)$ therefore goes to some nontrivial element of $\pi_1 (SO3/I)$.