Perfect subgroups of $S^3$

Question

Are there any finite perfect subgroups of the sphere $S^3\cong SU(2)$ different from $SL_2(\mathbb{F}_5)$ (I consider the Lie group structure of $S^3$)?

Answer 1

No. I am not aware of any "insightful" proofs. I just go to the list here and abelianize all the group presentations.

For fun, here is the result of this calculation. Whenever I say "the abelianized presentation" I am giving a description of the group as a quotient of $\Bbb Z^n$ by some relations, as opposed to the free group $F_n$.

1) If $\pi_1$ is cyclic, then of course $H_1$ is the same cyclic group.

2) If $\pi_1 = \langle x,y \mid xyx^{-1} = y^{-1}, x^{2m} = y^n\rangle$ is the fundamental group of one of the 'prism manifolds', where $m$ and $n$ are coprime and $n \geq 2$, then $H_1 = \langle x, y \mid 2y = 0, 2mx = ny\rangle$. If $n$ is even, we obtain $\Bbb Z/2 \oplus \Bbb Z/2m$; if $n$ is odd we obtain $\Bbb Z/4m$.

3) In the tetrahedral case, $\pi_1 = \langle x, y, z \mid (xy)^2 = x^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3m} = 1\rangle$ where $m$ is a positive odd integer. The abelianized presentation is $$\langle x, y, z \mid 2x = 0, 2y = 0, x = y, x = 0, 3mz = 0\rangle;$$ clearly we may simplify this to $\Bbb Z/3m$, where $m$ is positive and odd.

4) The octahedral case is $\pi_1 = \langle x, y \mid (xy)^2 = x^3 = y^4\rangle$. The abelianized presentation is $$\langle x, y \mid 2x = 2y, x = 2y\rangle = \Bbb Z/2.$$

5) Finally, the icosahedral case is $\pi_1 = \langle x,y \mid (xy)^2 = x^3 = y^5\rangle$, and the abelianized presentation is $$\langle x, y \mid x = 2y \mid 2x = 5y\rangle$$ but because $(1,2)$ and $(2,5)$ generate $\Bbb Z^2$, this is the trivial group, as desired.

Coincidentally, this says that every spherical 3-manifold which is an integer homology sphere is either $S^3$ or the Poincare sphere $\Sigma(2,3,5)$.