I have two questions.

  1. What is the quickest way to see from scratch that $\text{SL}_2(5)/\{\pm I\}$ is isomorphic to the alternating group $A_5$?
  2. Does $\text{SL}_2(5)$ have any subgroups isomorphic to $A_5$?
  • 2
    $\begingroup$ For 2, it would be an index-two subgroup, so the kernel of a morphism to $\{\pm 1\}$, but $SL_2(5)$ is perfect, so that can't happen. For 1, I would look at the action on the projective line: that embeds $PSL_2(5)$ as an index-six subgroup of $A_6$, and then you have to show that $A_5$ is the only possibility (up to isomorphism, not up to conjugacy). $\endgroup$ – PseudoNeo Jul 13 '16 at 22:52
  • $\begingroup$ Can you show that $SL_2(5)$ has five Sylow 2-subgroups? Conjugation action on those then gives a homomorphism $\phi:SL_2(5)\to S_5$. Obviously the center will be in the kernel. It's probably not too hard to show that the image is contained in $A_5$.... $\endgroup$ – Jyrki Lahtonen Jul 13 '16 at 22:53
  • 1
    $\begingroup$ "not too hard" = obvious, once you know that $PSL_2(5)$ is simple (the sign morphism $PSL_2(5) \to S_5 \to \{\pm 1\}$ has to be trivial) $\endgroup$ – PseudoNeo Jul 13 '16 at 22:54
  • 1
    $\begingroup$ Yup. @PseudoNeo I wasn't sure whether we can assume simplicity of $PSL_2(5)$. :-) $\endgroup$ – Jyrki Lahtonen Jul 13 '16 at 22:56
  • $\begingroup$ Also here. $\endgroup$ – Jyrki Lahtonen Jul 13 '16 at 23:11