TITLE: Alternative proofs that $A_5$ is simple
QUESTION [51 upvotes]: What different ways are there to prove that the group $A_5$ is simple?
I've collected these so far:

By directly working with the cycles: page 483 of http://www.math.uiowa.edu/~goodman/algebrabook.dir/algebrabook.html
Because it has order 60 and two distinct 5-Sylow groups: https://crypto.stanford.edu/pbc/notes/group/sylow.xhtml
There's no way to sum up the conjugacy classes 1, 15, 20, 12, 12, to get a normal subgroup.
By Iwasawa's lemma on $PSL_2(5) \simeq A_5$.
It's not solvable since it's perfect, and every group of order $<60$ is solvable, so it cant have normal subgroups.

REPLY [2 votes]: Viewing $A_5$ as the rotational symmetries of an icosahedron, one can easily verify that each conjugacy class generates the group. There are really only three cases, as while there are two conjugacy classes of five cycles, they are evidently power equivalent, so we lump them together. It then suffices to see that any vertex can be carried to any other by rotations through vertex axes, any face can be taken to any other by rotations through face axes, and any edge can be taken to any other by rotations through edge axes (as each respective case clearly has the relevant stabilizers, just transitivity of the action establishes that everything is generated.
But each of those cases is very straightforward: once you're familiar with it, there's no need for even a model!
Of course, you may want to prove that this group of rotations is $A_5$, but using this established simplicity and the action on the compound of five cubes (which I think is beautiful enough to warrant thinking about), that can be established without issue.