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
issolvable, so it cant have normal subgroups.

**Answer**

I'm very happy to show this proof, which makes use of a technique frequently employed in a recent paper of mine. It's my favorite proof that A_5 is not solvable, which as you pointed out in your last bullet proves that A_5 is simple.

Definition.For n\in \mathbb{N}, denote by \pi(n) the set of prime divisors of n. Theprime graphof a finite group G, denoted \Gamma_G, is the graph with vertex set \pi(|G|) with an edge between primes p and q if and only if there is an element of order pq in G.

By Lucido (1999), Prop. 1, the complement of the prime graph of a solvable group G is triangle-free. It is obvious from cycle types that A_5 contains no elements of order 6,10, or 15, so \Gamma_{A_5} is the empty graph on three vertices. Therefore, A_5 is not solvable.

**Attribution***Source : Link , Question Author : Community , Answer Author : Alexander Gruber*