Why is $A_5$ a simple group?

Solution 1:

Q1: There are $\frac{5\cdot4\cdot 3}3=20$ three-cycles in $A_5$, and each is in a $3$-Sylow group, whereas each $3$-Sylow group consists of two elements of order three (plus the identity). Hence the number of $3$-Sylow groups is exactly $10$.

Likewise, each $5$-Sylow group contains four elements of order $5$ and there are $\frac{5!}5$ such elements.

Each choice of four numbers in $\{1,2,3,4,5\}$ gives rise to a copy of the Klein 4-group, hence there are $5$ such subgroups.

Q2: As $A_5$ acts transitively on the set of its $p$-Sylow groups by conjugation and a normal subgroup is invariant under conjugation, we have this all-or-nothing situation: If $N$ contains one $p$-Sylow group, it must contain all its conjugates.

Q3: If $\#N=2$ then the nontrivial element must be fix under conjugation, i.e. central. If $\#N=4$ it must be one of those Klein 4-groups and $\#N=8,16,\ldots$ is not possible

Q4: See Q2. Any subgroup is normal iff it is invariant under conjugation.

Q5 We have seen in Q2 that $N=A_5$ if $3|\#N$ or $5|\#N$, hence from $1<\#N|60$ we are left only with $\#N\in\{2,4\}$ to check. This was shown impossible in Q3, Q4.

Solution 2:

An answer to Question 1: Count the numbers of cycles of length 2, 3 and 5.

An answer to Question 2: Since all Sylow $p$-subgroups are conjuncted.

An answer to Question 3: If $|N|=4$ then $N$ coincides with a Sylow $2$-group; if $|N|=2$ then the non-identity element of $N$ is central.

An answer to Question 5: "Why does N have to be a p-Sylow group?" -- See Question 2.

Solution 3:

Another way to prove this is by looking at the rotational symmetries of an icosahedron (or dodecahedron) - this group is $A_5$. Perhaps there the question of when two elements are conjugate is more easily seen. Have a look at Artin's Algebra for a proof.