Sylow's theorem

If one look at Lagrange's theorem naturally the question arises: "Are there subgroups for any divisor of the group order ?"
The answer is:"Maybe." Not good, but Sylow's theorems are helping here a little.

Sylow 1: Let G be a finite group. For any prime p and non negative integer k with p^k divides the order of G, there exists at least one subgroup with order p^k.

Sylow 2: Let G be a finite group. A maximal subgroup with order p^k, p prime, k non negative integer, is called a Sylow p-subgroup. A p-group, p prime, is a group, whose elements have order p^k, k non negative integer.
The following statements hold:

1. Two Sylow p-subgroups P,Q for the same p are conjugate. (P = gQg^-1 for a g of G)
2. Any p-group contained in G is contained in a Sylow p-group
3. The number of Sylow p-subgroups (for the same p) divides the index of a Sylow p-subgroup and equals 1 modulo p

It flollows that all Sylow p subgroups for the same p have the same order.
The first and the third statement of Sylow 2 can be used to check if a group is simple. (Usually proving that a group is non simple is easier here: you just proof that there is only a single Sylow p subgroup for a p via 3. From 1 follows that this subgroup is normal)
Note that there are alternative definitions of Sylow p subgroup (It is sufficient to demand a maximal p-group), but they are all equivalent for finite groups.
For the proof see proof of Sylow's Theorem

Example: The alternating group A₅ is simple.
Let's write Sp for the number of Sylow p-subgroups of A₅.
|A₅| = 3 * 4 * 5 = 60. By Sylow 2.3. we know that Sp must divides the index of a Sylow p-subgroup. Therefore we have the possiblities:
S2: 1 3 5 15
S3: 1 2 4 5 10 20
S5: 1 2 3 4 6 12
By Sylow 2.3 Sp = 1 modulo p, this reduces the possibilities to:
S2: 1 3 5 15 nothing is happing here, that's a usual problem with Sylow 2-subgroups
S3: 1 4 10
S5: 1 6
Any Sylow 5-subgroup has order 5 and there are more than 4 5-cycles, thus we have s5=6.
The intersection of 2 Sylow 5-subgroups is trivial (= 1), because they are cyclic group of prime order. (Any non 1 element of a cyclic group of prime order generates the group, therefore 2 of these groups with non trivial intersection must be identical)
Any normal subgroup must contain all Sylow p-subgroups for a p or none. This follows from Sylow 2.1 and the fact that a normal subgroup has no conjugate subgroups.
Any normal subgroup containing an element of order 3 or 5 must contain all Sylow 3- or Sylow 5-subgroups, because the element generates a Sylow 3- or 5- subgroups.
If a normal subgroup contains all Sylow 5-subgroups, it has at least 4*6 + 1 = 25 Elements.
The divisors of 60 are: 60, 30, 20, 15, 12,5,4,3,2,1.
Due to Lagrange's Theorem the order of the normal subgroup must divide 60 and we get an order of 30 or 60.
60 is pointless here, so let's examine 30.
By Sylow 1 the normal subgroup has a Sylow 3-subgroup, contains an element of order 3 and thus contains all Sylow 3-subgroups.
Then it contains all 3-cycles and is equal to A₅ which is a contradiction to the order 30. (The A_n is generated by the 3-cycles, I'm not going to prove that here.)
By the same argument we can rule out that the normal subgroup contains any elements of order 3.
By Sylow 1 it follows that the normal subgroup must have a order of 2ⁿ, otherwise there would be a Sylow 3- or Sylow 5-subgroup. This leaves 2 and 4 as group order. Group order 4 would imply that there is only 1 Sylow 2-subgroup. Therefore there would be only 3 elements of order 2ⁿ and there are more than 3 such elements. (e.g. (12)(34),(12)(35),(12)(45),(13)(25) )
This leaves 2 as the order of the normal subgroup, by this it equals {1,a}, where a is an element of order 2. a can be represented a the product of disjoint cycles. (This is possible for any permutation) Let a be = bc...x, where b,c,...,x are cycles. We have:
1 = a a = bc...x bc...x = b b c c ... x x (cycles disjoint !)
If b,c,...,x are disjoint then bb,cc,...,xx are disjoint. Therefore bb=1, cc=1 ,...,xx=1. It follows, that the cycles must be 2-cycles = transpositions (the cycle length is the order of a cycle). Because a is element of A_n the number of the transpositions must be even. Because the transpositions are disjoint a must be the product of 2 transpositions (4 numbers get used up for 2 transpositions -> only 1 left, we are in A₅ !)
Let a be = (b c)(d e). Then (b c)(e f) is conjugate to a with f= {1,2,3,4,5}\{b,c,d,e} via (d e)(e f)=(d e f).
Therefore {1,a} is not normal in A₅.

Note: The structure of A₅ was used here, Sylow and the group order won't be sufficient.
It's an interesting question, if the facts used about the group structure already define A₅, making the whole proof pointless. (Finite calculation would do the same.)