Simple Groups

Since every subgroup of any Abelian group is normal, it occurs there is always a decomposition series of the group into prime order cyclic factors. Every cyclic prime order abelian group is therefore both "simple" and "soluble", with only one simple subgroup, {e}. Cyclic (sub)groups of order p^n (a power of a prime) have n prime (p) order factors and are therefore "soluble" (but not simple) since their subgroups are all normal. The fundamental theorem of finitely generated abelian groups states that every abelian group is formed in a fashion that is isomorphic to a cartesian product of prime-power order cyclic groups. (With addition done component wise) to the tune of;

(a+b, c+d, e+f ...) = (a+b = x mod pⁿ, c+d = y mod q^m, e+f = z mod r^s,... ) etc.

Then each component of the cartesian product factors into a set of prime order cyclic simple groups, or {e} in the trivial case.

However in the non-abelian case things are much more complicated. In particular we need to show that A5 is a simple group. That its only factors are {e} and itself.

There is a not too difficult proof that A5 is simple. First, it is a normal subgroup of S5 (readily checked, as each conjugation gHg^-1 adds an even number of transpositions, and is therefore also in A5.) Also important to note that every element of S5 not in A5 may be found in the coset of A5 in S5, i.e. g = xH where H = A5, and x is some transposition. Then {e, x} is also a subgroup of S5 and is a cyclic prime order group which itself is simple and soluble. Sn for n>4 has only three normal subgroups Sn, An and {e}.

Sn has (n!) = n(n-1)(n-2)(n-3)...(2)(1) elements, so An has n!/2 elements in every case by the "Theorem of Lagrange".

Next we need to show that An is generated by three cycles. We begin with a product of transpositions X of length 2k and induce on k.

Note that every 3-cycle is a product of an even number of transpositions so every three cycle generates a subgroup of A5.

Inducing on the number of transpositions in A5 ( on k in pairs as 2k)

if {a,b} = {c,d} then (a,b)(c,d) = e which is fine eX = X.

Now suppose {a,b}{c,d} have one common member (a,b)(b,c) = (a,b,c)

So (a,b,c)X is generated by 3 cycles by induction as X is so generated by three cycles by the induction hypothesis.

suppose {a,b}{c,d} are disjoint

(a,b)(c,d) = (a,b,c)(b,c,d) So (a,b,c)(b,c,d)X is also generated by three cycles.

The induction is correct, 'An' for n>3 is generated by three cycles.

Suppose that H is a normal subgroup of An, n > 4 and H contains one three cycle. We prove that H contains every three cycle in An.

let (a,b,c) be in H. Then as n>4 there is a three cycle (a,d,e) so that (a,d,e)(a,b,c)(a,e,d) is in H. (because (a,d,e) is in An and H is normal in An)

then (a,d,e)(a,b,c)(a,e,d) = (a)(b,c)(c,d) = (a)(b,c,d)(e) = (b,c,d)

Now, noting that "d" was arbitrary we can repeat this step and transform (a,b,c) to (x,y,z) for any (x,y,z)

(a,b,c) becomes (b,c,x) becomes (c,x,y) becomes (x,y,z)

Hence any (x,y,z) belongs in H. H contains every three cycle.

Suppose H is normal in An and H is not {e}, and n>4 Then H contains a three cycle.

Assume an element not equal to {e} from H as x = y₁y₂...y_s Is a product of disjoint cycles y_i and we may assume that the length of the cycles decrease so that, length(y₁) >= length(y₂) >= length(y₃ .... >= length(y_s)

we will write y₁ = (a₁,a₂,a₃,....,a_m)

CASE 1) if m>3

put W = (a₁,a₂,a₃)

then Wx(W^-1)(x^-1) is in H. I.e.

(W)(y₁y₂...y_s)(W^-1)(y_s^-1)(y_s-1^-1)...(y₂^-1)(y₁^-1) = (W)(y₁y₂...y_s)(W^-1)(y₁^-1)(y₂^-1)...(y_s^-1)

(ie the y_i commute)

Now, W contains only symbols from y₁ so W commutes with y₂ to y_s

Wx(W^-1)(x^-1) = Wy₁(W^-1)(y₁^-1) = (a₁,a₂,a₃)(a₁,a₂,a₃,...,a_m)(a₃,a₂,a₁)(a_m,...a₂,a₁) = (a₁,a₂,a₄)(a₃) = (a₁,a₂,a₄), and therefore H contains a three cycle. //

CASE 2) length(y₁) = length(y₂) = 3

write y₁ = (a₁,a₂,a₃)

y₂ = (a₄,a₅,a₆)

put W = (a₂,a₃,a₄)

Then Wx(W^-1)(x^-1) = (W)(y₁)(y₂)(W^-1)(y₁^-1)(y₂^-1) as before

= (a₂,a₃,a₄)(a₁,a₂,a₃)(a₄,a₅,a₆)(a₄,a₃,a₂)(a₃,a₂,a₁)(a₆,a₅,a₄) = (a₁,a₄,a₂,a₃,a₅) So H contains a cycle of length 5, so we reapply CASE 1 above, and therefore H contains a three cycle. //

CASE 3) length(y₁) = m=3 and length(y₂) = ... = length(y_s) = 2

then x² = (y₁²)(y₂²)...(y_s²) = y₁², which is also a three cycle in H. //

CASE 4) y_i for all i are transpositions

y₁ = (a₁,a₂), y₂ = (a₃,a₄)

put W = (a₂,a₃,a₄)

then Wx(W^-1)(x^-1) = (a₂,a₃,a₄)(y₁...y_s)(a₄,a₃,a₂)(y₁...y_s) = (a₂,a₃,a₄)y₁y₂(a₄,a₃,a₂)y₁y₂

(since y₃...y_s commute with (a₄,a₃,a₂) as these only appear in y₁ and y₂ and the cycles are ALL disjoint.)

Wx(W^-1)(x^-1) = (a₂,a₃,a₄)(a₁,a₂)(a₃,a₄)(a₄,a₃,a₂)(a₁,a₂)(a₃,a₄) = (a₁,a₄)(a₂,a₃) which is in H

Now, we can take x = (a₁,a₄)(a₂,a₃) and since n>4 we can choose a₅ not equal to a₁, a₂, a₃ or a₄ and we may then put W=(a₁,a₄,a₅) and compute Wx(W^-1)(x^-1) in H

Wx(W^-1)(x^-1) = (a₁,a₄,a₅)(a₁,a₄)(a₂,a₃)(a₁,a₅,a₄)(a₁,a₄)(a₂,a₃) = (a₁,a₅,a₄)(a₂)(a₃) = (a₁,a₅,a₄) which is a three cycle in H. //

Proof so far complete. //

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