Metamath

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 primepower order cyclic groups. (With addition done component wise) to the tune of;
(a+b, c+d, e+f ...) = (a+b = x mod p^{n}, 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 nonabelian 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(n1)(n2)(n3)...(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 3cycle 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 = y_{1}y_{2}...y_{s} Is a product of disjoint cycles y_{i} and we may assume that the length of the cycles decrease so that, length(y_{1}) >= length(y_{2}) >= length(y_{3} .... >= length(y_{s})
we will write y_{1} = (a_{1},a_{2},a_{3},....,a_{m})
CASE 1) if m>3
put W = (a_{1},a_{2},a_{3})
then Wx(W^{1})(x^{1}) is in H. I.e.
(W)(y_{1}y_{2}...y_{s})(W^{1})(y_{s}^{1})(y_{s1}^{1})...(y_{2}^{1})(y_{1}^{1}) = (W)(y_{1}y_{2}...y_{s})(W^{1})(y_{1}^{1})(y_{2}^{1})...(y_{s}^{1})
(ie the y_{i} commute)
Now, W contains only symbols from y_{1} so W commutes with y_{2} to y_{s}
Wx(W^{1})(x^{1}) = Wy_{1}(W^{1})(y_{1}^{1}) = (a_{1},a_{2},a_{3})(a_{1},a_{2},a_{3},...,a_{m})(a_{3},a_{2},a_{1})(a_{m},...a_{2},a_{1}) = (a_{1},a_{2},a_{4})(a_{3}) = (a_{1},a_{2},a_{4}), and therefore H contains a three cycle. //
CASE 2) length(y_{1}) = length(y_{2}) = 3
write y_{1} = (a_{1},a_{2},a_{3})
y_{2} = (a_{4},a_{5},a_{6})
put W = (a_{2},a_{3},a_{4})
Then Wx(W^{1})(x^{1}) = (W)(y_{1})(y_{2})(W^{1})(y_{1}^{1})(y_{2}^{1}) as before
= (a_{2},a_{3},a_{4})(a_{1},a_{2},a_{3})(a_{4},a_{5},a_{6})(a_{4},a_{3},a_{2})(a_{3},a_{2},a_{1})(a_{6},a_{5},a_{4}) = (a_{1},a_{4},a_{2},a_{3},a_{5}) So H contains a cycle of length 5, so we reapply CASE 1 above, and therefore H contains a three cycle. //
CASE 3) length(y_{1}) = m=3 and length(y_{2}) = ... = length(y_{s}) = 2
then x^{2} = (y_{1}^{2})(y_{2}^{2})...(y_{s}^{2}) = y_{1}^{2}, which is also a three cycle in H. //
CASE 4) y_{i} for all i are transpositions
y_{1} = (a_{1},a_{2}), y_{2} = (a_{3},a_{4})
put W = (a_{2},a_{3},a_{4})
then Wx(W^{1})(x^{1}) = (a_{2},a_{3},a_{4})(y_{1}...y_{s})(a_{4},a_{3},a_{2})(y_{1}...y_{s}) = (a_{2},a_{3},a_{4})y_{1}y_{2}(a_{4},a_{3},a_{2})y_{1}y_{2}
(since y_{3}...y_{s} commute with (a_{4},a_{3},a_{2}) as these only appear in y_{1} and y_{2} and the cycles are ALL disjoint.)
Wx(W^{1})(x^{1}) = (a_{2},a_{3},a_{4})(a_{1},a_{2})(a_{3},a_{4})(a_{4},a_{3},a_{2})(a_{1},a_{2})(a_{3},a_{4}) = (a_{1},a_{4})(a_{2},a_{3}) which is in H
Now, we can take x = (a_{1},a_{4})(a_{2},a_{3}) and since n>4 we can choose a_{5} not equal to a_{1}, a_{2}, a_{3} or a_{4} and we may then put W=(a_{1},a_{4},a_{5}) and compute Wx(W^{1})(x^{1}) in H
Wx(W^{1})(x^{1}) = (a_{1},a_{4},a_{5})(a_{1},a_{4})(a_{2},a_{3})(a_{1},a_{5},a_{4})(a_{1},a_{4})(a_{2},a_{3}) = (a_{1},a_{5},a_{4})(a_{2})(a_{3}) = (a_{1},a_{5},a_{4}) which is a three cycle in H. //
Proof so far complete. //
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