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Strongs:  Mathematics 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 pn (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 qm, e+f = z mod rs,... ) 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 I need to show that A5 is a simple group. That its only factors are {e} and itself. There is a none too difficult proof that A5 is simple. First, An is a normal subgroup of Sn (readily checked, as each conjugation gHg-1 adds an even number of transpositions, and is therefore also in An.) Also important to note that every element of Sn not in An may be found in the coset of An in Sn, i.e. g = xH where H = An, and x is some transposition. Then {e, x} is also a subgroup of Sn and is a cyclic prime order group which itself is simple and soluble. Sn for n>4 has only the normal subgroups Sn, An, C2 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".Continue To Next Page Return To Section Start '