Solvable GroupsDate: 12/10/2002 at 13:48:44 From: Aynur Bulut Subject: Solvable groups Dear Dr. Math, I know that An (alternating group of degree n) is not solvable for n>=5 because of simplicity of An for n>=5. What is the proof that shows An is simple for n>=5 ? I would like to ask one more question: What is the connection between the solvability of polynomials for degree>=5 and the solvability of An for n>=5 ? Thanks, Aynur Bulut Date: 12/23/2002 at 12:18:48 From: Doctor Nitrogen Subject: Re: Solvable groups Hi, Aynur: I will answer your second question first. Here are some background concepts: First, loosely defined, a group is called "solvable" if all the factor groups in its composition series are Abelian. A "composition series," loosely speaking, is a sequence of factor groups of a group G, each factor group in the sequence smaller than the one following right after it in the sequence, and these factor groups are all simple. Mind you, you should look up the more formal definitions of these terms. Now let f(x) be a polynomial of degree 5 in F[x], with coefficients in a field F. There is a theorem in Galois theory stating: THEOREM: Let f(x) be a polynomial that is an element of the ring of polynomials F[x] and with coefficients taken from a field F. Let E be the splitting field of f(x) over F. Then f(x) is "solvable by radicals over F" if and only if (iff) the splitting field E for f(x) over F has a Galois group G(E/F) that is solvable. Note that you should also look up the definition of "solvable by radicals over F." For a polynomial of degree 5 an element of F[x] and with coefficients from a field F, its Galois group, G(E/F), is isomorphic to S_5. The composition series for S_5 is: {id} < A_5 < S_5, where "{id}" denotes the singleton set and subgroup of S_5 consisting only of the identity in S_5 as an element. The only problem with the series {id} < A_5 < S_5, is A_5 is indeed simple, but it is not Abelian. This means S_5 is not solvable, and since G(E/F) is isomorphic to S_5, G(E/F) is not solvable either, meaning by the above Theorem that f(x) cannot be solvable by radicals over F. A similar argument is used when f(x) has degree n > 5 and A_n has degree > 5 (n > 5). Now for your first question. I could not find any online proofs that A_n is simple for n => 5. Your best bet for finding this proof would be in a university library. There are at least two possible ways to prove it, or I really should say a way to prove A_5 is simple and a way to prove A_n is simple for n > 5. The first involves a proof by contradiction, using the fact that A_5 has no element of order 15, and trying to prove A_5 is not simple and that it does have a nontrivial normal subgroup, and you would do this by investigating any proper candidates N as normal subgroups of A_5 and their orders, as well as the orders of the elements of A_5. The possible orders are the factors of 60. You would look at the possible orders of A_5/N. Doing this would lead to a contradiction that | A_5/N | = either 15 or 30, meaning A_5 would have an element of order 15, which is impossible, as A_5 has no such element. That would be the contradiction. The way to prove A_n is simple involves looking at the 3-cycles in A_n. This is a little more involved, but here are some books you could find in your university or local library to help you get started on proving A_n, n > 5, is simple: Abstract Algebra, Sixth Edition, John B. Fraleigh, 1999, ISBN #0-201-33596-4. In my opinion, this book is the absolute Grand Bible of Abstract Algebra, at least at the undergraduate level. It does not have a proof, but there is an exercise in Chapter Three that gives a pretty extensive outline for how to construct this 3-cycle proof. A book in which you will find a proof that A_5 is simple is: Contemporary Abstract Algebra, Joseph Gallian, 1986, ISBN #0-669-09325-4. There is also a famous little book on Modern Algebra/Galois Theory by Emil Artin, and another by Van der Waerden, which could be helpful. If you are interested in learning more on A_5 not having an element of order 15, note that A_5 is isomorphic to the group of symmetries (isometry group) of the regular dodecahedron (and icosahedron), and this symmetry group has no element of order 15. I hope this helped answer the questions you had concerning your mathematics problem. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ Date: 12/30/2002 at 14:08:23 From: Aynur Bulut Subject: Thank you (Solvable groups) I was so grateful for the answers to my questions. Thanks very much, Dr. Nitrogen. Best wishes, Aynur Bulut |
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