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### Solvable Groups

Date: 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:

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

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

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

| 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
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.

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
Associated Topics:
College Modern Algebra

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