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How many structure theorems do we have in Abstract Algebra for finite algebraic structures? I know some of the following theorems:

  • If $G$ is a finite abelian group, then $G$ is a product cyclic groups of prime power order.
  • If $R$ is a finite commutative ring, then $R$ is Artinian and hence a direct product of local rings.

I am wondering if there are other structure theorems for finite algebraic structures. Is it possible to write a non-commutative or a non-abelian group in this way?

Why are the theorems available for abelian cases but not for non-abelian ones?

If someone can please help me find some references where I can get structure theorems for non-commutative or a non-abelian group/rings, I will be thankful.

Please help.

Charlotte
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    We can classify finite fields, and say there are no other finite division algebras. – anon Mar 15 '21 at 06:15
  • The Jordan-Hölder theorem plus the classification of finite simple groups can be viewed as a structure theorem for finite groups. – Erick Wong Mar 15 '21 at 06:20
  • Note that the classification theorem for finite simple groups is much much much much much more complicated & difficult to prove than the classification theorem for finite abelian groups. That's just how it goes. Commutativity makes life easier. – Gerry Myerson Mar 15 '21 at 06:33
  • @ErickWong CFSG is a certainly a classification theorem, but its proof is impossibly difficult, so I am guessing it's not what the OP has in mind. It is certainly not true, however, that Jordan-Hölder plus CFSG provide any sort of classification of finite groups. Compare the following: the number and structure of abelian groups of order $p^{100}$, the number and structure of finite fields of size $p^{100}$ and the number and structure of groups of order $p^{100}$. The first two questions admit easy answers. If you can answer the last one, I'll pay you $10$k out of my own pocket. – the_fox Mar 15 '21 at 06:33
  • @the_fox Your point is well-taken, but I do believe looking up the CFSG will would give the OP some appreciation for why the non-abelian group case does not have an obvious structure theorem. I should have stated it as more like the closest thing we have given the complexity of group structures, rather than making it sound like an equivalent to the abelian case. – Erick Wong Mar 15 '21 at 07:47
  • @ErickWong; I got through your comments, I am still confused about how Jordan-Holder Theorem and the classification theorem can be used as a structure theorem for finite groups. – Charlotte Mar 15 '21 at 10:29
  • @ErickWong; Can you kindly illustrate your point with an example. Take for example a group $G$ of order 100. How shall we write $G$ using the two theorems. I apologise but I am confused. – Charlotte Mar 15 '21 at 10:31
  • @GerryMyerson; So regarding my question on how to write a finite non-abelian group (if possible) as a decomposition of other groups, do you know of any recent progress which has been made in this direction – Charlotte Mar 15 '21 at 10:33
  • @Math_Freak I suggest you read diracdeltafunk’s answer. It does an excellent job of explaining in great detail what I meant in my comment, and I agree with its content. In the case of $|G|=100$, the group is necessarily solvable, so the composition factors are all $C_2$ and $C_5$ (two of each). This means $G$ can be written as a series of extensions by those four factors in some order, but there are several different ways to chain them together (if you haven’t already read it, you should start by looking at semi-direct products). – Erick Wong Mar 15 '21 at 11:02
  • What @Erick said. – Gerry Myerson Mar 15 '21 at 11:19

1 Answers1

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Non-abelian finite groups are quite complicated, but we have completely classified the finite simple groups. Here's an overview from Wikipedia. Moreover, every finite group can be "built out of finite simple groups" in an essentially unique way, in the same sense that every nonzero integer can be written as a product of primes in an essentially unique way! If you want to know precisely what this means, this is the content of the Jordan–Hölder theorem (you will first need to learn what a composition series is). In contrast to prime factorizations, where we just multiply the primes to reproduce the original integer, there are many complicated ways that simple groups can be put together to form a non-simple group. The general problem of understanding the possible isomorphism types of a group from its simple factors is what's known as an "extension problem". In short, while we know all the finite simple groups, and we know that all finite groups are built from these, we are not close to understanding all finite groups.

Non-commutative finite rings are quite complex! Wedderburn's little theorem says that a finite ring with no zero divisors is a field. Besides this, I don't know of any general classification results for finite rings. Even the finite commutative rings are quite difficult to classify: we easily reduce to a product of finite local rings, but those are complicated to describe. The good news is that, by considering the possible underlying abelian groups, it is not too computationally expensive to compute all isomorphism classes of finite commutative local rings of any given cardinality $n$.

I also feel obligated to mention the $5/8$ theorem:

Theorem ($5/8$ Theorem for Finite Groups) Let $G$ be a finite group such that the probability that two randomly chosen elements of $G$ commute is greater than $5/8$. Then $G$ is abelian.

Theorem ($5/8$ Theorem for Finite Rings) Let $G$ be a finite ring such that the probability that two randomly chosen elements of $G$ commute is greater than $5/8$. Then $G$ is commutative.

See here for a proof, or here for a generalization to compact topological groups.

diracdeltafunk
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  • Thanks a lot for your answer but I am afraid I have many questions. Q1: Suppose $G$ is a finite non-abelian group. Then how can we built $G$ from simple groups and what are those simple groups which will build $G$. – Charlotte Mar 15 '21 at 10:40
  • Q2:Also JH Theorem states that any two composition series of a finite group $G$ are equivalent. Also we know that if $G$ is a finite simple group then $G$ is one of the one members of the 3 group families. However I am still confused how to use the two facts to classify any finite non-abelian group – Charlotte Mar 15 '21 at 10:42
  • Let me reiterate: no one knows how to classify all finite groups! I tried to emphasize this in my answer. See here and here for explanations of how finite groups are built from simple ones. I also want to correct your understanding of the classification: there are $18$ infinite families of simple groups, plus $26$ simple groups which do not lie in these families. It's a complicated picture! – diracdeltafunk Mar 15 '21 at 19:13
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    Here's the part which we do understand: for a given $n$, we can list out all simple groups whose order divides $n$. Then we can write down the possible lists of composition factors for a group of order $n$. For example, there are exactly four simple groups whose orders divide $120$, namely $\mathbb{Z}/2$, $\mathbb{Z}/3$, $\mathbb{Z}/5$, and $A_5$. This shows that any group of order $120$ has one of the following lists of composition factors:

    $$\mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/3, \mathbb{Z}/5$$

    $$\mathbb{Z}/2, A_5$$

     – diracdeltafunk Mar 15 '21 at 19:20
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    There are only two possible lists of composition factors, but there are $47$ isomorphism classes of groups of order $120$! So again: we understand the finite simple groups, but the ways they fit together to form non-simple groups are very complicated, and we are not close to being able to classify all finite groups.

    Just for completeness, $44$ of these groups are solvable (a.k.a have cyclic composition factors) and $3$ are not (having $\mathbb{Z}/2, A_5$ as their list of composition factors).

     – diracdeltafunk Mar 15 '21 at 19:22
  • Thanks a lot for your explanation, +1 – Charlotte Mar 16 '21 at 02:39