Classifying Groups of Small Order ()

Gerard Thompson^{*}

Department of Mathematics and Statistics, The University of Toledo, Toledo, OH, USA.

**DOI: **10.4236/apm.2016.62007
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Department of Mathematics and Statistics, The University of Toledo, Toledo, OH, USA.

The classification of groups of order less than 16 is reconsidered. The goal of the paper is partly historical and partly pedagogical and aims to achieve the classification as simply as possible in a way which can be easily incorporated into a first course in abstract algebra and without appealing to the Sylow Theorems. The paper concludes with some exercises for students.

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Thompson, G. (2016) Classifying Groups of Small Order. *Advances in Pure Mathematics*, **6**, 58-65. doi: 10.4236/apm.2016.62007.

Received 24 January 2015; accepted 24 January 2016; published 28 January 2016

1. Introduction

We shall say a few words about notation: means that H is a subgroup of G and means that H is a normal subgroup of G. The identity element in G will be denoted by e. We shall denote the center of the group by and the centralizer of an element, by, that is,. We shall use the word order frequently with different meanings. Thus denotes the order of G, that is, the number of elements in G. We shall also use the same notation if merely. On the other hand, is the order of an element, that is to say, the smallest positive power is r such that. The reason for using the same word is that is also the order of the cyclic subgroup generated by g and that cyclic subgroup will be denoted by. Another not very serious remark: I have consistently used numbers rather than writing numbers in words because there are a lot of them and I considered that to do so had a much of a more visual impact.

2. Some History

3. Elementary Facts

1) Groups of prime order are cyclic and unique up to isomorphism.

2) Conjugate elements have the same order.

3) If G is a group and is its center then the factor group cyclic implies that G is abelian.

4) If all elements of G except e are of order 2 then G is abelian.

5) If p is prime the number of elements of order p is a multiple of.

6) If a group G is generated by two normal subgroups H and K (so that every is of the form for some finite p) and H and K are complementary in the sense that, then, the direct product.

4. Conjugation and the Class Equation

4.1. Group Actions

Let G be a group and X a set. Then we say G acts on X if there is a group homomorphism from G to, the latter set denoting the group of permutations of X. Here are three examples of group actions where the group acts on itself, that is,.

・ For and let (left translation)

・ For and let (right translation)

・ For and let (conjugation): note for.

Define to be the orbit of x and define a relation on X according as elements belong to the same orbit. It is left as exercise to show that this relation is actually an equivalence relation. Define also or to be the stabilizer subgroup of x, that is,. In the case of the first two actions defined above there is only one orbit for each, that is G itself; in that case we say that the action is transitive. Also, for each we have. However, in the third example of the action of conjugation, the orbits are precisely the conjugacy classes of G and is the centralizer of g, that is,. It is easy to show that. Notice also that conjugation does nothing at all for abelian groups! In fact conjugation is a very special kind of action because G acts on itself not merely as a set but as a group; in other word the action maps not only to but to, the group of automorphisms of G.

**Theorem 1**. There is a one to one correspondence between elements of and, that is, the space of (left) cosets of in G.

Proof. Suppose that; then there exists such that. We obtain a map from to by mapping g to the left coset of in G. We leave the student to verify that this map is well-defined (the hardest part for students) and bijective.

4.2. Example

The group D_{6} or consists of the elements subject to the relations. The last relation used repeatedly enables us to “twist” products of x and y to work all the x’s to the front in a string of x’s and y’s and all the y’s to the back. Then the other relations may be used to reduce any string to one of the 6 elements that constitute the group. The conjugacy classes of are so that

. Now and so. Also

and and and

so.

4.3. Conjugation and Centralizers

The class equation expresses as a sum of the orders of the conjugacy classes. Thus the order of a conjugacy

class is the index of the centralizer of any element x in that conjugacy class, that is,. In particular the

order of a conjugacy class divides. Notice that if and only if if and only if z comprises its own conjugacy class and that for provided G is not the trivial group consisting of a single element. Hence on the right hand side of the class equation we will have as many 1’s as appear in and sums of various proper divisors of.

5. The Dihedral Group

The group of order 2n is generated by x and y subject to the relations and. We shall now determine the class equation for. Note that and and likewise

. If n is odd then whereas if n is even is a

singleton and ~ signifies “is conjugate to”.

Now and. Continuing in this way we find that if n is odd comprises a single conjugacy class, whereas if n is even

and constitute distinct conjugacy classes.

Assuming so that is not abelian, as a result of the preceding calculations we see that, apart from

, the only other singleton conjugacy class is in the case where n is even. Thus if n is odd

whereas if n is even.

A few more elementary calculations convince one that if n is odd, the class equation for is

the number of 2’s being, whereas if n is even, the class equation for is

the number of 2’s being this time.

Notice finally that if where r is odd then in fact: indeed and also being a subgroup of index 2 that is isomorphic to. Since the normal subgroups are complementary by Fact 6, it follows that.

6. Theorem

**Lemma 1**. If G is a non-abelian group of order pq where p and q are distinct primes then.

Proof. Since we must have that divides by Lagrange’s Theorem and hence can only be one of. Now if and only if G is abelian. If then or. Now and in either case the group is cyclic hence by Fact 3 G must be abelian.

**Lemma 2**. If G is a group of order where p is prime then.

Proof. If the class equation gives. Each of the terms on the right hand side except the first must be divisors of since they are indexes of centralizers and cannot be 1 otherwise - contradiction and hence is impossible.

**Theorem 2**. Suppose that G is a non-abelian group of order 2n where n is either an odd prime or 4 or 6. Suppose further that there exist such that and that. Then.

Proof. Clearly are distinct as too are. The only possibility for collapsing among these 2n elements is if y is a power of x and necessarily, which case we are excluding. Now: it certainly contains and if then and G would be abelian. Hence x belongs to a conjugacy class of 2 elements. Furthermore for some. Thus

and. Thus.

Suppose now that n is an odd prime. Then has only the solution.

Suppose that. Then the only solution of for is.

Suppose that. Then the only solution of for is.

Hence in all 3 cases covered by the Theorem.

**Corollary 1**. Suppose that G is non-abelian and that where p is an odd prime. Then.

Proof. By 1 and the only possibility for the class equation is. Consider an element y in the conjugacy class of order p. Now and so since.

Turning to an element x in a conjugacy class of order 2 then. Since p is prime it follows that

is cyclic of order p. We cannot have for then and whereas. Hence.

7. Case by Case Study

For each order bigger than 5 we are looking for a non-abelian group G. Since G is not abelian we have that. Since groups of prime order are cyclic we do not need to consider the cases. Furthermore Theorem 1 disposes of the cases so it remains to study the cases 8,9,12,15.

7.1.

It follows from Lemma 2 that. Hence we can only have that. Now and so is cyclic and hence G is abelian by Fact 3.

7.2.

By Lemma 1 and the class equation gives. The elements of the conjugacy class of order 5 are themselves of order 3: they cannot be of order 5 because their centralizers are of order 3 whereas an element of order 5 has a centralizer of order at least 5. On the other hand each of the elements in the 3 conjugacy classes of order 3 must have order 5: indeed, if x is such an element then and x must generate. Now the total number of elements of order 5 must be 9, which contradicts Fact 5, since it would have to be multiple of 4. Hence there cannot exist G for which and G is non-abelian.

7.3.

We know of course that is a non-abelian group of order 8. We investigate whether there are any others. According to Fact 4, G must have an element x of order 4. Suppose that but that. Then according to Theorem 2, if G is not isomorphic to we may assume both that and. It is then easy to see by cancelation that the elements of G must be. Now or else G will be abelian. The only possibilities for are 2 or 4; however, if then and again G would be abelian and so. Again by cancelation and elimination of all other possibilities we see that; furthermore since xy has order 2. We will now rewrite the elements

of G as. Note that and hence. Similarly.

Finally we can replace e by 1, by and by, respectively, so as to obtain the quaternion group Q usually written as.

7.4.

We have since that could be 1, 2, 3, 4 or 6. We may reject the cases or because of Fact 4. Suppose next that. Then in the class equation we must have another odd number, which can only be 3. Thus. Consider an element x in the conjugacy classes of order 3. Its centralizer is of order 4 and contains as a subgroup which is impossible since 3 does not divide 4. Thus or 2.

Suppose then that where. Then again by Fact 4 we can only have that. Now we know that contains a cyclic normal subgroup of order 3. According to the correspondence theorem [4] , which is really the deluxe version of the first isomorphism theorem, engenders a normal subgroup N of order 6 in G that contains. Since N itself has a non-trivial center it too must be cyclic. Suppose that. Since there exists such that and furthermore.

Notice that since and if any other positive power of x were in then. Furthermore, since commutes with x and y so or else in which case by Theorem 2.

So in order to avoid having we must have. Evidently the group G can be written as. So which element is yx? Clearly, since, we have and. Moreover or else G would be abelian. If then since

which would give and so. If then.

So and so by Theorem 2. If then so that and. According to Theorem 2 we must have unless, which is of course impossible.

Thus the only way to avoid having is to have and and of course and all the conditions for the existence of a group are now met. To recognize this group, introduce a new generator so that. The elements of G can be written as. Now where we have used the fact that. Hence. Thus the group generated by y, acts on generated by w via conjugation and engenders a group denoted by T that is a semi-direct product of and. Although and in fact neither of these extensions split: here.

In fact the group T belongs to a familiar class of finite groups of order 4n called the dicyclic groups and also known as the binary dihedral groups. Depending how one counts, the first such group is and the second the group Q studied in subsection 6.3. For further details see [13] which is another nice reference for more advanced material.

Now suppose that. The class equation must contain another odd number which can only be 3. Hence we have the following cases:

・

・

・

・ .

We consider the first three cases for which 2 appears in the class equation. We take x in one of the conjugacy classes of order 2. Then. Since has index 2 it is normal in G. Hence we may write where. If is a positive power of x, it will be a central element. However, we are considering now only cases where so the only possibility is. Now we can invoke Corollary 1 and conclude that. Since for we have that we conclude that none of the three cases for which 2 appears in the class equation actually do occur.

Now consider the last case of the class equation. Each of the elements in the conjugacy classes of order 4 must have order 3 because their centralizers are of order 3. Elements in the conjugacy class of order 3 could have order 2 or 4 because each of their centralizers are of order 4; however, if the elements have order four their squares would have order 2 and there are no conjugacy classes left to accommodate them. Hence the elements in the conjugacy class of order 3 have order 2.

Let x and y be elements of order 2 and 3, respectively. Then and are of order 2 since they are conjugate to x; if any two of are equal we would have that. In that case we have a cyclic subgroup of order 6 generated by xy and whose only element of order 2 is x. Now pick a second element of order 2 and appeal to Corollary 1 to conclude again that and again contradicting. Thus we may assume that are distinct. Now xy and yx must be of order 3. It follows that and. Besides the remaining elements of G are. Now obviously y and and and, respectively, are conjugate. Furthermore, and and are conjugate. Now implies that and that so and so is conjugate to. Hence the conjugacy classes have to be. It follows that. We can obtain the usual presen- tation of by mapping x and y to the permutations (12) (34) and (123), respectively.

Finally, a more sophisticated approach is to note that the conjugacy class of order 3 together with e forms a normal subgroup of order 4: indeed the centralizer of an element x in the conjugacy class of order 3 is of order 4. Since elements in the two conjugacy class of order 4 are each of order 3, the only possibility is to obtain the normal subgroup N described above. It follows that. From there it is not difficult to argue that acts on N as a semi-direct product and deduce that. Of course N turns out to be the Sylow 2-subgroup.

8. Conclusions

We have classified all groups of order < 16 without using Sylow theory and assuming we have known the classification of finite abelian groups. It seems remarkable to the author that for the classification of the non-abelian groups becomes almost routine and depends only on elementary facts: the case is the only one that is at all challenging. So why stop at? In fact, it turns out that up to isomorphism there are 14 distinct such groups! As an exercise (see next Section) try to find 8 of them. In fact, the groups that are the hardest to classify are p-groups where the order of and p is prime and m is a positive integer. Sylow’s powerful theorems tell us nothing for such groups. Furthermore, the smaller that p is, the harder are the groups to classify. Naively speaking, the smaller that p is the more combinatorial possibilities there are to satisfy the group axioms.

In [14] one can find descriptions of the non-abelian groups of order < 32 in terms of generators and relations. We should also mention the computer algebra system GAP that contains the “Small Groups Library”. In that system groups of order up to 2000 are listed up to isomorphism with the exception of groups of order: apparently there are an 49,487,365,422 non-isomorphic 2-groups of order 1024. At the time of writing, according to Wolfram, all groups have been classified up to isomorphism up to order. In addition to Wolfram the author has also gained a lot of information from Wikipaedia: search for “List of small groups”. However, I am warned to add the usual disclaimers about referring to websites.

9. Exercises for the Student

・ Supply proofs of the six Elementary Facts. As a hint for the sixth, note that it suffices to map generators to generators.

・ Show that the orbits defined in Section 4 according as elements belong to the same orbit actually is an equivalence relation.

・ Show that for the stabilizer subgroup of or defined in Section 4,.

・ Finish the details of Theorem 4.1.

・ Find 8 mutually non-isomorphic groups of order 16.

・ Find generators and relations for the group starting from its definition as the subgroup of the symmetric group consisting of even permutations.

・ Find an explicit isomorphism between and.

Acknowledgements

The author thanks Paul Hewitt for stimulating discussions and some valuable suggestions from the referees for improving many of the arguments.

Conflicts of Interest

The authors declare no conflicts of interest.

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