13.1: Finite Abelian Groups
( \newcommand{\kernel}{\mathrm{null}\,}\)
In our investigation of cyclic groups we found that every group of prime order was isomorphic to Zp, where p was a prime number. We also determined that Zmn≅Zm×Zn when gcd(m,n)=1. In fact, much more is true. Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order; that is, every finite abelian group is isomorphic to a group of the type
Zpα11×⋯×Zpαnn,
where each pk is prime (not necessarily distinct).
First, let us examine a slight generalization of finite abelian groups. Suppose that G is a group and let {gi} be a set of elements in G, where i is in some index set I (not necessarily finite). The smallest subgroup of G containing all of the gi's is the subgroup of G generated by the gi's. If this subgroup of G is in fact all of G, then G is generated by the set {gi:i∈I}. In this case the gi's are said to be the generators of G. If there is a finite set {gi:i∈I} that generates G, then G is finitely generated.
Example 13.1
Obviously, all finite groups are finitely generated. For example, the group S3 is generated by the permutations (12) and (123).
Solution
The group Z×Zn is an infinite group but is finitely generated by {(1,0),(0,1)}.
Example 13.2
Not all groups are finitely generated. Consider the rational numbers Q under the operation of addition. Suppose that Q is finitely generated with generators p1/q1,…,pn/qn, where each pi/qi is a fraction expressed in its lowest terms. Let p be some prime that does not divide any of the denominators q1,…,qn.
Solution
We claim that 1/p cannot be in the subgroup of Q that is generated by p1/q1,…,pn/qn, since p does not divide the denominator of any element in this subgroup. This fact is easy to see since the sum of any two generators is
pi/qi+pj/qj=(piqj+pjqi)/(qiqj).
Proposition 13.3
Let H be the subgroup of a group G that is generated by {gi∈G:i∈I}. Then h∈H exactly when it is a product of the form
h=gα1i1⋯gαnin,
where the giks are not necessarily distinct.
- Proof
-
Let K be the set of all products of the form gα1i1⋯gαnin, where the giks are not necessarily distinct. Certainly K is a subset of H. We need only show that K is a subgroup of G. If this is the case, then K=H, since H is the smallest subgroup containing all the gis.
Clearly, the set K is closed under the group operation. Since g0i=1, the identity is in K. It remains to show that the inverse of an element g=gk1i1⋯gknin in K must also be in K. However,
g−1=(gk1i1⋯gknin)−1=(g−knin⋯g−k1i1).
The reason that powers of a fixed gi may occur several times in the product is that we may have a nonabelian group. However, if the group is abelian, then the gis need occur only once. For example, a product such as a−3b5a7 in an abelian group could always be simplified (in this case, to a4b5).
Now let us restrict our attention to finite abelian groups. We can express any finite abelian group as a finite direct product of cyclic groups. More specifically, letting p be prime, we define a group G to be a p-group if every element in G has as its order a power of p. For example, both Z2×Z2 and Z4 are 2-groups, whereas Z27 is a 3-group. We shall prove the Fundamental Theorem of Finite Abelian Groups which tells us that every finite abelian group is isomorphic to a direct product of cyclic p-groups.
Theorem 13.4. Fundamental Theorem of FInite Abelian Groups
Every finite abelian group G is isomorphic to a direct product of cyclic groups of the form
Zpα11×Zpα22×⋯×Zpαnn
here the pi's are primes (not necessarily distinct).
Example 13.5
Suppose that we wish to classify all abelian groups of order 540=22⋅33⋅5.
Solution
The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities.
- Z2×Z2×Z3×Z3×Z3×Z5;
- Z2×Z2×Z3×Z9×Z5;
- Z2×Z2×Z27×Z5;
- Z4×Z3×Z3×Z3×Z5;
- Z4×Z3×Z9×Z5;
- Z4×Z27×Z5.
The proof of the Fundamental Theorem of Finite Abelian Groups depends on several lemmas.
Lemma 13.6
Let G be a finite abelian group of order n. If p is a prime that divides n, then G contains an element of order p.
- Proof
-
We will prove this lemma by induction. If n=1, then there is nothing to show. Now suppose that the lemma is true for all groups of order k, where k<n. Furthermore, let p be a prime that divides n.
If G has no proper nontrivial subgroups, then G=⟨a⟩, where a is any element other than the identity. By Exercise 4.5.39, the order of G must be prime. Since p divides n, we know that p=n, and G contains p−1 elements of order p.
Now suppose that G contains a nontrivial proper subgroup H. Then 1<|H|<n. If p∣|H|, then H contains an element of order p by induction and the lemma is true. Suppose that p does not divide the order of H. Since G is abelian, it must be the case that H is a normal subgroup of G, and |G|=|H|⋅|G/H|. Consequently, p must divide |G/H|. Since |G/H|<|G|=n, we know that G/H contains an element aH of order p by the induction hypothesis. Thus,
H=(aH)p=apH,
and ap∈H but a∉H. If |H|=r, then p and r are relatively prime, and there exist integers s and t such that sp+tr=1. Furthermore, the order of ap must divide r, and (ap)r=(ar)p=1.
We claim that ar has order p. We must show that ar≠1. Suppose ar=1. Then
a=asp+tr=aspatr=(ap)s(ar)t=(ap)s1=(ap)s.
Since ap∈H, it must be the case that a=(ap)s∈H, which is a contradiction. Therefore, ar≠1 is an element of order p in G.
Lemma 13.6 is a special case of Cauchy's Theorem (Theorem 15.1), which states that if G is a finite group and p a prime such that p divides the order of G, then G contains a subgroup of order p. We will prove Cauchy's Theorem in Chapter 15.
Lemma 13.7
A finite abelian group is a p-group if and only if its order is a power of p.
- Proof
-
If |G|=pn then by Lagrange’s theorem, then the order of any g∈G must divide pn, and therefore must be a power of p. Conversely, if |G| is not a power of p, then it has some other prime divisor q, so by Lemma 13.6, G has an element of order q and thus is not a p-group.
Lemma 13.8
Let G be a finite abelian group of order n=pα11⋯pαkk, where where p1,…,pk are distinct primes and α1,α2,…,αk are positive integers. Then G is the internal direct product of subgroups G1,G2,…,Gk, where Gi is the subgroup of G consisting of all elements of order pri for some integer r.
- Proof
-
Since G is an abelian group, we are guaranteed that Gi is a subgroup of G for i=1,…,k. Since the identity has order p0i=1, we know that 1∈Gi. If g∈Gi has order pri, then g−1 must also have order pri. Finally, if h∈Gi has order psi, then
(gh)pti=gptihpti=1⋅1=1,
where t is the maximum of r and s.
We must show that
G=G1G2⋯Gk
and Gi∩Gj={1} for i≠j. Suppose that g1∈G1 is in the subgroup generated by G2,G3,…,Gk. Then g1=g2g3⋯gk for gi∈Gi. Since gi has order pαi, we know that gpαii=1 for i=2,3,…,k, and gpα22⋯pαkk1=1. Since the order of g1 is a power of p1 and gcd(p1,pα22⋯pαkk)=1, it must be the case that g1=1 and the intersection of G1 with any of the subgroups G2,G3,…,Gk is the identity. A similar argument shows that Gi∩Gj={1} for i≠j.
Next, we must show that it possible to write every g∈G as a product g1⋯gk, where gi∈Gi. Since the order of g divides the order of G, we know that
|g|=pβ11pβ22⋯pβkk
for some integers β1,…,βk. Letting ai=|g|/pβii, the ai's are relatively prime; hence, there exist integers b1,…,bk such that a1b1+⋯+akbk=1. Consequently,
g=ga1b1+⋯+akbk=ga1b1⋯gakbk.
Since
g(aibi)pβii=gbi|g|=e,
it follows that gaibi must be in Gi. Let gi=gaibi. Then g=g1⋯gk∈G1G2⋯Gk. Therefore, G=G1G2⋯Gk is an internal direct product of subgroups.
If remains for us to determine the possible structure of each pi-group Gi in Lemma 13.8.
Lemma 13.9
Let G be a finite abelian p-group and suppose that g∈G has maximal order. Then G is isomorphic to ⟨g⟩×H for some subgroup H of G.
- Proof
-
By Lemma 13.7, we may assume that the order of G is pn. We shall induct on n. If n=1, then G is cyclic of order p and must be generated by g. Suppose now that the statement of the lemma holds for all integers k with 1≤k<n and let g be of maximal order in G, say |g|=pm. Then apm=e for all a∈G. Now choose h in G such that h∉⟨g⟩, where h has the smallest possible order. Certainly such an h exists; otherwise, G=⟨g⟩ and we are done. Let H=⟨h⟩.
We claim that ⟨g⟩∩H={e}. It suffices to show that |H|=p. Since |hp|=|h|/p, the order of hp is smaller than the order of h and must be in ⟨g⟩ by the minimality of h; that is, hp=gr for some number r. Hence,
(gr)pm−1=(hp)pm−1=hpm=e,
and the order of gr must be less than or equal to pm−1. Therefore, gr cannot generate ⟨g⟩. Notice that p must occur as a factor of r, say r=ps, and hp=gr=gps. Define a to be g−sh. Then a cannot be in ⟨g⟩; otherwise, h would also have to be in ⟨g⟩. Also,
ap=g−sphp=g−rhp=h−php=e.
We have now formed an element a with order p such that a∉⟨g⟩. Since h was chosen to have the smallest order of all of the elements that are not in ⟨g⟩, |H|=p.
Now we will show that the order of gH in the factor group G/H must be the same as the order of g in G. If |gH|<|g|=pm, then
H=(gH)pm−1=gpm−1H;
hence, gpm−1 must be in ⟨g⟩∩H={e}, which contradicts the fact that the order of g is pm. Therefore, gH must have maximal order in G/H. By the Correspondence Theorem and our induction hypothesis,
G/H≅⟨gH⟩×K/H
for some subgroup K of G containing H. We claim that ⟨g⟩∩K={e}. If b∈⟨g⟩∩K, then bH∈⟨gH⟩∩K/H={H} and b∈⟨g⟩∩H={e}. It follows that G=⟨g⟩K implies that G≅⟨g⟩×K.
The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly from Lemma 13.9. Suppose that G is a finite abelian group and let g be an element of maximal order in G. If ⟨g⟩=G, then we are done; otherwise, G≅Z|g|×H for some subgroup H contained in G by the lemma. Since |H|<|G|, we can apply mathematical induction.
We now state the more general theorem for all finitely generated abelian groups. The proof of this theorem can be found in any of the references at the end of this chapter.
Theorem 13.10. The Fundamental Theorem of Finitely Generated Abelian Groups
Every finitely generated abelian group G is isomorphic to a direct product of cyclic groups of the form
Zpα11×Zpα22×⋯×Zpαnn×Z×⋯×Z,
where the pi's are primes (not necessarily distinct).