Abstract Algebra, Lecture 7 Jan Snellman Abstract Algebra, Lecture 7 the Classiﬁcation of ﬁnite Abelian Groups

Abstract Algebra, Lecture 7 Jan Snellman Abstract Algebra, Lecture 7 The classiﬁcation of ﬁnite abelian groups

Direct products again Jan Snellman1 Torsion and p-groups 1Matematiska Institutionen Link¨opingsUniversitet The classiﬁcation Finitely generated (and presented) abelian groups

Link¨oping,fall 2019

Lecture notes availabe at course homepage http://courses.mai.liu.se/GU/TATA55/ Abstract Algebra, Lecture 7

Jan Snellman Summary

Direct products again Torsion and p-groups 1 Direct products again 3 The classification The classification Direct sums vs direct products 4 Finitely generated (and Finitely generated 2 Torsion and p-groups presented) abelian groups (and presented) abelian groups Abstract Algebra, Lecture 7 For this lecture, all groups will be assumed to be abelian, but will usually Jan Snellman be written multiplicatively. Definition

G1,..., Gr groups. Their direct product is Direct products again Direct sums vs direct G1 × G2 × · · · × Gr = { (g1,..., gr ) gi ∈ Gi } products Torsion and with component-wise multiplication. p-groups The classiﬁcation Lemma Finitely generated (and presented) Put Hi = { (g1,..., gr ) gj = 1 unless j = i }. Then abelian groups 1 Hi ' Gi ,

2 H1H2 ··· Hr = G,

3 Hi ∩ Hj = {1} if i 6= j. Abstract Algebra, Lecture 7

Jan Snellman

Deﬁnition

Let G be a group, and H1,..., HK ≤ G. Then G is the internal direct Direct products product of H1,..., H if G ' H1 × · · · × H . again k k Direct sums vs direct products Theorem Torsion and p-groups TFAE:

The classiﬁcation 1 G is the internal direct product of H1,..., Hk ,

Finitely generated 2 Every g ∈ G can be uniquely written as g = h1h2 ··· hk with hi ∈ Hi , (and presented) abelian groups 3 H1H2 ··· Hk = G and Hi ∩ Hj = {1} for i 6= j. Abstract Algebra, Lecture 7

Jan Snellman

Example 2 3 Direct products Let G = C6 = hgi, H1 = hg i, H2 = hg i. Then again Direct sums vs direct H H = 1, g 2, g 4 1, g 3 = 1, g 3, g 2, g 5, g 4, g 6 products 1 2 Torsion and p-groups and H ∩ H = {1} , The classiﬁcation 1 2

Finitely generated so G is the internal direct product of H1 and H2. We have that H1 ' C3, (and presented) abelian groups H2 ' C2, so C6 ' C3 × C2. Abstract Algebra, Lecture 7

Jan Snellman Deﬁnition Let I be an inﬁnite index set.

• If Gi is a group for each i ∈ I , then the direct product Direct products again Gi Direct sums vs direct products i∈I Y Torsion and p-groups has the cartesian product of the underlying sets of the Gi as its The classiﬁcation underlying set, and componentwise multiplication Finitely generated • The direct sum (and presented) ⊕ Gi abelian groups i∈I is the subgroup of the direct product consisting of all sequences where all but ﬁnitely many entries are the corresponding identities Abstract Algebra, Lecture 7

Jan Snellman

Lemma

Direct products If Hi ≤ G, Hi ∩ Hj = {1} for i 6= j, and for each g ∈ G there is a ﬁnite again subset of S ⊂ I such that Direct sums vs direct products g = h , h ∈ H , Torsion and j j j j∈S p-groups Y The classiﬁcation then Finitely generated G ' ⊕ H . (and presented) i∈I i abelian groups Abstract Algebra, Lecture 7

Jan Snellman

Deﬁnition

Direct products • The torsion subgroup of G is the subset of all elements of G of finite again order Torsion and p-groups • If p is any prime number, then the p-torsion subgroup of G is defined The classification as a Finitely generated G[p] = { g ∈ G o(g) = p for some a ∈ N } (and presented) abelian groups Of course, any finite group is equal to its torsion subgroup. Abstract Algebra, Lecture 7

Jan Snellman Deﬁnition G is a p-group if G = G[p].

Direct products Lemma again If G is finite, then G is a p-group iff |G| = pa for some a. Torsion and p-groups Proof. The classification If |G| = pa, then by Lagrange o(g)|pa for all g ∈ G. But the only divisors Finitely generated (and presented) of pa are pb with b ≤ a. abelian groups Conversely, suppose that all G is finite, of size n, where p, q are two distinct prime factors of n. By Cauchy’s thm, which we’ll prove later, G contains elements of order p, and elements of order q. It is thus not a p-group. Abstract Algebra, Lecture 7

Jan Snellman

Example Direct products again The torsion subgroup of the circle group T consists of all complex numbers Torsion and p-groups m z = exp( 2πi), m, n ∈ Z, n 6= 0, gcd(m, n) = 1 The classiﬁcation n Finitely generated and its p-torsion subgroup are those complex numbers where n = p. (and presented) abelian groups Abstract Algebra, Lecture 7

Jan Snellman

Lemma Direct products again • The torsion subgroup of G is a subgroup of G, which contains all Torsion and p-groups p-torsion subgroups as subgroups. The classiﬁcation • The torsion subgroup of G is the direct sum of the G[p] as p ranges Finitely generated over all primes. (and presented) abelian groups Abstract Algebra, Lecture 7

Jan Snellman Proof.

a1 ar ai Let o(g) = n < , with n = p1 ··· pr . Let, for 1 ≤ i ≤ r, mi = n/pi , r mi xi and write 1 = i=1 mi xi . Put hi = g . Then

Direct products ∞ a P p i ai again i mi xi pi nxi n xi hi = g = g = (g ) = 1, Torsion and p-groups ai and hi ∈ G[pi ], since o(hi )|pi . Furthermore, The classiﬁcation 1 m1x1+···+mr xr m1x1 mr xr Finitely generated g = g = g = g ··· g = h1 ··· hr (and presented) abelian groups If u ∈ G[p] ∩ G[q] then o(u) = pa = qb, forcing a = b = 0. Thus G[p] ∩ G[q] = {1}. The previous lemma now shows that G is the internal direct sum of the G[p]’s. Abstract Algebra, Lecture 7 Jan Snellman Example

Let again C6 = hgi, with o(g) = 2 ∗ 3 = n. Put m1 = n/2 = 3, m2 = n/3 = 2, and write Direct products again 1 = 1 ∗ m1 + (−1) ∗ m2 = 1 ∗ 3 + (−1) ∗ 2. Torsion and 3 −2 p-groups Put h1 = g , h2 = g . Then

The classification 1 1 h2 = g 3∗2 = 1, h3 = g −2∗3 = 1, Finitely generated 1 2 (and presented) abelian groups so h1 has 2-torsion, and h2 has 3-torsion. (In fact, the 2-torsion subgroup is hh1i, and the the 3-torsion subgroup is hh2i.) We also see that 3 −2 1 h1h2 = g g = g = g Abstract Algebra, Lecture 7 Theorem (Classification of the finite abelian groups) Jan Snellman

a1 ar Let G be an abelian group of size n = p1 ··· pr . r 1 G ' j=1 G[pj ].

Direct products 2 EachQpj -group G[pj ] is isomorphic to a direct product of cyclic again pj -groups. Torsion and 3 A ﬁnite abelian p-group with pa elements is isomorphic to p-groups The classiﬁcation Cpb1 × · · · × Cpbs , b1 ≥ b2 ≥ · · · ≥ br , a = b1 + ··· + br Finitely generated (and presented) and the b ’s are uniquely determined. abelian groups i 4 Alternatively,

G ' Cd1 × · · · Cd` , d1|d2|d3 ··· |d`,

also uniquely. Abstract Algebra, Lecture 7

Jan Snellman Example What are the isomorphism classes of abelian groups with 36 = 22 ∗ 32

Direct products elements? We can list them as again Torsion and C4 × C9, C4 × C3 × C3, C2 × C2 × C9, C2 × C2 × C3 × C3, p-groups The classification or as C , C × C , C × C , C × C . Finitely generated 36 3 12 2 18 6 6 (and presented) abelian groups Recall that Cm × Cn ' Cmn when gcd(m, n) = 1. Abstract Algebra, Lecture 7 The part of the classification theorem that we have not proved is Jan Snellman Theorem Every finite p-group is the internal direct sum of cyclic p-groups.

The main step is to establish that Direct products again Lemma Torsion and a b p-groups If |G| = p and g ∈ G has maximal order p , then there is a “complement” H ≤ G, such that The classiﬁcation Finitely generated hgiH = G, hGi ∩ H = {1} (and presented) abelian groups

Proof. See Judson (but note that the desired subgroup is K in his proof).

a−b Given the lemma, G ' Cpb × H, where H is a p-group with p elements; by induction, it can be written as a direct product of cyclic p-groups. Abstract Algebra, Lecture 7

Jan Snellman

For the remainder of the lecture, groups are abelian, and written additively. Definition Direct products r again Z = Z × Z × · · · Z is a free abelian group of rank r, as is any group r Torsion and isomorphic to Z . p-groups The classification Theorem Finitely generated The rank is well-defined, i.e., (and presented) abelian groups r s Z ' Z r = s

⇐⇒ Abstract Algebra, Lecture 7

Jan Snellman Example

• Free abelian groups are akin to vector spaces. Direct products 2 again • Consider Z = Z × Z: it has an (ordered) basis Torsion and p-groups e = [e1, e2] = [(1, 0), (0, 1)] The classiﬁcation w.r.t. which all group elements can be uniquely expressed as integral Finitely generated (and presented) linear combinations. abelian groups • We have similar results regarding subspaces and dimensions as in linear algebra. Abstract Algebra, Lecture 7 Example (contd.) Jan Snellman • Take for example

1 1 1 u = (1, 1) = e , v = (1, 2) = e , w = (1, 3) = e Direct products 1 2 3 again 2 Torsion and • Then hu, v, wi = Z , but the vectors are not linearly independent: p-groups indeed 1 1 The classiﬁcation w = u + v, 2 2 Finitely generated (and presented) so abelian groups u + v − 2w = 0.

2 • This means that hv, wi = Z . • Furthermore, one can check that there are no non-trivial linear relations between v and w, which hence constitute another basis for 2 Z . Abstract Algebra, Lecture 7 Theorem Jan Snellman r If H ≤ G with G ' Z then H is free abelian of rank ≤ r. Proof (sketch). Let G have basis x ,..., x (to lazy for boldface). Put Direct products 1 r 0 r−1 0 0 0 again G = hx1,..., xr−1i ' Z and H = H ∩ G . By induction, H is free Torsion and abelian of rank ≤ r − 1. p-groups Using the second isomorphism theorem, written additively, we have The classiﬁcation H H H + G 0 G Finitely generated = ' ⊆ ' Z (and presented) H 0 H ∩ G 0 G 0 G 0 abelian groups We know that any non-trivial subgroup of Z is isomorphic to Z, so H = hv + H 0i ' H 0 Z for some v ∈ H. That v, together with the basis for H 0, makes a basis for H. Abstract Algebra, Lecture 7

Jan Snellman

Recall that an (abelian or not) group G is finitely generated if there is a finite generating set such that G = hg1,..., gr i. Direct products again Theorem Torsion and A finitely generated abelian group G can be written as p-groups The classification r G ' GT × Z , Finitely generated (and presented) abelian groups where GT is the torsion subgroup, which will be a finite abelian group, and where the rank r is well defined. Abstract Algebra, Lecture 7

Jan Snellman Example A Klein bottle

Direct products is a two-dimensional closed surface that can be 4 again conveniently embedded into R , but not (without 3 Torsion and self-intersection) into R . It can be constructed p-groups by gluing together a left-twisted Möbiusstrip The classification with a right-twisted on along their common edge. Finitely generated It is an “non-orientable” surface; this induces (and presented) abelian groups torsion in its first “homology group”, which is

Z2 × Z.

Figure: Klein Bottle Abstract Algebra, Lecture 7

Jan Snellman Stolen — from the internet!

Direct products again Torsion and p-groups The classification Finitely generated (and presented) abelian groups Abstract Algebra, Lecture 7 Jan Snellman Definition A finitely presented abelian group G is given by a finite list of generators g1,..., gn, and a finite list of relations among the generators, from which Direct products all other relations can be derived. again Torsion and Example p-groups In the Klein bottle example, G is generated by a, b, c, and all relations The classification among the generators are integral linear combinations of Finitely generated (and presented) abelian groups a + b − c = 0 a − b + c = 0

For instance, by adding the relations, we conclude that 2a = 0, so there is 2-torsion in G. Abstract Algebra, Lecture 7 A ﬁnitely presented abelian group can be analyzed using the so-called Jan Snellman Smith normal form of integer matrices: Example The relations from the previous example can be summarized as Direct products again 1 1 −1 A = Torsion and 1 −1 1 p-groups The classiﬁcation By simultaneous (independent) change of bases in the source and in the Finitely generated target, we get (and presented) abelian groups D = LAR  1 1 0  1 0 0 0 1 1 1 −1 = 0 1 1 0 2 0 1 −1 1 −1 1   0 0 1 Abstract Algebra, Lecture 7

Jan Snellman

Example (contd.)

Direct products The normal form 1 0 0 again D = Torsion and 0 2 0 p-groups shows that, in the (ﬁrst) homology group G of the Klein bottle, The classiﬁcation • there is one factor isomorphic to , Finitely generated Z (and presented) • there is one factor isomorphic to 2, abelian groups Z • the last generator is not needed.