Lecture Notes on Abelian Groups and the Structure Theorem, Study notes of Mathematics

Download Lecture Notes on Abelian Groups and the Structure Theorem and more Study notes Mathematics in PDF only on Docsity!

Lecture Notes Math 371: Algebra (Fall 2006)

by Nathanael Leedom Ackerman September 14, 2006

0.1 TALK SLOWLY AND WRITE NEATLY!!

0.2 Abelian Group

0.2.1 Definitions

Definition of Abelian Group

Definition 0.2.1.1. We say a group 〈A, +, 0 > is Abelian if (∀a, b ∈ A)a + b = b + a

Spanning Notation

Definition 0.2.1.2. Let G be a group and let g 1 , g 2 , · · · ∈ G. We then define 〈g 1 , g 2 , · · ·〉 = subgroup of G gener- ated by g 1 , g 2 , · · ·

Torsion Subgroup

Definition 0.2.1.3. Let 〈G, +, 0 〉 be an abelian group and let a ∈ G. We say a is a torsion element if it has finite order.

0.2.2 Theorems

Subgroups of Free Groups are Free

Theorem 0.2.2.1 (Dedekind). Let F be a free abelian group of rank r and let G be a subgroup of F. Then G is a free abelian group of rank s ≤ r. Further F has a set of generators 〈ui, · · · , ur〉 such that G is generated by v 1 = a 11 u 1 + a 12 u 2 +... + a 1 rur v 2 = + a 12 u 2 +... + a 2 rur ... ... ... vs = assus +... + asrur for some aij such that all aii which are positive.

Proof. Let 〈ui, · · · , ur〉 be a basis for F for notational convenience we will say that x = Σi≤rc(x, i)ui for all x ∈.

Now choose some non-zero element b = Σi≤rc(b, i)ui of

G.

By reordering the indexes and possibly taking inverses of some of the basis elements we can assume that c(b, 1) >

Now lets look at all the values of c(b, 1) for b ∈ G and choose a minimum positive value (which we know must exist because c(b, 1) > 0). Call the smallest such integer a 11 and pick some element v such that c(v, 1) = a 11

We therefore have that if x = Σi≤rc(x, i)ui ∈ G then a 11 divides c(x, 1). This is because c(x, 1) = p ∗ a 11 + a where a < a 11. And so if a 11 didn’t divide c(x, 1) then −pv 1 + x would have a coordinate in u 1 with coefficient less than a 11.

further

v 2 = a 22 u 2 + a 23 u 3 +... + a 2 rur v 3 = + a 33 u 3 +... + a 3 rur ... ... ... vs = assus +... + asrur with a 22 · · · ass all positive. Hence we know that G = Span〈v 1 , · · · , vs〉 by the in- ductive hypothesis.

So all that is left is to check that v 1 , · · · , vn are inde- pendent. If not then there are some Σi≤rdivi = 0 where di ∈ Z

But we know that if r = 1 this can’t happen. And hence by induction we that 〈vs, · · · , vr〉 are independent. So this means that there must be some linear combination of vi which adds to 0 and what is more this must have

the coeffecient of v 1 non-zero.

−d 1 v 1 = d 2 v 2 + · · · + dnvn

But, putting any such linear combination in terms of the ui we see that d 1 = 0 (as v 1 is the only element with u 1 component and the ui’s are linearly independent). ⇒⇐.

Hence the 〈v 1 , · · · , vs〉 are a linear independent basis for F as a free abelian group.

Theorem about basis

Theorem 0.2.2.2. Let F = 〈u 1 , · · · , ur〉 and let v = b 1 u 1 + · · · + brur with gcd(b 1 , · · · , br) = 1. Then there exists v 2 ,... vr ∈ F such that F = 〈v, v 2 ,... , vr〉.

Proof. Set s = |b 1 | + |b 2 | + · · · + |br|. If s = 1 then the result is trivial as v = ±ui for some i.

Now lets assume this theorem is true for all 0 ≤ r < s.

rank s with 0 < s ≤ r. Then there exists v 1 ,... , vr ∈ F such that F = 〈v 1 ,... , vr〉 G = 〈h 1 v 1 ,... , hrvr〉

where h 1 ,... , hr are all positive and satisfy hi divides hi+1.

Proof. Let u 1 ,... , ur be a set of generators for F. Take x ∈ G and write x = x 1 u 1 + · · · + xrur. We then define δ〈u^1 ,...,ur〉(x) = gcd(x 1 , · · · , xr).

Lemma 0.2.2.4. δ〈u^1 ,...,ur〉(x) is independent of the choice of generators.

Proof. We know by the definition of x that

x = δ〈u^1 ,...,ur〉(x)(y 1 u 1 + · · · + yrur)

for some integers r.

Now let w 1 , · · · , wr generate F. Then we can represent each ui in terms of w 1 , · · · , wr. Hence

x = δ〈u^1 ,...,ur〉(x)(y 1 ∗ w 1 + · · · + y r∗ wr)

and so δ〈u^1 ,...,ur〉(x) divides δ〈w^1 ,...,wr〉(x).

And so by symmetry we must have δ〈u^1 ,...,ur〉(x) = δ〈w^1 ,...,wr〉(x)

We will then simply write δ(x) for δ〈u^1 ,...,ur〉(x).

Now take any nonzero y 1 ∈ G such that δ(y 1 ) is min- imal. Set h 1 = δ(y 1 ). We then have

y 1 = h 1 (z 1 u 1 + · · · + zrur)

where gcd(z 1 , · · · , zr) = 1. So in particular by the previ- ous lemma we know that if we set v 1 = z 1 u 1 + · · · + zrur then there exist v 2 ∗ , · · · , v r∗ such that

〈v 1 , v∗ 2 ,... , v∗ r 〉 = F

But we therefore know that

F = 〈v 1 , v 2 ,... , vr〉

Now lets consider a y ∈ G. Well then we know we can write y = a 1 v 1 + a 2 v 2 ∗ + · · · + arv∗ r

But we know a 1 = qh 1 + m for some m < h 1 and further δ(y − qy 1 ) ≤ m. Hence m = 0 because we choose y 1 to be such that δ(y) was minimal. So h 1 divides a 1.

So we have y − qh 1 v 1 = t 2 v 2 ∗ + · · · + trv r∗

Hence G 1 = 〈h 1 v 1 , h 2 v 2 ,... , hrvr〉

and all that is left to show is that h 1 divides h 2.

let h 2 = ah 1 + b where 0 ≤ b < h 1. Then if we let α = h 1 v 1 + h 2 v 2 ∈ G we have gcd(y 0 ) = gcd(h 1 , h 2 ) = gcd(h 1 , b). Hence by the minimality of h 1 we have b = 0.

So by induction we are done.

0.3 The Structure Theorem for Finitely Gen- erated Abelian Groups

0.3.1 Theorem

From the previous theorem we have

Existence of Decomposition By Increasing Div

Theorem 0.3.1.1. Every finitely generated Abelian group A can be expressed as the direct sum of cyclic groups A ∼= Zm^ ⊕ Z/h 1 ⊕ · · · ⊕ Z/hj

where hi divides hi+

and we are done.

But what is more, is that this decomposition is unique. Specifically we have

Uniqueness of Decomposition By Increasing D

Theorem 0.3.1.2. Let A be a finitely generated Abelian group. Let

A = Zr^ ⊕ Z/e 1 ⊕ · · · ⊕ Z/hn

where ei divides ei+1 and

A = Zs^ ⊕ Z/d 1 ⊕ · · · ⊕ Z/dm

where di divides di+1. Then s = r, m = n and ei = di for all i

But before we show this we need another theorem.

Existence of Decomposition By Sylow Subgrou

Theorem 0.3.1.3. Let G be a finite abelian group of order pa 11 pa 2 2 · · · where the pi’s are distinct primes. Then G = P 1 ⊕ P 2 ⊕ · · ·

Where Pi are the subgroups of elements of G with order a power of pi.

Proof. Let x ∈ G be an element of order pα 1 f 1 where p 1 and f 1 are relatively prime. We can then find u, v such that uf 1 + vpα 1 = 1.

We then have x = uf 1 x + vpα 1 x but we know that uf 1 x has order pα 1 and vpα 1 x has order f 1.

So we know we can break x = b 1 + x 1 where b 1 has order pα 1 and x 1 has order f 1. And so by iterating this process we can break x = b 1 + b 2 +... where bi has order pa ii

Corollary 0.3.1.4. Let e = pa 11 · · · pa mm. Then

Z/(e) ∼= Z/(pa 1 1 ) ⊕ · · · ⊕ Z/(pa mm )

Proof. We know

Z/(e) = P 1 ⊕ P 2 ⊕ · · ·

where Pi = {z ∈ Z/(e) : order(z) is a power of pi}. In particular if x generates Z/(e) then we have

x = q 1 ⊕ · · · ⊕ qn

where each qi has order a power of pi.

but every y ∈ Z/(e) = mx and so in particular this means that every z ∈ Z/(e) of order pi = mzx.

So in particular (n/pa i i)z = (n/pa i i)mzx = (n/pa i i)mzqi.

But as n/pa i i is relatively prime to pa i i there is an in- teger s such that s ∗ n/pa i i ≡ 1 mod pa i i.

So s(n/pa i i)z = z = (n/pa i i)mzqi = mzqi (because or- der of (n/pa i i)z = pa i i).

In particular this means that Pi is generated by a sin- gle element and hence Pi ∼= Z/(pa i i) and we have

Z/(e) ∼= Z/(pa 1 1 ) ⊕ · · · ⊕ Z/(pa mm )

Now we can return to the proof of the uniqueness of the decomposition.

Proof Uniqueness of Decomposition By Increas

Proof. First to see that r = s notice that

A ∼= Zs^ ⊕ T ∼= Zr^ ⊕ T

where T is the torsion group of A.