Finite abelian groups

Having seen a little example of how Sylow's theorem (which applies to all nite groups) can be used to examine the possibilities for groups of a given order, let's add an important hypothesis, that of abelian-ness.

As it happens, the class of nite abelian groups can be described completely in terms of elementary divisibility and the groups

Z

^=N, as follows.

Given a nite abelian groupG, there is a uniquely-determined integerm >0 and uniquely-determined sequence of numbersd1;d2;:::;dmwith the divisibility property

d1^jd2^jd3^j:::^jdm

and so that

G

Z

^=d1

Z

^=d2

Z

^=d3:::

Z

^=dn

The positive integersd1;d2;:::;dm occurring in such an expression for a nite abelian group are the

elementary divisors

of the group.

Thus, every nite abelian group can be `decomposed' or `broken up' into simpler pieces, each of which is one of the relatively elementary groups

Z

^=N.

This reduces classication of all nite abelian groups of a given order to yet another question in elemen-tary arithmetic.

For example, let's nd all the abelian groups of order 12. We must nd a sequence of integers, each dividing the next, whose product is 12. In this simple example we can `see' the possibilities for the elementary divisors: the only possibilities are

Z

⁼2

Z

⁼6

and

Z

⁼¹²

Let's nd all abelian groups of order 48. This is still easy to do without a systematic approach:

Z

⁼⁴⁸

Z

⁼²

Z

⁼²⁴

Z

⁼⁴

Z

⁼¹²

Z

=2

Z

=2

Z

=12

Z

⁼²

Z

⁼²

Z

⁼²

Z

⁼⁶

#24.182

Letpbe a prime. Suppose that a groupGhaspelements. Prove thatGis cyclic.

#24.183

Suppose that a nite groupGhas no subgroups but^fe^gandG. ShowGis cyclic.

#24.184

^LetGbe a group of ordern. Show that for allg²Gwe havegⁿ=e.

#24.185

^{Let m;n}be relatively prime. LetH;K be subgroups of a groupGwhere ^jH^j=mand ^jK^j=n. Show thatH^\K=^fe^g.

#24.186

Fix a primep. Suppose all proper subgroups of a nite groupGhave orders powers ofp. Prove that^jG^jis a power of p.

#24.187

^{Let p;q}be distinct primes. Show that any abelian group of order pq has an element of orderpq, so is cyclic.

#24.188

^LetN be a normal subgroup of groupG, and show that not only is it thatgNg^,1N, but in fact gNg^,1=N.

#24.189

Show that a subgroupN of a groupGis normal if and only ifgN =Ngfor allg²G.

#24.190

Show that in an abelian group every subgroup is normal.

#24.191

Show that for any subgroupH of a groupG (withH not necessarily normal in G), the relation dened by

xy if and only if xH =yH is an equivalence relation.

#24.192

Show that a homomorphism f : G ^! H is injective if and only if its kernel is trivial, that is, if and only if its kernel is the trivial subgroup ^fe^g of G. (Hint: On one hand, if f(g) = f(g⁰) then eH =f(g)^,1f(g⁰) =f(g^,1g⁰) sog^,1g⁰ is in the kernel. Ifg⁶=g⁰ then this gives a non-trivial (not equal to eG) element of the kernel. On the other hand, reversing this argument you can show that if the kernel is trivial (is just^feG^g) thenf(g) =f(g⁰) implies thatg=g⁰).

#24.193

Letf :G^!H andg:G^!K be two group homomorphisms. Let F:G^!HK

be dened by

F(x) = (f(x);g(x)) Show that

kerF = kerf ^\kerg

#24.194

Letm;nbe relatively prime positive integers. Let f :

Z

^=mn!

Z

^=m

be dened by

f(x^,mod^,mn) =xmodm and also dene

g:

Z

=mn^!

Z

=n

by g(x^,mod^,mn) =xmodn

Then also dene

F :

Z

^=mn!

Z

^=m

Z

⁼ⁿ

by f(x^,mod^,mn) = (f(x);g(x))

First show that the intersection of the kernels off and ofgis trivial. From this conclude that the kernel ofF is trivial, so thatF is injective. Then, by counting, deduce thatF is also surjective, so is an isomorphism).

#24.195

^{Let N} be a normal subgroup of a group G. Let f : G ^! H be a group homomorphism whose kernel containsN. Show that the map

f :G=N ^!H dened by

f(gN) =f(g)

is well-dened and is a group homomorphism. (Hint: To prove well-denedness usually amounts to showing that the denition of something does not depend excessively on the notation, but really only on the underlying thing).

#24.196

LetGbe a cyclic group of (nite) orderN, with generatorg. Show thatGis isomorphic to

Z

=N, by showing that the mapf :G^!

Z

=N dened byf(gⁿ) =n-mod-N is such an isomorphism.

#24.197

LetGbe a cyclic group of innite order, with generatorg. Show thatGis isomorphic to

Z

, by showing that the mapf :G^!

Z

dened byf(gⁿ) =nis such an isomorphism.

#24.198

Letf1:G^!H1andf2:G^!H2 be group homomorphisms. Dene f :G^!H1H1

by f(g) = (f1(g);f2(g))

Show that thisf is a group homomorphism.

#24.199

^Letm;nbe relatively prime positive integers. Dene f :

Z

=mn^!

Z

=m

Z

=n by (using additive notation)

f(x+mn

Z

) = (x+m

Z

^{; x}+n

Z

)

Show that this is an isomorphism. (Hint: In eect, this says that a system of congruences xamodm xbmodn

can always be solved forxfor anya;b, and that the solutionxis uniquely determined modulomn. Use the fact that there are integerss;tso thatsm+tn= 1. Tryx=bsm+atn?)

#24.200

Find an integer x so thatx2 mod 10 andx7 mod 11. Then nd a dierent integerx⁰ with the same property.

#24.201

Fix an elementg_oof a groupG. Show that both mapsL;Rdened by L(g) =g_og R(g) =gg_o

are bijections ofGto itself.

#24.202

^Letx;y be xed elements in a group G. Fix a subset S ofG, and let T =xSy. Show that the

map f(g) =xgy

(which is dened on all ofG) does indeed mapS to T, and gives a bijection fromS toT. (Compare to the previous exercise!)

#24.203

Fix a subgroupH of a groupG and x g ²G. Show that the conjugategHg^,1 of H byg is a subgroup. (And observe that the previous exercise shows thatgHg^,1 has the same order as doesH).

#24.204

Show that a group homomorphism f : G ^! H is injective if and only if its kernel is `trivial', meaning that kerf =^fe^g.

#24.205

^{Let G} be a group. For h ² G dene fh : G ^! G by fh(g) = hgh^,1. Prove that fh is an automorphism ofG. (Such automorphisms are called

inner automorphisms

^).

#24.206

LetGbe an abelian group. Prove thatf(g) =g^,1is an automorphism ofG.

#24.207

Let gbe a generator for a cyclic group G. Letf :G^!Gbe an automorphism ofG. Show that f(g) is also a generator ofG. (Hint: Automorphisms are bijections, so every element inGcan be written as f(h) for someh²G. So to prove ^hf(g)ⁱ=Git suces to prove that for eachh²Gthere is`²

Z

so that f(g)^`=f(h).)

#24.208

Grant that automorphisms send generators to generators. Prove that all automorphisms of

Z

^=N

are of the form

f(xmodN) =rxmodN for somer²

Z

=N. (Hint: we know all possible generators of

Z

=N).

#24.209

Grant that automorphisms send generators to generators. Prove that there are exactly 2 auto-morphisms of

Z

: the identity map and the mapf(x) =^,x.

#24.210

Show that every group of order 33 is cyclic. (Hint: Use Sylow's theorem).

#24.211

Show that every group of order 85 is cyclic. (Hint: Use Sylow's theorem).

#24.212

Using the Sylow Theorem, show that in a group of orderpqwith two primesp;qandp < q, there is only one subgroup of orderq(by a counting argument).

#24.213

Granting that in a group of orderpq with two primes p;qand p < q, there is only one subgroup of orderq, show that this subgroup is necessarily normal.

#24.214

Show that the order of a product groupGH is the product ^jG^jjH^jof the orders^jG^j;^jH^jof the two groups.

#24.215

Find all abelian groups of order 12.

#24.216

Find all abelian groups of order 125.

#24.217

Find all abelian groups of order 127.

#24.218

Find all abelian groups of order 64.

#24.219

Find all abelian groups of order 2²3²5².

#24.220

Find all abelian groups of order 2²3²5³.

No documento Intro Abstract Algebra c (páginas 158-162)