Capítulo II Marco teórico
2.3.5. Ruptura de la relación conyugal: violencia familiar y divorcio
Exercise 4. LetGbe a group and letH andK be two subgroups ofG. 1. Is H∩Ka subgroup ofG? If your answer is yes, prove it. If your answer
is no, provide a counterexample.
2. Is H∪Ka subgroup ofG? If your answer is yes, prove it. If your answer is no, provide a counterexample.
Answer.
1. This is true. It is enough to check that xy−1 ∈H∩K forx, y ∈H∩K.
But sincex, y∈H, we havexy−1∈HsinceHis a subgroup, and likewise, xy−1∈K forx, y∈K sinceKis a subgroup.
2. This is false. For example, take the groups of integers modulo 3 and 2, namelyZ/3ZandZ/2Z. Then 2 and 3 are in their union, but 5 is not.
Exercise 5. Show that ifGhas only one element of order 2, then this element is in the center ofG(that is the elements ofGwhich commute with every element in G).
Answer. Letxbe the element of order 2. Thenyxy−1 has also order 2. Thus
it must be either 1 or x. If yxy−1 = 1, then x = 1 a contradiction. Thus
yxy−1=x.
Exercise 6. LetGbe a group andH be a subgroup ofG. Show that NG(H) ={g∈G, gH=Hg}
and
CG(H) ={g∈G, gh=hgfor allh∈H}
are subgroups ofG.
Answer. Takex, y ∈NG(H). We have to check that xy−1∈NG(H), that is,
that xy−1H =Hxy−1. ButHxy−1=xHy−1 sincex∈N
G(H), and xHy−1=
xy−1H sinceyH =Hy ⇐⇒ y−1H=Hy−1.
Now take x, y ∈ CG(H). We have to check that xy−1h = hxy−1 for all
h ∈H. But hxy−1 = xhy−1 because x ∈C
G(H), and xhy−1 =xy−1hsince
yh=hy ⇐⇒ y−1h=hy−1.
2.2
Cyclic groups
Exercise 7. LetG=Z∗24be the group of invertible elements in Z24. Find all
cyclic subgroups ofG.
Answer. We have that the size ofGis
Gis given by the elements that are invertible mod 24, that is, those that are coprime to 24: G={1,5,7,11,13,17,19,23}. Now h1i={1}, h5i={5,52= 1}, h7i={7,72= 49 = 1}, h11i={11,112= 121 = 1} h13i={13,132= 169 = 1} h17i={17,172= (−7)2= 1} h19i={19,192= (−5)2= 1} h23i={23,232= (−1)2= 1
and there are 8 cyclic subgroups ofG, including the trivial subgroup{1}.
Exercise 8. LetG=Z∗20 be the group of invertible elements inZ20. Find two
subgroups of order 4 inG, one that is cyclic and one that is not cyclic.
Answer. As in the exercise above,Gcontains
|G|=ϕ(20) =ϕ(4)ϕ(5) = 2·4 = 8. These 8 elements are coprime to 20, that is
G={1,3,7,9,11,13,17,19}. The subgroup
h3i={3,32= 9,33= 7,34= 21 = 1} is cyclic of order 4. We have that
11,112= 121 = 1,19,192= (−1)2= 1,11·19 = (−11) = 9,92= 81 = 1
and
{1,11,19,9} is a group of order 4 which is not cyclic.
Exercise 9. Letϕ be the Euler totient function. Let Gbe a cyclic group of ordern.
1. First show that the order ofgk is
|gk|=n/gcd(k, n).
2. Show that ifm|n, thenhgn/miis the unique subgroup ofGof orderm.
3. Prove that for every factormofn, the number of elements inGwith order mis exactlyϕ(m).
2.2. CYCLIC GROUPS 69
4. Furthermore, show that Pm|nϕ(m) =n.
Answer. LetG=hgibe a cyclic group of ordern, so that every element inG is of the form
gk, 1≤k≤n.
1. Set m =gcd(k, n), so thatk =mk′, n =mn′. If (gk)r = 1, then n|kr,
and
n m|
kr m.
By definition of m, n/m and k/m are coprime, so that n/m divides r. Hence n/m is the smallest power of gk such that (gk)n/m = 1 showing
thatn/m=|gk|.
2. LetH be a subgroup of orderm, thenH =hgkiwith|H|=mand some
k >0. We will show first thatH can be also generated by an elementgd
whered=gcd(k, n), and in particular, we can always write
H =hgdi, d|n.
Since d|k, k = dq and gk = gdq ∈ hgdi and hgkni ⊆ hgdi. Conversely,
d=gcd(k, n) =kr+nsfor some r, sand
gd=gkr+ns=gkr ∈ hgki
andhgdi ⊆ hgki. Now m=|H|=|gk|=|gd|=n/gcd(d, n) by the above,
and sinced|n, we get that m=n/d, ord=m/n.
3. Now form|n, an element is of order mif and only if it is the generator of the only subgroup ofGof orderm. Now there are as many generators for this subgroup as elements coprime tom, that isϕ(m).
4. To show that Pm|nϕ(m) =n, we can sort the elements of Gaccording to their order. Since the order of each element dividesn, we have
n=X
m|n
nb of elements of orderm=X
m|n
ϕ(m).