猫论HW1Solutions
27.)A subgroup H is conjugate to a subgroup K of a group G if there exists an inner automorphism i g of G such that i g[H]=K.Show that conjugacy is an equivalence relation on the collection of subgroups of G.
So again,this problem is saying to show that the relation on subgroups defined by
H∼K if and only if there exists g∈G such that i g[H]=gHg−1=K is an equivalence relation.So we need to show the three properties of an equivalence relation.
•reflexive:Here we want to show that H∼H.Consider the identity e∈G.Then
i e[H]=eHe−1=H.Therefore the relation is indeed reflexive.
•symmetric:Here we want to show that if H∼K then we also have K∼H.By assumption,there exists g∈G such that i g[H]=gHg−1=K.Therefore we have that
H=eHe−1=(g−1g)H(g−1g)=g−1(gHg−1)g=g−1Kg and so i g−1[K]=H giving us that K∼H and so the relation is indeed symmetric.
•transitive:Assume that H∼K and K∼L then we wish to show that H∼L.
By assumption,there exists g1,g2∈G such that i g
1[H]=g1Hg−1
1
=K and
i g
2[K]=g2Kg−1
2
=L.Therefore by substitution we have that
L=g2Kg−1
2
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=g2(g1Hg−1
1
)g−1
2
=(g2g1)H(g2g1)−1=i g
1g2
[H]
and so indeed H∼L.
35.)Show that if H and N are subgroups of a group G,and N is normal in G,then
H∩N is normal in H.Show by an example that H∩N need not be normal in G.
You showed for homework last mester that the interction of two subgroups is itlf a subgroup.B
ut let’s go over that again for practice.To show that H∩N is a subgroup, it is sufficient to show closure and that a−1is an element of H∩N for all a in H∩N.
•closure:Let a,b∈H∩N.Then since both H and N are themlves subgroups we have that ab∈H and∈N and therefore ab∈H∩N.
•invers:Let a∈H∩N.Again,since both H and N are themlves subgroups, we must have that a−1∈H and a−1∈N and so a−1∈H∩N.
Now we need only show that H∩N is normal in H.To do this,we will show that for all h∈H and g∈H∩N we must have hgh−1∈H∩N.First note that since
g∈H∩N we have that in particular g∈H and so since H is a subgroup we must have that hgh−1∈H.Furthermore,since N is normal in G we have that hgh−1∈N. Therefore hgh−1∈H∩N giving us that the interction is normal in H.
37.)Let Aut(G)denote the t of automorphisms of a group G.
a Show that Aut(G)is a group under function composition.
Here we need to show that Aut(G)satisfies the properties of a group.
•closure:Letφ,ψ∈Aut(G).Then we need to show thatφ◦ψis also an
automorphism.There are then three things to check.
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–homomorphism:We need to make sure thatφ◦ψis a homomorphism.
Let g1,g2∈G.Then becau bothφandψare themlves
homomorphisms,we have
φ◦ψ(g1g2)=φ(ψ(g1g2))=φ(ψ(g1)ψ(g2))=φ(ψ(g1))φ(ψ(g2))=φ◦ψ(g1)φ◦ψ(g2).
And so the composition also satisfies the homomorphism property.
–one-to-one:Here we need to show thatφ◦ψis also one-to-one.Note
that sinceφandψare injective then we have kerφ=kerψ={e}where
e∈G is the identity.By definition
kerφ◦ψ={g∈G|φ◦ψ(g)=φ(ψ(g))=e}.
So ifφ(ψ(g))=e then we must have thatψ(g)=e since kerφ={e}and
similarly we must then have that g=e since kerψ={e}.And so we
indeed have that kerφ◦ψ={e}giving us that the composition is an
injection.
–onto:Here we want to make sure that under the composition,for every
g1∈G there exists g ∈G such thatφ◦ψ(g )=g1.Note that Sinceφis
怀孕能吃竹笋吗surjective,there exists g2∈G such thatφ(g2)=g1.Furthermore,since
ψis also surjective,there exists g3such thatψ(g3)=g2.Therefore
φ◦ψ(g3)=φ(ψ(g3))=φ(g2)=g1
giving us that the composition is also surjective.
And so we have thatφ◦ψ∈Aut(G).
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b Show that the inner automorphisms of a group G form a normal subgroup of
Aut(G)under function composition.
For ea of notation,let Inn(G)denote the t of inner automorphisms of G,
Inn(G)={i g|g∈G}.
The name of the maps tells us that they are indeed automorphisms and
therefore Inn(G)⊂Aut(G).(We should have shown this is true in class but that’s
okay.)We need to show two things for this problem.First we need to show that
Inn(G)is a subgroup of Aut(G).Then we will show that it is a normal subgroup.
•subgroup:Again,to show that it is a subgroup,we need to show that Inn(G)is clod under function composition and for all i g∈Inn(G)that (i g)−1∈Inn(G).
–closure:By the proof in part(a)we know that for all g1,g2∈G that
i g
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1◦i g
2
is again an automoprhism and so we just need to show that it is
an inner automorphism.Note that for all h∈G
i g
1◦i g
2
学生思维(h)=i g
1
(i g
2
(h))=i g
1
(g2hg−1
2
)=g1(g2hg−1
2
)g−1
青椒鱿鱼1
=(g1g2)h(g1g2)−1=i g
1g2
(h)
giving us that Inn(G)is clod under composition.
–invers:Now let g∈G and consider g−1∈G.Then for all h∈G
i g◦i g−1(h)=i g(i g−1(h))=i g(g−1hg)=g(g−1hg)g−1=h=i e(h)
Where i e is just the identity map which is the identity element in
Aut(G).A similar argument can be made to show i g−1◦i g=i e and so
every inner automorphism as an inver inner automorphism.Therefore Inn(G)is indeed a subgroup of Aut(G).
•normality:So the easiest way to show that Inn(G)is normal in Aut(G)is to show thatφ◦i g◦φ−1∈Inn(G)for allφ∈Aut(G),i g∈Inn(G).Letφ∈Aut(G)and i g∈Inn(G)and h∈G.Using the homomorphism property ofφwe then have
φ◦i g◦φ−1(h)=φ(i g(φ−1(h)))=φ(gφ−1(h)g−1)
=φ(g)φ(φ−1(h))φ(g−1)=φ(g)h[φ(g)]−1
=iφ(g)(h)
which is an inner automorphism of G sinceφ(g)∈G.
