Download Nilpotent groups

Answers to Homework 4
due 2/28/02
Nilpotent groups
Lemma 8.10 Any finite product of nilpotent groups is nilpotent.
HW4.ex01: Prove this lemma (in your own words) and discuss the question:
Is this true for infinite products?
Ans: [by Georgi] It is enough to show that the product of two nilpotent
groups is nilpotent. So suppose that G and H are nilpotent of class a, b resp.
Let c = max(a, b). Then Z c (G) = G and Z c (H) = H and we claim that
Z c (G × H) = Z c (G) × Z c (H) = G × H. This follows from the following
lemma.
Lemma 4.1. Z i (G × H) = Z i (G) × Z i (H) for any groups G, H.
Proof. Since [(g, h), (x, y)] = ([g, x], [h, y]), (g, h) is central in G × H iff g is
central in G and h is central in H, i.e.,
Z(G × H) = Z(G) × Z(H).
Thus the lemma holds for i = 1. If the lemma hold for i then
G×H
G×H
G
H
=
=
×
Z i (G × H)
Z i (G) × Z i (H)
Z i (G) Z i (H)
which has center
µ
¶
µ
¶
µ
¶
G
G
H
H
Z i+1 (G) Z i+1 (H)
Z
×
=Z
×Z
=
×
Z i (G) Z i (H)
Z i (G)
Z i (H)
Z i (G)
Z i (H)
so Z i+1 (G × H) = Z i+1 (G) × Z i+1 (H).
Q In the case of an infinite product, if Gi is a nilpotent group of class i then
Gi is not nilpotent since
Y
Y
Y
Z k ( Gi ) =
Z k (Gi ) 6=
Gi
since it doesn’t contain Gk+1 .
HW4.ex02: (6.45 in Rotman 1st ed.)[Wielandt] A finite group G is nilpotent
iff every maximal subgroup is normal.
Ans: If G is not nilpotent then there must be a nonnormal Sylow subgroup
P . This means that N (P ) 6= G. Then any maximal subgroup containing
N (P ) cannot be normal. Since every subgroup of G containing N (P ) is
selfnormalizing.
Conversely, if G is nilpotent then the Sylow subgroups
Q Pi are all normal
and every maximal subgroup of G has the form N × i6=j Pi where N < Pj
is maximal. But we proved that maximal subgroups of p-groups are normal
(with factor group Z/p).
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Automorphisms
HW4.ex03: (7.11 in Rotman 4th ed) If G is a finite group and Aut(G) acts
transitively1 on the set G − {1} then G is elementary abelian.
Ans: First we claim that every nontrivial element of G has order p for some
fixed prime p. To see this let p be any prime dividing |G|. By Cauchy G has
an element g 6= 1 of order p. If h is another nontrivial element of G then
there is an automorphism φ of G so that φ(g) = h. Thus o(h) = o(g) = p.
Now we just have to prove that G is abelian. Since G is a p-group it has
a nontrivial center Z(G). But Z(G) is characteristic so φ(Z(G)) = Z(G)
for all φ ∈ Aut(G). If Z(G) < G then the automorphisms of G could not
send the elements of Z(G) to the elements of G − Z(G). This would be a
contradiction. Thus G = Z(G) is abelian. Since every nontrivial element has
order p it is elementary abelian.
HW4.ex04: (7.15 in Rotman 4th ed) Show that Aut(D8 ) ∼
= D8 but D8 is not
complete.
Ans: Since D8 = hs, t|s4 , t2 , ststi has a central element s2 , it is not complete.
To find Aut(D8 ) we look at the 8 elements of D8 . (We are using the newer
notation from the 4th ed of Rotman: D2n is the dihedral group of order 2n.)
D8 = {1, s, s3 , s2 , t, ts, ts2 , ts3 }
Of these only s, s3 have order 4 and t, ts, ts2 , ts3 are noncentral of order 2.
Thus there are at most 2 × 4 = 8 automorphisms φ of D8 since φ(s) = s, s3
has only two choices and φ(t) = tsi has 4 choices.
Georgi and Anish point out that D16 = ha, b|a4 , b2 , ababi has a subgroup
ha2 , bi isomorphic to D8 . Since the index is 2 this subgroup is normal so
D8 C D16 . The conjugation action gives a homomorphism
φ : D16 → Aut(D8 )
The kernel of φ is the centralizer of D8 in D16 . This is {1, a4 } since these are
the only elements of D16 which commute with both b and a2 . Thus
D16
Aut(D8 ) ∼
= 4 ∼
= D8
ha i
since there is an epimorphism ψ : D16 → D8 with kernel ha4 i given by
ψ(a) = s, ψ(b) = t.
Semidirect Products
HW4.ex05: Prove that Q (the quaternion group of order 8) is not a semidirect
product of Z/4 by Z/2.
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G acts transitively on X if for any x, y ∈ X there is a g ∈ G so that gx = y
2
Ans. Q cannot be a semidirect product because it has only one element of
order 2. [If Q were a semidirect product of A and B it would have complementary subgroups isomorphic to A and B. But every nontrivial 2-group
has an element of order 2 by Cauchy’s theorem so Q would have at least two
elements of order 2.]
HW4.ex06: (7.20 in Rotman 4th ed) If G is complete prove that Hol(G) ∼
=
G × G.
Ans: Since G is complete, we have an isomorphism:
φ : G → Aut(G)
sending g ∈ G to the inner automorphism φg . Thus the holomorph of G is
the semidirect product G o Aut(G) = G o G with multiplication given by:
(a, g)(b, h) = (aφg (b), gh) = (agbg −1 , gh).
≈
An isomorphism ψ : G × G −
→ G o G is given by
ψ(a, g) = (ag −1 , g).
This is obviously a bijection and it is easy to verify that it is a homomorphism:
ψ(a, g)ψ(b, h) = (ag −1 , g)(bh−1 , h) = (ag −1 gbh−1 g −1 , gh) = ψ(ab, gh)
OK, so that is the proof but how did I get the equation for ψ?
My method was to look for two commuting normal subgroups of G o G
which are isomorphic to G. One such subgroup is G o 1 C G o G. The other
subgroup must consist of elements (g, h) which commute with all elements
of G o 1:
(a, 1)(g, h) = (g, h)(a, 1)
(ag, h) = (ghah−1 , h)
In other words a = ghah−1 g −1 which holds if gh = 1. So g = h−1 and
H = {(h−1 , h)}. The isomorphism with G is given by projection to the
second factor so I guessed that
ψ(g, h) = (g, 1)(h−1 , h) = (gh−1 , h).
Note that the computation (in black above) is still needed to verify that this
is a homomorphism. The argument in red assumes that the result is true.
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