Download Solutions 8 - D-MATH

D-MATH
Prof. Brent Doran
Algebra I
HS 2013
Solutions 8
1. We start by eliminating non-simple groups of order < 60. First, recall that pgroups (pn , n > 1) have non-trivial center and that the center of a group is a
normal subgroup. Recall from exercise sheet 7, that there are no simple groups of
order pq or p2 q. The candidates left are then the groups of order
24, 30, 36, 40, 42, 44, 48 and 56.
For the groups of order 24 and 36, the argument is the following. Here s3 is either
1 or 4. If s3 = 1 then the group is not simple and we are done. If s3 = 4, then
let X be the set of Sylow 3-subgroups, and let G act on X by conjugation. This
is equivalent to the existence of a homomorphism ϕ : G → S4 . If |G| = 36, then
this homomorphism is not injective and the kernel is a normal subgroup in G.
If |G| = 24 and we assume ϕ to be an isomorphism, then G ' S4 with normal
subgroup A4 .
Let |G| = 30. Then s5 divides 6 and s5 ≡ 1 mod 5. Hence s5 is either 1 or 6.
If s5 = 1 then the subgroup in question is normal. If s5 = 6, then the six Sylow
5-subgroups must be pairwise disjoint - they have prime order. Then G contains
6·4 = 24 elements of order 5. On the other hand, s3 is either 1 or 10. If s3 = 1 then
the Sylow 3-subgroup is normal. If s3 = 10 and there are 20 elements of order 3.
But this accounts for 24 + 20 distinct elements in the group, which is not possible.
For groups of order 40, 42 and 44, the argument is the following. Say |G| = 42 =
6 · 7, then s7 divides 6 and s7 ≡ 1 mod 7. Hence s7 = 1, which is equivalent to say
that that G has a normal Sylow 7-subgroup, hence it is not a simple group.
If |G| = 48, then s2 = 1 or 3. If s2 = 3 and we again consider the conjugation action
of G on S3 . Then G has a normal subgroup of 48/6 elements, by the isomorphism
theorem.
Let |G| = 56 = 23 · 7. Then s7 = 1 or 8. Assume there are 8 Sylow 7-subgroups.
Then G contains 8 subgroups of order 7. These subgroups must be disjoint - they
have prime order. Thus G has 8 · 6 = 48 elements of order 7. Now if G has also 7
Sylow 2-groups we again get a contradiction as in the case |G| = 30.
2. Let H be the subgroup generated by ab. Note that (ab)−1 = ba in G. It is clear
that H is a proper subgroup since no word on {ab, ba} can have odd length, and
that it has index two as H ∪ aH.
3. A finite list of generators is given by taking all the elements of the group - there
are only finitely many. The composition of two elements g1 and g2 of the group
under the group’s law of composition coincides with an element g3 of the group. In
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particular, this is g1 g2 g3−1 = 1. All relations in the group can be described in this
manner. Since the multiplication table for a group of order n contains n2 entries,
the set of relations determining the group is also finite.
4. We show that S = {x, y, z} freely generate the free group F3 . Let w be a reduced
word in S ∪ S −1 , given by w = xε1 y ε2 z ε3 xε4 · · · z εn where each εi ∈ {−1, 0, 1}. Then
w may be rewritten a2ε1 b2ε2 (ab)ε3 · · · (ab)εn and w 6= 1 in {a, b}.
5. (a) The free group F is the group of reduced words in S ∪ S −1 . Let w ∈ F . The
homomorphism ϕ described on the exercise sheet must satisfy
ε
ε
ϕ(w) = ϕ(ai1i1 · · · ainin ) = ϕ(ai1 )εi1 · · · ϕ(ain )εin = f (ai1 )εi1 · · · f (ain )εin .
This indeed defines a homomorphism :
ϕ(vw−1 ) = f (ai1 )εi1 · · · f (ain )εin f (ajm )−εjm · · · f (aj1 )−εj1 = ϕ(v)ϕ(w)−1
ε
ε
jm
and this holds also if ainin = ajm
. In particular, ϕ is also well-defined.
(b) Set ϕ0 (gN ) = ϕ(g). We start by showing that this is well-defined. Assume
that gN = g 0 N . In particular, there is an element n ∈ N such that gn = g 0 .
Then ϕ(g) = ϕ(g 0 ) because ϕ is a homomorphism of groups that contains N
in its kernel. That is to say, ϕ0 (gN ) = ϕ0 (g 0 N ) if gN = g 0 N . The condition
ϕ0 ◦ π = ϕ forces ϕ0 to be unique. It is left to show that ϕ0 is actually a
homomorphism. In fact,
ϕ0 (g1 N g2−1 N ) = ϕ0 (g1 g2−1 N ) = ϕ(g1 g2−1 ) = ϕ0 (g1 N )ϕ0 (g2−1 N ).
6. (a) Both statements are shown as direct applications of the mapping property of
free groups.
First, let S be a set of m elements and let f : S → Fn be an injective function
mapping the m elements of S to a choice of m distinct generators in Fn . By
the mapping property, there exists a homomorphism ϕ : F → Fn . Because F
is generated by S, we have F ' Fm and by the commutativity property ϕ is
injective but not surjective. Hence Fm is isomorphic to the subgroup ϕ(F ) of
Fn .
Second, let S be a set of n elements and a map f : S → Fm as above. Observe
that f is surjective but not injective, and that then the homomorphism ϕ has
image Fm and a non-trivial kernel K. By the first isomorphism theorem, it
follows that Fm is isomorphic to Fn /K.
(b) The product Z × Z is an abelian group while any free group of rank n > 2 is
non-abelian. But F1 × F1 is also not isomorphic to Z or {1}.
(c) We consider the arbitrarily chosen element x ∈ Fn as a (unique) reduced word
in the generators a1 , . . . , an . We can show that x doesn’t commute with any
other reduced word of Fn . In fact, this follows from the fact that [ai , aj ] 6= 1
for i 6= j.
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7. By ex. 3(a) of this problem set, there is a unique homomorphism ϕ : F2 → A that
extends the map of sets f : {x, y} → A with f (x) = u and f (y) = v. Since ϕ is
a homomorphism taking values in an abelian group, it vanishes on commutators
(ex. 2 of the midterm). Let R denote the normal subgroup of F2 generated by
[x, y]. Then by ex. 3(b), there is a unique homomorphism ϕ0 : F2 /R → A that
extends ϕ. That is, ϕ0 (x) = ϕ(x) = f (x) = u and ϕ0 (y) = v.
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