H1 Solutions and H2

Homework 2 due Fri Oct 10, and Homework 1 solutions
Reading. Continue with last week’s reading if you haven’t had a chance to finish it.
Then read about solvable groups in section 9 of the notes, plus the first proposition (9.8)
on nilpotent groups (so you understand that finite p-groups are nilpotent, hence solvable).
Regarding the reading assigned last week, there are a few topics I haven’t had time for in
lecture. In particular:
1. Balanced products and induced G-sets aren’t needed immediately, but you could still
take a look. One of the main reasons I’ve included these topics here is that they provide a
simpler predecessor of “tensor products of modules” and “induced representations”, to be
discussed later. I personally always found the G-set prototype easier than tensor products,
and therefore useful as a guide. But you could just peruse this material for now if you want.
2. Products and coproducts of G-sets. I think these are pretty straightforward, except
for my references to the “categorical product” and “categorical coproduct”. You can get
by without knowing the categorical interpretation, but I highly recommend it (see the cat.
theory notes). The point is that the products of sets, groups, rings, G-sets, topological
spaces etc. etc. all have something in common, and category theory neatly pins down what
that something is. Coproducts are completely analogous (although perhaps less familiar);
one just follows the example of products but with “arrows reversed”.
Homework 2. Note: Use of Feit-Thompson or Burnside’s pa q b theorem not allowed!
1. Let G be a finite group, p a prime dividing |G|; say |G| = spk with s prime to p. A
p-complement in G is a subgroup H of order s. The three parts below are independent of
each other.
a) Suppose G has a normal p-complement H. Show that the extension H−→G−→G/H
splits.
b) Show that GL3 F2 has a 2-complement.
Remarks: GL3 F2 is an interesting group; for one thing, it is the second smallest nonabelian simple group (A5 being the smallest). However, you don’t need the fact that it’s
simple for this problem. In fact part (b) is typical of “real-life” mathematics in that often
it isn’t clear how much of the given data is relevant for the desired conclusion (in real-life
results don’t come to you with tailor-made hypotheses!). For example, could it be true that
any group of order 168 has a 2-complement?
c) Give an example of G, p where no p-complement exists.
2. Exercise M1 in the notes.
3. Exercise E1 in the notes.
Homework 1 solutions.
C1. a) In an invertible n × n matrix over Fq , the first column can be any nonzero vector.
So there are q n − 1 choices for the first column. The second column can be any vector which
1
is not a multiple of the first, so there are q n − q choices for it. The third column can be any
vector not in the span of the first two, so there are q n − q 2 choices, and so on. Hence
|GLn Fq | =
n−1
Y
n
(q n − q i ) = q ( 2 )
i=0
n
Y
(q i − 1).
i=1
b) Since the elements of Un F are matrices whose above diagonal entries are arbitrary, we
n
have |Un Fq | = q ( 2 ) . So by part (a) it is a p-Sylow subgroup.
c) Lemma. NGLn F Un F = Bn F .
Proof: By direct inspection, Dn F ⊂ NGLn F Un F . Hence Bn F ⊂ NGLn F Un F , since Dn F, Un F
generate Bn F . For the reverse inclusion, suppose g ∈ NGLn F Un F . By the Bruhat decomposition (see the notes), we can write g = b1 wb2 with bi ∈ Bn F and w ∈ Sn . Then
U = gU g −1 = b1 wU w−1 b−1
1 and hence
wU w−1 = b−1
1 U b1 = U,
i.e. w normalizes U . Since conjugation by w just permutes the rows by w and the columns
by w−1 , this forces w = e (the identity matrix), proving the lemma.
Then by the second Sylow theorem and parts a-b above, the number of p-Sylow subgroups
is
[GLn F : NGLn F Un F ] = [GLn F : Bn F ] =
n
Y
qi
i=1
−1
.
q−1
d) Note that for any field F , the homomorphism det : GLn F −→F × is surjective, indeed
split surjective (think of diagonal matrices with aii = 1 for i > 1). So we get
|SLn Fq | =
|GLn Fq |/|F×
q |
n−1
n Y
(
)
2
=q
(q i − 1).
i=1
Now for any field F , C(SLn F ) consists of scalar matrices of determinant 1, and so is
isomorphic to H := {a ∈ F × : an = 1}. Thus |P SLn Fq | = |SLn Fq |/|H|, but what is the
order of H? Well, since F×
q is cyclic, the order is d := gcd(n, q−1). So |P SLn Fq | = |SLn Fq |/d.
Example. The center of SL3 F5 is trivial, since gcd(3, 4) = 1. In other words, there are
no non-trivial cube roots of 1 in F5 . On the other hand, F7 has three cube roots of 1, and
C(SL3 F7 ) = C3 .
A1. Suppose Q8 acts on a set X with |X| < 8. Then all orbits have size < 8, so all
isotropy groups are non-trivial. But every non-trivial subgroup of Q8 contains the center C2
(note the six elements of order 4 all have square = −1), so C2 is contained in every isotropy
group. Thus if ρ : Q8 −→P erm X is the corresponding homomorphism, C2 ⊂ Ker ρ and ρ is
not injective. QED.
2
A2. a) GL2 F2 acts on the set F22 − 0, which has three elements. The action is clearly
faithful, so that after ordering the three elements we have an injective homomorphism φ :
GL2 F2 −→S3 . Since source and target have the same order 6, φ is an isomorphism.
Af f2 F2 acts on the set F22 , which has four elements. Again we get a faithful action and
an injective homomorphism ψ : Af f2 F2 −→S4 . Since source and target have the same order
24, ψ is an isomorphism.
b) Note that |P(F23 )| = 4 (there are 8 nonzero vectors in F23 ; now take F×
3 orbits to get
2
|P| = 8/2 = 4). The action of SL2 F3 on F3 yields an action on P, and since the center (scalar
matrices) acts trivially on P, we get an action of P SL2 F3 on P and hence a homomorphism
φ : P SL2 F3 −→S4 . If a linear transformation stabilizes every line, it must be a scalar matrix.
This shows that φ is injective. On the other hand since
|P SL2 F3 | =
|SL2 F3 |
= 24/2 = 12,
|F×
3|
and A4 is the unique subgroup of S4 of order 12, φ is an isomorphism onto A4 .
B3. For each player x, let Pk (x) denote the set of players that xR could possibly play
in the k-th round (should x be lucky enough to get that far). Then 7 C2 is the subgroup
leaving Pk (x) invariant for all k, x. For example, reversing the numbers of each initial pair
(i.e. 1 and 2, 3 and 4 etc.) obviously doesn’t change the draw in any essential way. Similarly
switching the pairs (1, 2) and (3, 4) has no effect on who player 1 might meet in the second
round. Or going to another extreme, switching the entire upper block of 64 players with the
lower block of 64 makes no difference either.
B4. Let Gk = k Cp , where G0 interpreted as the trivial group. Then I claim that
Gk ⊂ Spk . To see this, induct on k and use the inclusions
R
Gk = Gk−1
Z
Cp ⊂ Spk−1
Z
Sp ⊂ Spk ,
where the equality is by definition, the first inclusion is by induction, and the second is by
B1.
Thus we have inclusions
G := Ga00 × Ga11 × ... × Gakk ⊂ Sa0 × Spa1 × ... × Spakk ⊂ Sn .
Here the second inclusion comes from the evident partition of [n] determined by the p-adic
expansion. So all we need to show is that G has the right order, or νp |G| = νp n!.
Let ck = 1 + p + ... + pk−1 . Then by induction on k we have νp |Gk | = ck . Hence
νp |G| =
k
X
ai c i .
i=1
To compute νp (n!), first consider the case n = apk , 0 ≤ a < p. In n! the number of
factors divisible by p is apk−1 , the number divisible by p2 is apk−2 , and so on. So
νp (apk ) = ack .
3
k
k
In general write n = ak pk + m, so n! = (ak pk )!q where q = m
j=1 (ak p + j). Since m < p ,
νp (ak pk + j) = νp j for all j, and νp q = νp m!. Hence by induction on k we get
Q
νp (n!) = ak ck + νp (m!) = ak ck +
k−1
X
i=1
4
ai c i .

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