Download Algebra Qual Solutions September 12, 2009 UCLA ALGEBRA QUALIFYING EXAM Solutions

Algebra Qual Solutions
September 12, 2009
UCLA ALGEBRA QUALIFYING EXAM
Solutions
JED YANG
G. Groups
Convention. Conjugation on the right (such as P g ) is for typesetting convenience.
Errors. Qual Problem G6f2-b is false. We need n > 1 for Qual Problem G5w1-a.
G9s1. Let N be a normal subgroup of a finite group G and P a p-Sylow subgroup
of G. Prove that P ∩ N is a p-Sylow subgroup of N .
Proof. Repeat of Qual Problem G5f2.
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G9s2. Let A be an abelian group generated by n elements. Prove that any subgroup of A can be generated by n elements.
Proof. Repeat of Qual Problem G5f1.
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G9s3. Let G be a finite group and H < G a subgroup of index n. Suppose that
xH ∩ Hy 6= ∅ for any elements x, y ∈ G\H. Prove that |H| ≥ n2 − n. [Hint:
consider an action of G on (G/H) × (G/H).]
Proof.
G8f1. Let G be a finite group of order g and Z[G] ⊂ Q[G] be the group algebras
of G with integer and rational coefficients, respectively. Let
eZ[G] = {ea ∈ Q[G] : a ∈ Z[G]}
for e = g
−1
P
h∈G
h ∈ Q[G], and define a group
G′ = eZ[G]/(Z[G] ∩ eZ[G]).
Prove that G′ is a group of order g. Find a necessary and sufficient condition to
have G′ ∼
= G as groups.
Proof.
G8f2. Prove or disprove:
(a) The group GL2 (Q) of 2 × 2 matrices with rational coefficients has finite
cyclic subgroups of order bigger than any given positive integer N .
Proof. a
(b) The group GL2 (R) of 2 × 2 matrices with real coefficients has finite cyclic
subgroups of order bigger than any given positive integer N .
Proof. b
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G8f3. Let G be an additive abelian group such that multiplication by n: x 7→ nx
is surjective for all positive integers n. Let
G[n] = {x ∈ G : nx = 0}
and p be a prime.
(a) Prove that for a given integer m ≥ 1, there are only finitely many subgroups
H of order pm in G if G[p] is finite.
Proof. a
(b) Find a formula of the number of subgroups of G of order p if the order of
G[p] is p3 .
Proof. b
G8s1. Let p be a prime number. Show that a subgroup G of Sp which contains an
element of order p and which contains a transposition must be the whole of Sp .
Proof. Write elements of Sp in cycle notation. As p is prime, an element of order p
must necessarily be a p-cycle. WLOG, let it be g = (1, 2, 3, . . . , p). Let the transposition be (a, a + b), a, b ∈ Z/p and a 6= 0. It is obvious that Sp is generated by all
transpositions. Conjugating (a, a + b) by g repeatedly, we will get all transpositions
of the form (i, i + b) for all i ∈ Z/p. Conjugating (a, a + b) by (a + b, a + kb) yields
(a, a+kb). Thus by induction, we get all transpositions of the form (a, a+kb) for all
k ∈ (Z/p)× . Now conjugating by g repeatedly again, we will get all transpositions
of the form (i, i + kb) for all i, k ∈ Z/p, k 6= 0. This completes the proof.
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G8s2. Let G = D2n be the dihedral group of order 2n where n ≥ 3. Prove that
Aut(G) is isomorphic to the group of 2 × 2 matrices of the form
½µ
¶
¾
α β
H=
: α ∈ (Z/n)× , β ∈ Z/n .
0 1
­
®
Proof. Recall that D2n = ρ, σ : ρn = σ 2 = (ρσ)2 = 1 . This gives σρ = ρ−1 σ, and
more generally, σρk = ρ−k σ. Thus we may uniquely write¡ each
element of D2n as
¢
ρa σ b , with a ∈ Z/p and b ∈ Z/2. Identifying ρa σ b with ab , we obtain a natural
left action of H by left multiplication. By direct computation, we check that each
element of H actually act by automorphism. This yields a permutation representation, namely, a homomorphism ϕ : H −→ Aut(G), which is also easily checked.
Now ϕ is injective, as fixing ρ and σ force α = 1 and β = 0, respectively. It remains
to show that ϕ is surjective, or equivalently, |Aut(G)| ≤ |H|. An automorphism
necessarily sends ρ to an element of order n, namely ρk for (n, k) = 1, so has
|(Z/n)× | choices. It also sends σ to an element of order 2 that is not in the centre,
namely, ρk σ for any k ∈ Z/n. This gives us at most |H| choices of (potential)
automorphisms, completing the proof.
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G8s3. Let K be a normal subgroup of a finite group G. Let p be a prime. Let Np
be the number of p-Sylow subgroups of G and Np′ the number in K.
(a) Show that Np = [G : NG (P )] where P is any p-Sylow subgroup of G and
NG (P ) is the normaliser of P in G.
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Proof. Let P be a p-Sylow subgroup of G, and let S be the conjugates of
P by G. By the orbit-stabiliser theorem, the orbit containing P , namely S,
has the same cardinality as the index of the stabiliser of P , namely NG (P ).
Thus we obtain |S| = [G : NG (P )].
If a p-subgroup H normalises a p-Sylow subgroup P , then H < P .
[Indeed, by second isomorphism theorem, P ⊳ P H < G, and P H/P ∼
=
H/(H ∩ P ). But since P H is a p-group, P ⊂ P H and is p-Sylow, P H = P ,
hence H = H ∩ P , hence H < P .]
It remains to prove that all p-Sylow subgroups are conjugate. Take another p-Sylow subgroup P ′ . Let P ′ act on S by conjugation. Notice that
since P < NG (P ), we get [G : NG (P )] | [G : P ], which has no factor of p
by maximality of P . Hence |S| 6≡ 0 (mod p). On the other hand, by the
orbit formula, |S| is congruent to the number of fixed points of the action
modulo p (see Qual Problem G7f1). This means there is some conjugate
that is normalised by P ′ . Thus that conjugate is a subgroup of, hence equal
to, P ′ , as desired.
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(b) Prove that Np′ divides Np .
Proof. Let P ′ be a p-Sylow subgroup of K, and let S ′ be the conjugates of
P ′ by K. By part (a), we get Np′ = |S ′ |. On the other hand, conjugates of
P ′ by K and by G are the same, since K ⊳G. Thus |S ′ | = [G : NG (P ′ )]. But
P ′ < P for some p-Sylow subgroup P of G, yielding the desired result. ¤
G7f1. Let F be a finite field of characteristic p and let G be a subgroup of order
pa of the group GL(n, F ) of invertible n by n matrices with entries in F . Show that
there is a nonzero vector v in F n such that gv = v for every g ∈ G.
Proof. Let G act on F n naturally by coordinate-wise multiplication. By the orbitstabiliser theorem, if Gx is the orbit
P containingPx, and Gx is the stabiliser of x, then
|Gx| = [G : Gx ]. Thus |F n | = Gx |Gx| = Gx [G : Gx ]. Since G is a p-group,
[G : Gx ] is a power of p, thus a multiple of p save G = Gx . Thus, modulo p, this
counts the number of fixed points of F n . Since |F n | ≡ 0 (mod p) and 0 is a fixed
point, there is (at least p − 1) nonzero v that is also a fixed point.
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G7f2. Let M be the submodule of Z3 generated by elements (0, 3, 2), (6, 48, 24),
and (6, 24, 12). Describe the quotient group Z3 /M by giving a product of cyclic
groups to which Z3 /M is isomorphic.
α
Proof. Consider the short exact sequence 0 → M −
→ Z3 → Z3 /M → 0. We put the
matrix associated with α in Smith Normal Form. We get

 

0 6
6
1 0 0
3 48 24 ∼ 0 6 0  ,
2 24 12
0 0 12
yielding Z3 /M ∼
= Z/6 × Z/12.
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G7f3. A group G acts doubly transitively on a set X if for each pair (x1 , x2 ) and
(y1 , y2 ), with x1 6= x2 and y1 6= y2 of pairs of points of X, there exists a g ∈ G such
that gx1 = y1 and gx2 = y2 . Show that if a finite G acts non-trivially and doubly
transitively on X, then the stabiliser Sx of any point x in X is a maximal proper
subgroup of G. (Here a subgroup M of G is maximal proper if M is not equal to
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Yang 4
G and if, for any subgroup H of G which contains M , either H = M or H = G
holds.)
Proof. Obviously Sx is a proper subgroup of G. Let H < G contain Sx properly.
It remains to show that H = G.
Let h ∈ H\Sx . By definition, hx = y for some y 6= x. Let g ∈ G such that
gx = y. Then h−1 g ∈ Sx as it fixes x, hence g ∈ hSx ⊂ H.
Let z ∈ X\{x, y}. By double transitivity, there exists g ∈ G such that gx = y
and gy = z. By the above paragraph, g ∈ H, so gh ∈ H. But ghx = z, so by the
above paragraph applied to gh, anything mapping x to z is in H. Thus H = G. ¤
G.1. G7s1. Let G be a simple group containing an element of order 21. Prove that
every proper subgroup of G has index at least 10.
Proof. Let H < G be proper, so n = [G : H] > 1. Consider the natural action
of G on the coset space G/H by left multiplication. We obtain a homomorphism
ϕ : G −→ Sym(G/H) ∼
= Sn . Since G is simple this homomorphism is injective and
G embeds in Sn . Hence there is an element of order 21 in Sn . It is either a 21-cycle
or the produdct of disjoint 3- and 7-cycles. In either case, n ≥ 10.
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G7s2. Find the number of subgroups of Zn of index 5.
Proof. Let H < Zn be a subgroup of index 5. Since Zn is abelian, H is normal.
Thus H is the kernel of a surjective homomorphism Zn −→ Z/5. And conversely,
a kernel of such a homomorphism is a subgroup of index 5.
We first count the number of surjective homomorphisms. Notice that Zn has n
generators, each one being sent to one of 5 elements of Z/5. The homomorphism is
surjective if and only if not all of them are mapped to 0. As such, there are 5n − 1
different surjective homomorphisms.
Surjective homomorphisms Zn −→ Z/5 admit the same kernel if and only if they
differ by an automorphism of Z/5. Indeed, let σ ∈ Aut(Z/5), and ϕ : Zn −→ Z/5
a surjective homomorphism, then σ ◦ ϕ is also a surjective homomorphism with the
same kernel. Conversely, if two surjective homomorphisms ϕ, ϕ′ : Zn −→ Z/5 have
the same kernel, then we get this commutative diagram
ker ϕ
id
²
ker ϕ′
/ Zn
ϕ
/0
σ
id
²
/ Zn
/ Z/5
ϕ′
²
/ Z/5
/0
where the rows are exact. By part of the 5-Lemma, we get that σ is injective, and
thus an isomorphism as Z/5 is finite. Thus ϕ′ = σ ◦ ϕ, with σ ∈ Aut(Z/5). As such,
the number of different kernels is (5n − 1)/4, which is the number of subgroups of
Zn of index 5.
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G.2. G7s3. Let G be a group with cyclic automorphism group Aut(G). Prove that
G is abelian.
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Proof. Recall that conjugation by an element gives an inner automorphism. It is
easily seen that the set of such inner automorphisms form a group Inn(G), which
is a subgroup of Aut(G). Since a subgroup of a cyclic group is cyclic (cf. Qual
Problem G6f3-a), we conclude that Inn(G) is cyclic.
Consider the group homomorphism ϕ : G −→ Inn(G) given by g 7→ ϕg , where
ϕg : G −→ G is conjugation by g. The kernel of ϕ is evidently the centre Z(G).
Indeed, Z(G) is obviously in the kernel; and conversely, if ϕg is trivial, then gxg −1 =
x for all x, so g commutes with all x. Thus by the second group isomorphism
theorem, G/Z(G) ∼
= Inn(G). So G/Z(G) is cyclic with generator g, say. Take
ai ∈ G for i = 1, 2. We have that ai = g bi zi for some bi ∈ Z and zi ∈ Z(G).
Since the zi commute with everything and the g bi commute with each other, the
ai commute.
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G.3. G6f1. List all finite groups G whose automorphism group has prime order.
Proof. If Aut(G) has prime order, then it is cyclic. By Qual Problem G7s3, G
is abelian. By the fundamental theorem of finitely generated abelian groups, G ∼
=
Z/pa1 1 ⊕. . .⊕Z/pann , where pi are (not necessarily distinct) primes and ai ≥ 1. Recall
that the automorphism Aut(Z/pa ) is simply Z/φ(pa ), where φ is the Euler function.
Indeed, a generator, say, 1 ∈ Z/pa is sent to some number relatively prime to pa ,
and there are φ(pa ) = pa−1 (p − 1) choices. No matter what the other factors of G
are, the order of Aut(G) must have φ(pa ) as a factor. We see that p−1 is composite
if p ≥ 5. Thus p = 2 or 3. For φ(pa ) to be prime, if p = 2 then a = 1 or 2 and if p = 3
then a = 1. Thus G is a direct sum of mq copies of Z/q, for q = 2, 3, 4. The order
of Aut(G) then is a multiple of φ(2)m2 φ(3)m3 φ(4)m4 = 2m3 +m4 . It is obvious then
that we can have at most one (copy) of Z/3 and Z/4. An automorphism can also
send generators of different summands to each other. However, since order must be
preserved, this only happens when the summands are the same, namely, multiple
copies of Z/2. The order of Aut(G) is therefore m2 !2m3 +m4 . It is thus clear that
for this to be prime, the only possibilities are Z/2 ⊕ Z/2, Z/3, Z/2 ⊕ Z/3 ∼
= Z/6,
Z/4, and Z/2 ⊕ Z/4. TODO: This is utterly wrong, since G must be cyclic.
G.4. G6f2. Let G be a finite group, and H a non-normal subgroup of G of index
n > 1.
(a) Show that if |H| is divisible by a prime p ≥ n, then H cannot be a simple
group.
Proof. Let G act naturally on left cosets of H by left multiplication. We
thus obtain a homomorphism ϕ : G −→ Sym(G/H) ∼
= Sn , with kernel
K ⊳ G. By second group isomorphism theorem, we get G/K ∼
= im(ϕ),
hence [G : K] | n!. It is evident that K < H < G (those that fix H
are from H), so since K ⊳ G, we have K ⊳ H as well. Now K 6= H
otherwise H ⊳ G, a contradiction. Thus if H is simple, K = 1. Then
np | [G : H] · |H| = |G| = [G : K] | n!, a contradiction for p ≥ n.
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(b) Show that there is no simple group of order 504.
Proof. This is false, as PSL(2, 8) is a simple group of order 504.
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G.5. G6f3. Let
SL2 (Z/p) =
½µ
a b
c d
¶
¾
with a, b, c, d ∈ Z/p, ad − bc = 1
where p is an odd prime.
(a) Prove that any subgroup of a cyclic group is cyclic.
Proof. Let G = hgi be a cyclic group and let H < G. Let n > 0 be
minimal such that g n ∈ H. Then H = hg n i. Indeed, obviously hg n i < H.
Conversely, take g k ∈ H < G. There exists q ∈ Z and 0 ≤ r < n such that
k = nq + r. Thus g r ∈ H contradicting minimality of n unless r = 0, hence
n | k.
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(b) Compute the order of SL2 (Z/p).
Proof. By Qual Problem G6s2, we get SL2 (Z/p) = (p2 −1)(p2 −p)/(p−1) =
(p − 1)p(p + 1).
¤
(c) Prove that for any odd prime ℓ, the Sylow ℓ-subgroup of SL2 (Z/p) is cyclic.
Proof. Fix an odd prime ℓ. Recall by Sylow theorem that the ℓ-Sylow
subgroups are conjugate. Therefore it suffice to demonstrate one Sylow
ℓ-subgroup that is cyclic. Notice that the factors p − 1, p, and p + 1 do
not share odd prime factors. Therefore if a Sylow ℓ-subgroup has order ℓa ,
then ℓa must divide precisely one of those factors. Since a cylic group has
subgroups of all orders dividing the order of the group (cf. Qual Problem
G0f3), it remains to find cyclic subgroups of orders p − 1, p, and p + 1.
The subgroup
½µ
¶
¾
x
0
:
x
=
6
0
0 x−1
is cyclic of order p − 1, generated by any x 6= 1. The Sylow p-subgroup is
cyclic of order p. TODO: find a subgroup of order p + 1 (it does exist!)
G.6. G6s1. Let G be a finite group. Let K be a normal subgroup of G and P a
p-Sylow subgroup of K. Show that G = KNG (P ).
Proof. Let x ∈ G. As x is in some right coset Kg −1 of K, we may write x = kg −1
for k ∈ K and g ∈ G. Now as P < K, we also have P g < K g . But as K ⊳ G, we
have K g = K, hence P g is a p-Sylow subgroup of K. Recall that p-Sylow subgroups
(of K) are conjugates. Thus there exists h ∈ K such that (P g )h = P . That is,
P gh = P , hence gh ∈ NG (P ). We thus conclude that x = kg −1 = kh · (gh)−1 ∈
KNG (P ), as desired.
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G.7. G6s2. Let SL2 (F4 ) be the special linear group with coefficients in F4 .
(a) What is the order of SL2 (F4 )?
Proof. 60. Recall that SLn (Fq ) is the kernel of det : GLn (Fq ) −→ (Fq )× .
Therefore |SLn (Fq )| = |GLn (Fq )| / |(Fq )× |. Now we compute the order of
the general linear group. We choose elements for the matrix column by
column. There are q n −q t choices for the t+1st
Qncolumn, omitting the span of
the first t columns. Thus we get GLn (Fq ) = i=1 (q n −q i−1 ). Now |(Fq )× | =
q−1 thus we finally get, in particular, SL2 (F4 ) = (42 −1)(42 −4)/3 = 60. ¤
(b) Show there is an isomorphism from SL2 (F4 ) to A5 .
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Yang 7
Proof. There are five one-dimensional subspaces (“lines”) of the vector
space F24 of dimension two over F4 . Indeed, there are a total of 42 − 1 = 15
nonzero elements in F24 , each spanning a line; but we over count by a factor of 4 − 1 = 3, the number of nonzero elements in each line. Now it is
obvious that SL2 (F4 ) act by permuting the lines, thus we get a homomorphism SL2 (F4 ) −→ S5 . This is injective; indeed, a non-identity matrix of
dimension 2 can at most fix a 2-dimensional eigenspace. Thus by the first
group isomorphism theorem, SL2 (F4 ) is isomorphic to its image under the
monomorphism. But |S5 | = 5! = 120 = 2 |SL2 (F4 )|, thus it is isomorphic to
A5 , the only subgroup of index 2 by Qual Problem G3w3.
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G.8. G6s3. Let G be a group of order 2000 and suppose that P and P ′ are two
distinct Sylow 5-subgroups of G. Let I = P ∩ P ′ .
(a) Prove that |I| = 25.
Proof. Notice that |P | = |P ′ | = 125. Now I < P is a proper 5-subgroup
of P , hence has order 1, 5, or 25. Finally, |G| ≥ |P P ′ | = |P | |P ′ | / |I| gives
|I| ≥ 1252 /2000 > 5, as desired.
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(b) Show that the index of NG (I) is at most 2.
Proof. TODO: Check the following carefully. Many have claimed that this
is a hard problem.
Let S be the set of 5-Sylow subgroups. By Sylow, we have |S| ≡ 1
(mod 5) and |S| | 16, thus |S| = 16. Let I act on S by conjugation, and
S I be the fixed points of¯ this
¯ action. By orbit-stabiliser theorem (see Qual
Problem G7f1), |S| ≡ ¯S I ¯ (mod 5). If H ∈ S is fixed by I, then the pgroup I normalises the p-Sylow H, thus I < H (see Qual Problem G8s3).
Conversely, if I < H, obviously I fixes H. Since I is in P and P ′ , we
conclude that I is in precisely 6, 11, or 16 of the 5-Sylow subgroups. We
conclude that I is actually in all 16 of the subgroups by a combinatorial
argument (I am, after all, a combinatorialist).
Let K = K16 be a complete graph with vertex set S. Pick any edge
e = (P1 , P2 ) in K, let the colour of the edge be I(e) = P1 ∩ P2 . Let
V (J) = {H ∈ S : J < H}. Then V (I(e)) is a clique of colour I(e).
Conversely, if an edge (P3 , P4 ) has colour I(e), then P3 and P4 must contain
I(e), and hence are in V (I(e)). Thus the edge-colouring partitions the edges
of K into cliques of orders 6, 11, and 16.
Suppose, towards a contradiction, that K is not monochromatic.
¡ ¢ it
¡ ¢ Then
has a cliques of orders 6 and b cliques of order 11, satisfying a 62 + b 11
2 =
¡16¢
,
yielding
a
=
8
and
b
=
0.
Let
V
,
.
.
.
,
V
be
the
vertex
sets
of
1
8
2
the 8 cliques. Obviously |Vi ∩ Vj | ≤ 1, thus by the pigeonhole principle,
|(V1 ∪ . . . ∪ Vi ) ∩ Vj | ≤ i. This gives that |V1 ∪ V2 ∪ V3 ∪ V4 | ≥ 18, a contradiction.
Therefore K is monochromatic, thus I < Pi for all
T that
T i. Recalling
g
g
g
P
)
=
conjugation
by
g
∈
G
permutes
the
P
,
thus
I
=
(
i
i
i Pi =
i
T
¤
i Pi = I. Thus I is normal and [G : NG (I)] = 1, as desired.
G5f1. Let G be an abelian group generated by n elements. Prove that every
subgroup of G can also be generated by n elements.
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Yang 8
Proof. Proceed by induction on the number of generators. Base case: if G is
cyclic, then any subgroup is cyclic; see Qual Problem G6f3-a. Now suppose G =
hg1 , . . . , gn i is generated by n elements, and H < G a subgroup. Every element
g ∈ G can be written (not necessarily uniquely) as g = g1a1 · . . . · gnan for some
a1 , . . . , an ∈ Z. If every element admit such an expression with a1 = 0, then H is
a subgroup of hg2 , . . . , gn i, and we are done by induction. Otherwise, take h ∈ H
to be where |a1 | > 0 is minimal over all possible expressions of all elements g ∈ H.
Since not all elements admit an expression with a1 = 0, this h exists. Notice that
H is generated by h and H ∩ hg2 , . . . , gn i. Indeed, one inclusion is obvious. For the
other, take g ∈ H and notice that by multiplying a proper power of h, the exponent
of g1 is killed by minimality of |a1 |, exactly like the proof of the base case. Now
H ∩ hg2 , . . . , gn i is a subgroup of hg2 , . . . , gn i, hence can be generated by n − 1
elements by the induction hypothesis. Thus H can be generated by n elements, as
desired.
¤
Remark. This is false for nonabelian groups. The free group of rank 2 is generated
by 2 elements, but its commutator subgroup is not finitely generated. However, a
subgroup with finite index in a finitely generated group is finitely generated.
G.9. G5f2. Let N be a normal subgroup of G. Prove that for a Sylow p-subgroup
P of G, the intersection P ∩ N is a Sylow p-subgroup of N .
Proof. Let |G| = pa n, with p ∤ n, and N ⊳ G. Let P be a Sylow p-subgroup
of G, thus |P | = pa . Recall that as N is normal, P N is a subgroup of G. Let
|P N | = pb m, with p ∤ m. Since P N < G, we have b ≤ a. Moreover, its order
is |P N | = |P | · |N | / |P ∩ N |. Thus pb m/pa = |P N | / |P | = |N | / |P ∩ N | = [N :
P ∩ N ] ∈ Z. Thus it is evident that a = b and p ∤ [N : P ∩ N ], thus P ∩ N is a
p-Sylow subgroup of N .
¤
G.10. G5f3. Is there a nontrivial action of the alternating group A4 on a set of
two elements?
Proof. Negative. If a nontrivial action exists, then some g ∈ A4 must act nontrivially, namely, swap the two elements. But then g must have even order in
A4 . Recall that A4 has 8 elements of order 3, 3 elements of order 2 (the 2, 2permutations), and the identity element. Thus WLOG, (12)(34) acts nontrivially.
But then (12)(34)(123) = (243) also act nontrivially, a contradiction.
¤
G.11. G5w1. Recall a simple group is a group with no nontrivial normal subgroups.
(a) If G is a simple group that has a subgroup of index n, prove that the order
of G is a factor of n!.
Proof. [The statement is obviously false for n = 1.] Let H < G be a
subgroup with index [G : H] = n > 1. Consider the left coset space of H and
let G act naturally by left multiplication. We obtain a group homomorphism
ϕ : G −→ Sym(G/H) ∼
= Sn . Since G is simple, this homomorphism is
injective, hence G embeds in Sn , thus |G| | |Sn | = n!.
¤
(b) Prove that there is no simple nonabelian group of order pe m with e > 0
and p > m.
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Yang 9
Proof. Let G be a simple group of order pe m with e > 0 and p > m. We
will show that G is abelian. Let np be the number of p-Sylow subgroups in
G. By Sylow theorem, np ≡ 1 (mod p) and np | m. Together with p > m,
we obtain np = 1, thus P , the unique p-Sylow subgroup of G, is normal.
Since e > 0, |P | = pe is nontrivial. Since G is simple, we conclude that
P = G, so m = 1. But then G is a p-group, which has nontrivial centre.
A nontrivial p-group has nontrivial centre. [Indeed, the centre of a group
consists of the singleton conjugacy classes. All the conjugacy classes sum
to a power of p (in particular, a multiple of p). Since each conjugacy class
is a power of p (in particular, a multiple of p unless a singleton), and the
identity element gives a singleton conjugacy class, we must have at least
p − 1 other singleton conjugacy classes.] (See Qual Problem G2w3.)
Recall that the centre Z(G) of G is normal. Since it is nontrivial, and G
is simple, we conclude that G = Z(G), that is G is abelian, as desired. ¤
G.12. G5w2. An additive abelian group is called divisible if multiplication by n
for every positive n is a surjective endomorphism.
(a) Show that if G is divisible, G/H is divisible for any subgroup H of G.
Proof. Let ϕn : G −→ G be multiplication by n. As G is divisible, ϕn is
surjective. Notice that ϕn : G/H −→ G/H is clearly surjective as well.
Indeed, take g + H ∈ G/H. As ϕn is surjective, there exists g ′ ∈ G such
that ϕ(g ′ ) = g. Then ϕ(g ′ + H) = g + H.
¤
(b) Give an example (with proof) of a divisible group for which the multiplication by n is not an automorphism for every positive integer n.
Proof. It is obvious that Q is divisible, hence by (a), so is Q/Z. But multiplication by 2 sends both 1/4 and 3/4 to 1/2, hence it is not injective. ¤
(c) Prove or disprove that there is only one isomorphism class of finitely generated divisible groups.
Proof. Affirmative. Let G be a finitely generated divisible group. By the
fundamental theorem for finitely generated abelian groups, G ∼
= Zr ⊕ Gtors .
Multiplication by |Gtors | has trivial image in the torsion part, hence Gtors
must be trivial for this to be surjective. But Zr is obviously not division
for r ≥ 1. Thus G = {0}.
¤
G5w3. Let G be a free abelian group of finite rank r.
(a) Show that there are only finitely many homomorphisms of G into Z/n for
each positive integer n.
Proof. Recall that a homomorphism is completely determined by its action
on the generators. Since there is a generating set of r elements, each with
n possible images, the number of homomorphisms is bounded by rn .
¤
(b) Find a formula of the number of surjective homomorphisms of G onto Z/p
for a prime p if r = 2.
Proof. The number of surjective homomorphisms is p2 − 1. If both generators are sent to 0 then the homomorphism is not surjective as the image is
0. Otherwise, the image of (at least) one of the generators is nonzero, hence
generate Z/p. Thus all the other p2 − 1 homomorphisms are surjective. ¤
September 12, 2009
Yang 10
G.13. G4f1. Is there a finite group G such that G/Z(G) has 143 elements?
Proof. Negative. First, we prove that a group G with order 143 is cyclic. Let n11
and n13 be the number of 11- and 13-Sylow subgroups, respectively. By Sylow
theorem, n13 ≡ 1 (mod 13) and n13 | 11, hence n13 = 1. Similarly, n11 ≡ 1
(mod 11) and n11 | 13, hence n11 = 1. Let P11 and P13 be the unique 11- and
13-Sylow subgroups of G. Then both of them are normal, with trivial intersection
by order considerations. Therefore P11 P13 has order 143, and hence G = P11 P13 .
Since P11 and P13 are cyclic, by the fundamental theorem of finitely generated
abelian groups, G is cyclic.
Now let G be a finite group. Suppose G/Z(G) has 143 elements, then G/Z(G)
is cyclic. By the last four lines of the proof of Qual Problem G7s3, G is abelian.
Then G/Z(G) = 1, a contradiction.
¤
G.14. G4f2. Prove that every group of order 30 has a subgroup of order 15.
Proof. Let G be a group of order 30. Let np be the number of p-Sylow subgroups
of G, and Pp be one such p-Sylow subgroup. By order considerations P3 ∩ P5 is
trivial, thus |P3 P5 | = |P3 | |P5 | / |P3 ∩ P5 | = 15. It remains to show that one of P3
and P5 is normal, then P3 P5 is a group.
By Sylow theorem, n3 ≡ 1 (mod 3) and n3 | 10 so n3 = 1, 10; similarly, n5 = 1, 6.
If the 3- and 5-Sylow subgroups are both not normal, then n3 = 10 gives 20 elements
of order 3 and n5 = 6 gives 24 elements of order 5; exceeding the order of G.
Therefore, at least one of n3 and n5 is 1, as desired.
¤
G.15. G4f3. Find all finite groups that have exactly three conjugacy classes.
Proof. Let G be a finite group with 3 conjugacy classes. Obviously the identity
forms its own conjugacy class. We have two other conjugacy classes of size a and
b, respectively. Recall that the size of a conjugacy class divides the order of the
group. Thus a | 1 + a + b and thus a | 1 + b and similarly b | 1 + a. If a = b, then
a = b = 1 and we have G = Z/3. Otherwise, WLOG, a < b. But b | 1 + a means
b ≤ 1+a, hence b = 1+a. Finally, a | 1+b = 2+a gives a = 1 or 2, yielding |G| = 4
or 6, respectively. In the former case, G is abelian, giving 4 conjugacy classes, a
contradiction. So |G| = 6. Of course G must be nonabelian, which means G = S3 .
A simple check confirms that S3 has three conjugacy classes (one for each order).
Thus the finite groups with 3 conjugacy classes are exactly Z/3 and S3 .
¤
G.16. G4s1. A group G is said to act transitively on a set S if for any element
s ∈ S, the orbit Gs = S. Suppose G is finite and that G acts transitively on S. Let
f (g) be the number of elements of S fixed by the action of g ∈ G on S. Prove that
X
|G| =
f (g).
g∈G
Proof. Let us count the number of pairs (g, s) ∈ G × S where g.s = s two ways (I
am, after all, a combinatorialist). For each g ∈ G, the number of s fixed by g is
f (g) by definition. For each s ∈ S, the number of g ∈ G fixing s is the size of the
stabiliser Gs . By the
|Gs| = [G : Gs ]. Remembering that
P orbit-stabiliser
P theorem,P
|Gs| = |S|, we get g∈G f (g) = s∈S |Gs | = s∈S |G| / |S| = |G|, as desired. ¤
G.17. G4s2. Classify all groups of order 2 · 7 · 11.
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Yang 11
Proof. Let G be a group or order 2 · 7 · 11. Let Pp be a p-Sylow subgroup of G
for p = 2, 7, 11, which exists by Sylow theorem. Let np be the number of p-Sylow
subgroups of G. By Sylow, n11 ≡ 1 (mod 11) and n11 | 14, thus n11 = 1.
Let N = P7 P11 , which is a subgroup as P11 is the unique 11-Sylow subgroup
hence is normal. By order considerations, P7 ∩ P11 intersect trivially, hence |N | =
|P7 | |P11 | / |P7 ∩ P11 | = 77. We conclude that N is normal, as it has index 2.
A subgroup of index 2 is normal. [Indeed, let h ∈ H and g ∈ G; we need
ghg −1 ∈ H, which is obviously true if g ∈ H. Otherwise, g ∈
/ H, and hence the two
(left) cosets of H are H and gH. If ghg −1 ∈
/ H, then, ghg −1 ∈ gH, thus hg −1 ∈ H,
meaning g −1 ∈ H, a contradiction. Thus ghg −1 ∈ H, as desired.]
Again, since N is normal, N P2 is a subgroup, and N ∩ P2 is trivial by order
considerations, hence |N ∩ P2 | = 154 = |G|. Thus we have that G = N ⋊ϕ P2 is the
semidirect of N and P2 , where ϕ : P2 −→ Aut(N ) is a group homomorphism. Now
by the Chinese remainder theorem, Aut(N ) = Aut(Z/7 × Z/11) ∼
= Aut(Z/7) ×
Aut(Z/11) ∼
= Z/6 × Z/10. Let g ∈ P2 be the nontrivial
= (Z/7)× × (Z/11)× ∼
element. Under ϕ, g has order 1 (when ϕ is trivial) or order 2, namely (3, 0), (0, 5),
or (3, 5). Thus we have (at most) four non-isomorphic groups of order 154.
to list
­ Now nit remains
® four non-isomorphic groups of order 154. Let D2n =
ρ, σ : ρ = σ 2 = (ρσ)2 be the dihedral group of order 154. Recall that for n ≥ 3,
­
®
Z(D2n ) = 1 if (and only if) n is odd (otherwise it is ρn/2 ). Notice that, obviously,
if B = Z(B) is abelian, then Z(A × B) = Z(A) × B. With these observations, it is
evident that Z/154, D14 × Z/11, D22 × Z/7, and D154 are non-isomorphic groups
of order 154, as their centres have orders 154, 11, 7, and 1, respectively.
¤
G.18. G4s3. Let G be a finite group and H a subgroup of G. Let n = [G : H] be
the index of H in G.
(a) Show that
\
[G :
xHx−1 ]
x∈G
is a factor of n!.
Proof. Let G act by left multiplication on the left coset space G/H. We
obtain a group homomorphism
Sym(G/H) ∼
= Sn . It remains to
T ϕ : G −→
−1
show that the kernel is N = x∈G xHx . First notice that N is obviously
in the kernel, since (gHg −1 )gH = gH. Conversely, let x be in the kernel.
For any g, xgH = gH, hence xg ∈ gH and thus x ∈ gHg −1 , as desired.
Thus by the first group isomorphism theorem, G/N ∼
= im(ϕ) < Sn , so
[G : N ] | n!.
¤
(b) Suppose that the index [G : H] is the minimal prime factor of the order of
G. Show H is a normal subgroup.
Proof. Again let N as above, it is normal in G as it is the kernel of a group
homomorphism. It is also obvious that N < H, so we have [G : N ] = [G :
H] · [H : N ]. Let p = [G : H] be a minimal prime factor of the order of G.
From part (a) we have that [G : N ] | p!. So [H : N ] | (p − 1)!, but [H : N ]
is either 1 or at least p, which cannot divide (p − 1)!. Therefore H = N ,
hence H ⊳ G.
¤
G.19. G3f1. Let G be a finitely generated group, and n > 1 an integer. Show that
G has at most a finite number of subgroups of index n.
September 12, 2009
Yang 12
Proof. Let G be finitely generated by S, with |S| < ∞. Fix n and let H be a
subgroup of index n. Let G naturally act on the left coset space G/H by left
multiplication. This induces a group homomorphism ϕ : G −→ Sym(G/H) ∼
= Sn .
Since g ∈ H if and only if gH = H, we get H = {g ∈ G : ϕ(g)(H) = H}. Therefore
the group homomorphism completely determines the subgroup H. Moreover, this
homomorphism ϕ is completely determined by the images of s ∈ S under ϕ. Thus
|S|
there is at most |Sn | < ∞ different group homomorphisms, and hence subgroups
of G of index n.
¤
G3f2. Let G be a finite group, K a normal subgroup, and P a p-Sylow subgroup
of K for some prime p. Prove that G = KNG (P ).
Proof. Repeat of Qual Problem G6s1.
¤
G3f3. Suppose G is the free abelian group on generators x, y, z, and w, considered
as an additive group. Let a = x − z + 2w, b = x − y + w, c = 3x − y − 2z + 5w,
d = 2x − 2y + 4w. If H = ha, b, c, di, determine the structure of G/H.
Proof. Since G is abelian, H is normal, and we get the exact sequence
0 −→ H −→ G −→ G/H −→ 0.
To determine the structure of G/H, we put the matrix associated with the map
H −→ G into Smith Normal Form as follows. We get


 
1
0
0
0
1
1
3
2
 0 −1 −1 −2  0 −1 −1 −2


 
−1 0 −2 0  ∼ −1 1
1
2
2 −1 −1 0
2
1
5
4
by subtracting copies of the first column from the others. Then we may remove −1
and 2 from first column, the entire third column, the entire second row, and the 1
in second column. Permuting, we get


1 0 0 0
0 1 0 0


0 0 2 0 ,
0 0 0 0
∼
¤
yielding that G/H = Z ⊕ Z/2.
G3w1. List, up to isomorphism, all abelian groups A which satisfy the following
three conditions:
(a) A has 108 elements;
(b) A has an element of order 9; and
(c) A has no element of order 24.
Proof. Let A be an abelian group of order 108 = 22 33 . By the fundamental theorem
αn
1
of finitely generated abelian groups, A ∼
= Z/pα
1 ⊕ . . . ⊕ Z/pn , where pi are (not
necessarily distinct) primes and αi ≥ 1 (and the decomposition is unique up to
permutation). By Lagrange, order of elements divide the order of the group, hence
no element will have order 24. If we were to have an element of order 9, we must have
Z/9t for t ≥ 1 in the decomposition. This yields the four possibilities Z/4 ⊕ Z/27,
Z/2 ⊕ Z/2 ⊕ Z/27, Z/4 ⊕ Z/3 ⊕ Z/9, and Z/2 ⊕ Z/2 ⊕ Z/3 ⊕ Z/9.
¤
September 12, 2009
Yang 13
G.20. G3w2. Let n ≥ 1 be a positive integer. Show that a finitely generated group
G has only finitely many subgroups of index at most n.
Proof. By Qual Problem G3f1, there are finitely many subgroups of each index;
the result follows.
¤
G.21. G3w3. Let n ≥ 2 be an integer. Show that the only subgroup of index 2 of
Sn is An .
Proof. Let N < Sn be a subgroup of index 2, it must be normal (see Qual Problem
G4s2). Thus we have a homomorphism ϕ : Sn −→ Sn /N ∼
= Z/2. If σ has odd
order, then ϕ(σ) also has odd order, thus σ is in the kernel. In particular, all 3-cycles
are in N , which generate An . Indeed, An is generated by pairs of transpositions.
Notice that we have (a b c)(b c d) = (a b)(c d) for disjoint transpositions, and
(a b c) = (a b)(b c) for non-disjoint transpositions, as desired.
¤
G2f1. Let A be a free abelian group of rank n. If H is a subgroup of A, show that
H is free abelian of rank n if and only if A/H is finite.
Proof. A subgroup of a free abelian group of rank n is free abelian of rank ≤ n.
[Indeed, we proceed by induction on n. Let A = hg1 , . . . , gn i be free abelian
P of rank
n and H < A. Each element h ∈ H may be written uniquely as h = i ai gi . If
a1 = 0 for all h, then H ⊂ hg2 , . . . , gn i and we are done by induction. Otherwise,
let a > 0 be the minimal (in absolute value) of a1 , running over h ∈ H. By the
same argument as in Qual Problem G5f1, H is free abelian, generated by ag1 and
the generators of H ∩ hg2 , . . . , gn i.]
TODO: What follows is probably the most hand-waving proof in the document.
Let A = hg1 , . . . , gn i, and H < A. If H has rank < n, then some hgi i must not
intersect H. Otherwise, hhgi i ∩ H : i = 1, . . . , ni is a subgroup of H of rank n, a
contradiction. Thus (the image of) hgi i is an infinite subgroup of A/H.
Conversely, suppose A/H is infinite. Since it is finitely generated abelian group,
we conclude that some x + H ∈ A/H has infinite order. That is, hxi intersect H
trivially. Since hH, xi < A is free abelian with rank ≤ n, H has rank < n.
¤
G.22. G2f2. Let G be a finite group of order 108. Show that G has a normal
subgroup of order 9 or 27.
Proof. Let n3 be the number of 3-Sylow subgroups of G. By Sylow, n3 ≡ 1 (mod 3)
and n3 | 4, thus either n3 = 1 or 4. If n3 = 1 then the unique 3-Sylow subgroup
is normal of order 27, and we are done. Otherwise, let S = {P1 , P2 , P3 , P4 } be the
3-Sylow subgroups. Let I = P1 ∩ P2 . Being a proper subgroup of P1 of order 27,
I must be order 1, 3, or 9. From |G| ≥ |P1 P2 | = |P1 | |P2 | / |I|, we get |I| = 9.
It remains
¯ ¯ to show that I is normal in G. Let I act on S by conjugation. Then
|S| ≡ ¯S I ¯ (mod 3), where S I is the set of fixed points of S under I (see Qual
Problem G7f1). Now P1 and P2 are fixed, so by this observation, also P3 and
P4 are. This means I < NG (P3 ), and By Qual Problem G2f3, P3 is the unique
3-Sylow, hence contains I, a 3-group. This is true for P4 as well, hence we conclude
I = P1 ∩ P2 ∩ P3 ∩ P4 . Recalling that by Sylow, the Pi are conjugate, and that
conjugating by g ∈ G actually permutes the Pi . we get I g = P1g ∩ P2g ∩ P3g ∩ P4g = I,
as desired.
¤
September 12, 2009
Yang 14
G.23. G2f3. Let G be a finite group and P a p-Sylow subgroup. Let NG (P ) be
the normalizer of P in G. Show that:
(a) P is the unique p-Sylow subgroup of NG (P ).
Proof. It is evident that we have P < NG (P ) < G, so since [G : P ] is
coprime to p, [NG (P ) : P ] also is. Thus P is a p-Sylow subgroup of NG (P ).
But p-Sylow subgroups are conjugate (see Qual Problem G8s2), thus P is
normal if and only if there are no other p-Sylow subgroups.
¤
(b) NG (P ) is self-normalizing in G.
Proof. If g ∈ G normalizes N = NG (P ), then P g is a p-Sylow of N g = N ,
which is P . Thus g normalizes P , hence is in N .
¤
G.24. G2s1. Show that a group G of order 2m, where m is odd, has a normal
subgroup of order m.
Proof. Let G be a group of order 2m, where m is odd. By Cauchy, there is an
element x ∈ G of order 2. The action of G on itself by left translation induces a
homomorphism ϕ : G −→ Sym(G) ∼
= S2m . As x has order 2, ϕ(x) is also order
2, thus is a product of disjoint involutions. Since x 6= e, ϕ(x) does not have fixed
points, thus is a product of m disjoint involutions. Consider the usual signature
homomorphism sgn : S2m −→ {±1}. Since ϕ(x) is an odd permutation, sgn ◦ϕ is
surjective. The kernel is therefore a normal subgroup of index 2 in G, as desired. ¤
G2s2. List, up to isomorphism all finite abelian groups A satisfying the following
two conditions:
(a) A is a quotient of Z2 , and
(b) A is annihilated by 18, i.e., 18a = e for all a ∈ A.
Proof. Let A be a finite quotient of Z2 . By the fundamental theorem of finitely
generated abelian groups, A ∼
= Z/a ⊕ Z/b (see Qual Problem G5f1). If we require
a | b, then this representation is furthermore unique. Moreover, b | 18 if and only if
A is annihilated by 18. Thus this gives 17 non-isomorphic groups. The details are
left as an exercise to the reader.
¤
G.25. G2s3. Prove that a group of order 120 is not simple.
Proof. Let G be a group of order 120 = 23 · 3 · 5. Suppose, towards a contradiction,
that G is simple. Let S be the set of 5-Sylow subgroups of G and n5 = |S|. By
Sylow, we have n5 ≡ 1 (mod 5) and n5 | 24, yielding n5 = 1 or 6. If n5 = 1 then
P ∈ S is normal, a contradiction. Let G act on S by conjugation. We obtain a
homomorphism ϕ : G −→ Sym(S) ∼
= S6 . Since G is simple, the kernel is trivial,
hence G embeds in S6 . Let H < G be the preimage of A6 < S6 under ϕ. If
g, h ∈ G\H acts oddly, then gh acts evenly, thus gh ∈ H, thus g ∈ Hh−1 . Thus
H has index 2, and is normal (see Qual Problem G4s2), a contradiction. Hence G
actually embeds in A6 . It remains to show that A6 does not have a subgroup of index
|A6 | / |G| = 3. Suppose otherwise, that H < A6 is of index 3. Let A6 act naturally
on the left coset space of H. We obtain a homomorphism A6 −→ Sym(A6 /H) ∼
= S3 .
Since A6 is simple (An is simple for n ≥ 5), this is injective, giving |A6 | | |S3 |, a
contradiction.
¤
September 12, 2009
Yang 15
Remark. Alternatively, one can, without using simplicity of An , conclude that a
group of order 120 has a subgroup of index 3 or 5, yielding a homomorphism
(hence isomorphism, if simple) to S5 , which has A5 as a normal subgroup.
G2w1. Let G be a free abelian group of rank n for a positive integer n (therefore
G∼
= Zn as groups).
(a) Prove for a given integer m > 1, there are only finitely many subgroups H
of index m in G.
Proof. This is a special case of Qual Problem G3f1.
(b) Find a formula of the number of subgroups of G of index 3.
Proof. (3n − 1)/2. See Qual Problem G7s2.
¤
¤
G2w2. Prove or disprove. There exists a finite abelian group G whose automorphism group has order 3.
Proof. Negative. See Qual Problem G6f1.
¤
G.26. G2w3. Let S and G be p-groups (with G 6= {e}), and assume that S acts on
G by automorphisms. Show that the fixed subgroup GS = {g ∈ G : s(g) = g, ∀s ∈
S} is non-trivial.
Proof. By the orbit-stabiliser theorem (see Qual Problem G7f1), if Sg and Sg are
the orbit and stabiliser of g by S, respectively, then |Sg| = [S : Sg ]. Notice that as
S is a p-group,
by p, unless S = Sg , that is, g ∈ GS . Therefore
¯ S¯
P [S : Sg ] is divisible
¯
¯
0 ≡ |G| = Sg [S : Sg ] ≡ G , modulo p. But GS contains the identity, so it has
size at least p, and hence is non-trivial.
¤
G.27. G1f1. Let G be a finite group whose centre has index n. Show that every
conjugacy class in G has at most n elements.
Proof. Let S be a conjugacy class and let G act on S by conjugation. Then by
orbit-stabiliser theorem, |S| = [G : NG (S)]. But Z(G) < NG (S), so [G : NG (S)] ≤
[G : Z(G)] = n, as desired.
¤
G.28. G1f2. Let G be a subgroup of Sn that acts transitively on the set [n] =
{1, 2, . . . , n}. Let H be the stabiliser in G of an element x ∈ [n]. Prove that
\
gHg −1 = {e}.
g∈G
Proof. Let a ∈ G\{e}. It suffice to find g ∈ G such that a ∈
/ gHg −1 . Since a 6= e,
there exists y ∈ [n] such that a.y 6= y. As G act act transitively on [n], there
exists g ∈ G such that g.x = y. For any h ∈ H, (ghg −1 ).y = y, but a.y 6= y, so
a∈
/ gHg −1 , as desired.
¤
G.29. G1f3. Let G be the group of matrices of the form
µ
¶
a b
0 1
where a ∈ (Z/p)∗ and b ∈ Z/p. Describe all normal subgroups of G.
September 12, 2009
Yang 16
Proof. For notational convenience, we simply write (a, b) for the element in G that
has (a, b) as its first row. Notice that (a, b)−1 = (a−1 , −a−1 b) and (a, b)(c, d) =
(ac, ad + b). Let N ⊳ G be a non-trivial normal subgroup.
Let (a, b) ∈ N be a nontrivial element. Then N contains (a, b) for b ∈ Z/p.
Indeed, conjugation by (c, d) yields (a, c−1 (d(a − 1) + b)). If a 6= 1, then a − 1 6= 0
is invertible, thus when d runs through Z/p, d(a − 1), and hence d(a − 1) + b, runs
through the same, as desired. Otherwise, if a = 1, then b 6= 0, as (a, b) is nontrivial.
Notice that conjugation by (c, d) yields (1, c−1 b). As c runs through (Z/p)× , c−1 b
runs through the same, as desired. (Notice (1, 0) ∈ N , of course.)
Let C be the set of top-left corner entries of N . It is evident that C < (Z/p)× .
Therefore N has the form {(a, b) : a ∈ C, b ∈ Z/p} for some C < (Z/p)× . Conversely, if C < (Z/p)× , then this set is indeed a normal subgroup, as can be easily
seen from the analysis above.
¤
G1s1. Determine a complete set of groups of order eight up to isomorphism and
show that every group of order eight is isomorphic to one of these.
Proof. Let G be of order 8. By the fundamental theorem of finitely generated
abelian groups, if G is abelian, then G is isomorphic to precisely one of Z/8, Z/2 ⊕
Z/4, or Z/2 ⊕ Z/2 ⊕ Z/2.
Now suppose G is non-abelian. Then G is an extension of Z/2 by Z/4. Indeed,
the order of the elements are factors of 8. If there exists an element with order 8,
then G is cyclic and is abelian. If all elements are of order 2, then each element is
its own inverse. Then ab = (ab)−1 = b−1 a−1 = ba, so G is abelian. Therefore there
exists an element g of order 4. Let ι : Z/4 −→ G be given by ι(n) = g n . This is
obviously a monomorphism by construction. Since im(ι) is a group of index 2, it is
normal and we get a cononical projection π : G −→ Z/2 to the only group of order
2, and such that ker(π) = im(ι), as desired.
The image im ι gives a normal subgroup N ⊳ G generated by ρ = ι(1) of order 4.
The other 4 elements are of orders 2 or 4. If there is an element of order 2, call it
σ, Then σρσ −1 ∈ N by normality. Furthermore, it is necesarily of order 4. Indeed,
σρσ obviously cannot be identity. If it has order 2, then σρσσρσ = σρ2 σ = 1,
implying that ρ2 = 1, a contradiction. Therefore σρσ = ρ or σρσ = ρ3 . If we get
the former, it is easy to check that this makes G abelian. Hence we have the latter.
This exactly gives us an explicit isomorphism with D8 . The details can be easily
checked by hand.
Otherwise, all the elements not in N have order 4. Call one of them j. Now let
i = ι(1). Since N is a subgroup of index 2, we can write the other 4 elements as
nj for n ∈ N . Each nj has order 4, so when squared equals i2 , the only element of
such order. Finally, i(1j)(ij) = i2 . We have shown that i2 = j 2 = k 2 = ijk = −1,
where i2 = −1 and k = ij. So we get an explicit isomorphism with Q8 .
¤
G1s2. A finite group G acts on itself by conjugation. Determine all possible G if
this action yields precisely three orbits.
Proof. See Qual Problem G4f3.
G.30. G1s3. Let G be a finitely generated group.
(a) Show for each integer n there exists finitely many subgroups of index n.
¤
September 12, 2009
Yang 17
Proof. See Qual Problem G3f1.
¤
(b) Suppose that there exists a subgroup of finite index in G. Prove that G
contains a characteristic subgroup of finite index.
Proof. Suppose [G : H] = n. Let S be the set
T of subgroups of index n. By
part (a), S is a finite collection. Let I = S be the (finite) intersection
of subgroups of index n. Recall that the index of a finite intersection of
subgroups of finite index is finite. Thus I < G is of finite index. It remains
to show that I is characteristic. Let ϕ be an automorphism of G. Then as
I < H, we have ϕ(I) < ϕ(H). But ϕ(H) also has index n (automorphisms
preserve indices), thus ϕ(H) < I. This shows ϕ(I) < I, as desired.
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G0f1.
G.31. G0f2. A group of order a power of a prime p is called a p-group. Let G
be a finite group. Prove that for any given prime p, there exists a unique normal
subgroup N of G such that
(a) G/N is a p-group, and
(b) any homomorphism π of G into a p-group is trivial on N (that is, π(N ) = 1).
Proof.
G.32. G0f3. Let G be a finite group of order n. Suppose that G has a unique
subgroup of order d for each positive divisor d of n. Prove that G is a cyclic group.
Proof.
G0s1. State a theorem which classifies all finite abelian groups up to isomorphism.
This means that each finite abelian group should be isomorphic to exactly one group
of your list. Use your classification to list abelian groups of order 24.
Proof. By the fundamental theorem of finitely generated abelian groups, a finite
abelian group is isomorphic to Z/a1 × Z/a2 × . . . × Z/an for some n ≥ 0 and ai ≥ 2,
and a1 | a2 | . . . | an . The ai are unique up to a sign. If |G| = 24 = 23 · 3, then G
is isomorphic to precisely one of Z/2 × Z/2 × Z/6, Z/2 × Z/12, or Z/24.
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G0s2. Let S5 be the symmetric group on 5 letters. For each positive integer n, list
the number of elements of S5 of order n.
Proof. Recall that each element of Sn can be written as the product of disjoint
cycles. The least common multiple of the lenghts of the cycles give the order of the
element. It is a trivial combinatorial exercise to count the multiplicity of each cycle
type. We present the result:
cycle type
15
21 13
22 11
31 12
31 21
41 11
51
number of elements
¡15¢
¡52¢¡=
¢10
3
¡25¢ 2 /2 = 15
¡53¢ · 2 = 20
¡53¢ · 2 = 20
4 · 3! = 30
4! = 24
order
1
2
2
3
6
4
5.
September 12, 2009
Yang 18
Adding 10 + 15 = 25 gives the number of elements of order n = 2. Other values of
n can be read from the table directly.
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G0s3. Let F4 be the field with 4 elements. Let G = SL(2, F4 ) be the group of
2 by 2 invertible matrices with entries in F4 . What is the order of G? Show, by
analysing the action of G on the lines containing the orign in (F4 )2 , that G is a
simple group.
Proof. See Qual Problem G6s2.
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