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Homework 2
due 2/07/02
2.1
Remak-Krull-Schmidt.
(4.47) If K E G then show that G satisfies both chain conditions (for normal
subgroups) if both K and G/K satisfy both chain conditions.
We can assume that G is infinite since the chain conditions are obviously
both satisfies for finite groups.
Ans: Let φ : G → G/K be the quotient map and suppose that
{Gi } = {G1 ≥ G2 ≥ G3 · · · }
is a decreasing sequence of normal subgroups of G . Then {φ(Gi )} is a
decreasing sequence of normal subgroups of G/K and {Gi ∩K} is a decreasing
sequence of normal subgroups of K. So both of these become stationary
for large enough i, say i ≥ n. By the following lemma this implies that
Gn = Gn+1 = Gn+2 = · · · . This proves the DCC. The ACC is similar.
Lemma 2.1 (Kyle’s Lemma). If A ≥ B, φ(A) = φ(B) and A∩K = B ∩K
then A = B.
Proof. It suffices to show that A ≤ B. So take any a ∈ A. Then φ(a) ∈
φ(A) = φ(B) so there is a b ∈ B ≤ A such that φ(a) = φ(b). This implies that
φ(ab−1 ) = 1, i.e., ab−1 ∈ ker φ = K. But ab−1 is also an element of A since a, b
both lie in A. Thus ab−1 ∈ K ∩ A = K ∩ B so ab−1 ∈ B ⇒ a ∈ Bb = B.
Partially conversely, if G satisfies both chain conditions then show that any
quotient also satisfies both chain conditions.
This is obvious since sequences of normal subgroups of G/K correspond
to sequences of normals subgroups of G which contain K.
(4.48) If G satisfies both chain conditions and G × G ∼
= H × H then show
that G ∼
= H. [Note: use RKS but notice that you are not assuming that H
satisfies the chain conditions.]
Ans: First we need to show that H satisfies both chain conditions. This
follows from (4.47) above. Since G satisfies both chain conditions, so does
G × G since G is a normal subgroup of G × G with quotient G. Since H is
a quotient of H × H ∼
= G × G, H also satisfies both chain conditions by the
second part of (4.47). By the same argument H × H also satisfies both chain
conditions.
The RKS theorem now applies to H and H × H. So both G and H are
products of indecomposable groups. In G×G ∼
= H×H, these indecomposable
groups occur twice. By RKS, they occur the same number of times in each
decomposition. Dividing these numbers by 2 we conclude that G ∼
= H.
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2.2
p-Groups
Exercise 1: If P is a nonabelian group of order p3 then show that P/Z(P ) is
elementary abelian of order p2 .
Ans: We know that P/Z(P ) is not cyclic so it’s order must be at least p2 .
But Z(P ) is nontrivial so it must have order p and index p2 . The quotient
group P/Z(P ) has order p2 so we know that it is abelian. Since it is not
cyclic it must be Z/p × Z/p.
Exercise 2: Let Φ(P ) be the intersection of all maximal subgroups of a pgroup P . [Φ(P ) is called the Frattini subgroup P .] Show that P/Φ(P ) is
elementary abelian.
Ans: The Frattini subgroup Φ(P ) is the kernel of
Y
Y
P −→ H =
G/Mi ∼
Z/p
=
Mi <G max
Q
Thus P/Φ(P ) is a subgroup of H ∼
= Z/p. But any subgroup of an elementary abelian group is obviously elementary abelian.
[Anton’s proof] For any maximal subgroup M of G, G/M ∼
= Z/p. So:
1. M contains G0 .
2. M contains g p for any g ∈ G.
Since this is true for each maximal M we have:
T
1. Φ(P ) = M contains G0 .
2. Φ(P ) contains g p for any g ∈ G.
The first statement implies that P/Φ(P ) is abelian. The second statement
implies that every element has order p so it is elementary abelian.
2.3
Applications of Sylow
Ex. 3: Suppose that G is a group of order p2 q where q ≡
/ 1 mod p and p2 ≡
/1
mod q. Then show that G is abelian.
Ans: The index of N (Q) (where Q is a q-Sylow subgroup of G) cannot
be p or p2 since neither of these numbers is congruent to 1 modulo q. Consequently, the index is 1 and Q C G. Similarly, q ≡
/ 1 mod p implies P C G
where P is the p-Sylow subgroup of G. But P ∩ Q = 1 so G = P × Q. Since
both P and Q are abelian (groups of order p2 are abelian), G = P × Q is
abelian.
Ex. 4: Find all subgroups of D12 (the dihedral group of order 12).
Ans: The dihedral group D12 = ha, b|a6 = 1 = b2 , bab = a−1 i has a cyclic
normal subgroup C = hai = {1, a, · · · , a5 } of order 6.
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1. C has four subgroups 1, ha3 i , ha2 i , C of order 1, 2, 3, 6. Any other subgroup H of D12 must intersect C in one of these subgroups.
2. Every element bai of D12 not in C has order 2 and generates a subgroup
Hi = {1, bai }. These are the subgroups satisfying Hi ∩ C = 1
3. If H ∩ C = ha3 i (and HC 6= D12 ) then H = Hi ha3 i. There are three
of these subgroups:
(a) H0 ha3 i = H3 ha3 i = {1, a3 , ba3 , b}
(b) H1 ha3 i = H4 ha3 i = {1, a3 , ba, ba4 }
(c) H2 ha3 i = H5 ha3 i = {1, a3 , ba2 , ba5 }
4. If H ∩ C = ha2 i (and HC 6= D12 ) then H = Hi ha2 i. There are two of
these subgroups:
(a) H0 ha2 i = H2 ha2 i = H4 ha2 i = {1, a2 , a4 , b, ba2 , ba4 }
(b) H1 ha2 i = H3 ha2 i = H5 ha2 i = {1, a2 , a4 , ba, ba3 , ba5 }
5. Finally, if H ∩ D12 = C (and H 6= C) then H = D12 .
This makes a total of 4 + 6 + 3 + 2 + 1 = 16 subgroups of D12 . With a
similar analysis we see that, in general, D2n has
X
(d + 1)
d|n
­ ®
­
®
subgroups and they have the form ad where d|n and ad , bai where d|n
and 0 ≤ i < d.
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