Download HW4

Algebra I Homework Four
Name:
Instruction: In the following questions, you should work out the solutions in a clear and concise manner.
Three questions will be randomly selected and checked for correctness; they count 50% grades of this
homework set. The other questions will be checked for completeness; they count the rest 50% grades of the
homework set. Staple this sheet of paper as the cover page of your homework set.
1. (Section 2.5) If G is an infinite p-group (p prime), then either G has a subgroup of order pn for each
n ≥ 1, or there exists m ∈ N∗ such that every finite subgroup of G has order ≤ pm .
Suppose that G has no subgroup of order pn for some n ≥ 1. Let m be the least positive integer such that G
has no subgroup of order pm+1 . Then G cannot have a finite subgroup of order > pm . Otherwise, let H be a finite
subgroup of G of order > pm , then |H| = pk for some k > m. As an abelian group, H contains a subgroup of order
pm+1 , which leads to a contradiction.
2. (Section 2.5) Find the Sylow 2-subgroups and Sylow 3-subgroups of S3 and S4 and describe each by a set
of generators.
The number of Sylow p-subgroups in a finite group satisfies the conditions in Third Sylow Theorem. We describe
these groups by a set of generators.
S3 : There are 3 Sylow 2-subgroups (of order 2) and 1 Sylow 3-subgroup (of order 3):
i. Sylow 2-subgroups: h (12) i, h (13) i, h (23) i.
ii. Sylow 3-subgroups: h (123) i.
S4 : There are 3 Sylow 2-subgroups (of order 8) and 4 Sylow 3-subgroups (of order 3):
i. Sylow 2-subgroups: h (1234), (12)(34) i, h (1243), (12)(43) i, h (1324), (13)(24) i.
ii. Sylow 3-subgroups: h (123) i, h (124) i, h (134) i, h (234) i.
3. (Section 2.5) Every group of order 12, 28, 56, and 200 must contain a normal Sylow subgroup, and hence
is not simple.
Let nG (p) denote the number of Sylow p-subgroups in a group G, where p is a prime.
12: Assume |G| = 12 = 22 · 3. By Third Sylow Theorem, nG (3) = 1 or 4. If nG (3) = 1, then G contains a normal
Sylow 3-subgroup; if nG (3) = 4, then G contains 2 · 4 = 8 elements of order 3, and thus the remaining 4
elements of G form a normal Sylow 2-subgroup of G.
28: Assume |G| = 28 = 22 · 7. By Third Sylow Theorem, nG (7) = 1. So G contains a normal Sylow 7-subgroup.
56: Assume |G| = 56 = 23 · 7. By Third Sylow Theorem, nG (7) = 1 or 8. If nG (7) = 1, then G contains a normal
Sylow 7-subgroup; if nG (7) = 8, then G contains 6 · 8 = 48 elements of order 7, and thus the remaining 8
elements of G form a normal Sylow 2-subgroup of G.
200: Assume |G| = 12 = 23 · 52 . By Third Sylow Theorem, nG (5) = 1. So G contains a normal Sylow 5-subgroup.
4. (a) (Section 2.5) If p is a prime, prove that every group of order p2 is abelian.
Let G be a group of order p2 . As a p-group, G has nontrivial center. If C(G) = G, then G is abelian. If
C(G) 6= G, then |C(G)| = p. Suppose C(G) = hai. Then G/C(G) has order p. So G/C(G) is generated by
an element, say bC(G) for b ∈ G. Then ba = ab and G is generated by a and b. Therefore, G is abelian. This
contradicts to C(G) = hai =
6 G.
(b) (Section 2.6) Describe all groups G of order pq ≤ 60 where p ≥ q are (not necessarily distinct) odd
primes.
3 × 3:
3 × 5:
3 × 7:
3 × 11:
3 × 13:
Z3 × Z3 , Z9 .
Z15 .
Z21 , ha, b | a3 = b7 = e, aba−1 = b2 i.
Z33 .
Z39 , ha, b | a3 = b13 = e, aba−1 = b3 i.
1
3 × 17:
3 × 19:
5 × 5:
5 × 7:
5 × 11:
7 × 7:
Z51 .
Z57 , ha, b | a3 = b19 = e, aba−1 = b7 i.
Z5 × Z5 , Z25 .
Z35 .
Z55 , ha, b | a5 = b11 = e, aba−1 = b3 i.
Z7 × Z7 , Z49 .
5. (Section 2.6) Let G and H be groups and θ : H → Aut G a homomorphism. Let G ×θ H be the set G × H
with the following binary operation: (g, h)(g 0 , h0 ) = (g[θ(h)(g 0 )], hh0 ). Show that G ×θ H is a group with
identity element (e, e) and (g, h)−1 = (θ(h−1 )(g −1 ), h−1 ). G ×θ H is called the semidirect product of
G and H.
[(g, h)(g 0 , h0 )](g 00 , h00 )
(g, h)[(g 0 , h0 )(g 00 , h00 )]
=
(g[θ(h)(g 0 )], hh0 )(g 00 , h00 ) = (g[θ(h)(g 0 )][θ(hh0 )(g 00 )], hh0 h00 )
=
(g[θ(h)(g 0 )][θ(h)(g 00 )][θ(h0 )(g 00 )], hh0 h00 ),
=
(g, h) (g 0 [θ(h0 )(g 00 )], h0 h00 ) = (g[θ(h)(g 0 [θ(h0 )(g 00 )])], hh0 h00 )
=
(g[θ(h)(g 0 )][θ(h)(g 00 )][θ(h0 )(g 00 )], hh0 h00 ).
Hence G ×θ H satisfies the associativity law.
(e, e)(g, h) = (e[θ(e)(g)], eh) = (g, h),
(g, h)(e, e) = (g[θ(h)(e)], he) = (g, h).
So (e, e) is the identity in G ×θ H.
(g, h)(θ(h−1 )(g −1 ), h−1 )
=
(g[θ(h)(θ(h−1 )(g −1 ))], hh−1 ) = (e, e),
(θ(h−1 )(g −1 ), h−1 )(g, h)
=
([θ(h−1 )(g −1 )][θ(h−1 )(g)], h−1 h) = (e, e).
Therefore, every (g, h) ∈ G ×θ H has the inverse, given by (θ(h−1 )(g −1 ), h−1 ).
6. (Section 2.7) Let C0 (G) ≤ C1 (G) ≤ C2 (G) ≤ · · · be the ascending central series of a group G. Prove that
(a) Ci (G) C G for all i;
Proceed by induction on i. C0 (G) = heiCG. Assume that Ci (G)CG, then Ci+1 (G)/Ci (G) = C(G/Ci (G))C
G/Ci (G). For x ∈ Ci+1 (G) and y ∈ G,
yxy −1 Ci (G) = [yCi (G)][xCi (G)][yCi (G)]−1 = xCi (G) ∈ Ci+1 (G).
Thus yxy −1 ∈ Ci+1 (C). Therefore Ci+1 (G) C G. The proof is completed.
(b) Ci+1 (G) = {x ∈ G | xyx−1 y −1 ∈ Ci (G) for all y ∈ G}.
We shall use the result Ci (G) C G for all i.
If x ∈ Ci+1 (G), then for any y ∈ G,
yxy −1 Ci (G) = [yCi (G)][xCi (G)][yCi (G)]−1 = xCi (G).
So yxy −1 x−1 ∈ Ci (G).
Conversely, if x ∈ G satisfies that xyx−1 y −1 ∈ Ci (G) for all y ∈ G, then
[yCi (G)][xCi (G)][yCi (G)]−1 = yxy −1 Ci (G) = xCi (G).
Thus xCi (G) ∈ C(G/Ci (G)) so that x ∈ Ci+1 (G).
7. (Section 2.7) Let G(i) be the i-th derived subgroup of a group G. Prove that G(i) C G for all i.
Proceed by induction on i.
When i = 0, G(0) = G C G.
Assume that G(j) C G for j = i − 1. To prove that G(i) C G, it suffices to show that for any element a in a
set of generators of G(i) and any element b ∈ G, bab−1 ∈ G(i) . Notice that G(i) is generated by xyx−1 y −1 for all
x, y ∈ G(i−1) . For any x, y ∈ G(i−1) and any b ∈ G,
b(xyx−1 y −1 )b−1 = (bxb−1 )(byb−1 )(bxb−1 )−1 (byb−1 )−1
where bxb−1 , byb−1 ∈ G(i−1) by induction hypothesis. Therefore, b(xyx−1 y −1 )b−1 ∈ G(i) . Thus G(i) C G.
Overall, the statement holds for all i ∈ N.
2
8. (Section 2.7) Prove that a finite group G is nilpotent if and only if every maximal proper subgroup of
G is normal. Conclude that every maximal proper subgroup has prime index. [Hint: if P is a Sylow
p-subgroup of G, show that any subgroup containing NG (P ) is its own normalizer; see Theorem 2.5.11.]
If G is nilpotent, then for any maximal proper subgroup H G we have H NG (H) (See Lemma 2.7.4).
Therefore NG (H) = G by the maximality of H. Then H C G. There is a one-to-one correspondence between the
subgroups in G/H and the subgroups of G that contains H. Since H is a maximal proper subgroup of G, G/H
must be of prime order. So H has prime index in G.
Conversely, if every maximal proper subgroup of G is normal, we show that every Sylow subgroup of G is normal,
which implies that G is the direct product of its Sylow subgroups so that G is nilpotent.
Let P be any Sylow p-subgroup of G. Suppose on the contrary, P 6 G. Then NG (P ) G. Let M be
a maximal proper subgroup of G that contains NG (P ). On one hand, M C G by assumption. If y ∈ G, then
yP y −1 ≤ yM y −1 = M . Thus every Sylow p-subgroup of G is a Sylow p-subgroup of M . On the other hand,
NG (P ) ≤ M so that NM (P ) = NG (P ) ∩ M = NG (P ). The number of Sylow p-subgroup in M equals to [M :
NM (P )] = [M : NG (P )], which is strictly less than [G : NG (P )], the number of Sylow p-subgroups in G. This is a
contradiction. Therefore, P C G.
9. (Section 2.8) Find all composition factors of S4 and D6 .
S4 : The group S4 has a composition series:
hei < {e, (12)(34)} < {e, (12)(34), (13)(24), (14)(23)} < A4 < S4 .
The composition factors of S4 are: Z2 , Z2 , Z3 , Z2 .
D6 : The group D6 is generated by a = (123456) and b = (16)(25)(34). D6 has a composition series:
hei < ha2 i < hai < D6 .
The composition factors of D6 are: Z3 , Z2 , Z2 .
10. (Section 2.8) Any group of order p2 q (p, q primes) is solvable.
Let G be a group with |G| = p2 q. If p = q, then G is a p-group, which must be nilpotent and thus solvable.
Assume that p 6= q. If G contains a normal Sylow subgroup N , then both N and G/N are solvable. Thus G is
solvable.
If p > q, then by Third Sylow Theorem, G contains a unique Sylow p-subgroup of G, which is normal.
If p < q, and G does not contain a normal Sylow subgroup. Then by Third Sylow Theorem, G contains p2
Sylow q-subgroups. Therefore, there are p2 (q − 1) elements of order q in G. The remaining p2 elements form the
unique Sylow p-subgroup of G. This contradicts to the assumption that G has no normal p-subgroup. Thus G has
a normal p-subgroup, which implies that G is solvable.
3