Name:
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Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Friday May 13, 2016 - 08:30 am to 09:20 am
Instructions:
Prob.
Points Score
possible
1
22
2
10
3
18
TOTAL
50
• Read each problem carefully.
• Write legibly.
• Show all your work on these sheets. Feel free to
use the opposite side.
• This exam has 6 pages, and 3 problems. Please
make sure that all pages are included.
• You may not use books, notes, calculators, etc.
Cite theorems from class or from the texts as appropriate.
• Proofs should be presented clearly (in the style
used in lectures) and explained using complete
English sentences.
Good luck!
Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Page 2 of 6
Question 1. (Total of 22 points)
a) (8 points) State the definitions/statements of:
• A composition series of a group G;
• A solvable group G.
• The Jordan-Hölder theorem.
b) (6 points) Let G be a group and N E G a proper normal subgroup. Supposing G
admits a composition series, show there exists a composition series containing N .
c) (8 points) Let G be a group and N E G. Prove that if G is solvable, then N and
G/N are both solvable.
Solution. a) The definitions/statements are as follows.
• A subnormal series (Hi )ni=0 of a group G is a composition series if all the
factor groups Hi+1 /Hi are simple.
• A group G is solvable if it admits a composition series (Hi )ni=0 for which all
the factors Hi+1 /Hi are abelian.
• Any two composition (or principal) series of a group G are isomorphic.
b) By the Schreier refinement theorem, any composition series for G must be isomorphic to some refinement of the series {e} ≤ N ≤ G (recall, a composition series
admits no proper refinement). Since a subnormal series which is isomorphic to a
composition series must itself be a composition series, it follows that there exists
a refinement of {e} ≤ N ≤ G which is a composition series.
c) We may assume without loss of generality that N is a proper normal subgroup.
Supposing G is solvable, by part b) there exists a composition series which contains
N , say
{e} = H0 E H1 E H2 E · · · E Hd = N E Hd+1 E · · · E Hn = G.
Since G is solvable, by the Jordan-Hölder theorem the factors Hi+1 /Hi of this
series must be abelian. It follows that {Hi }di=0 is a composition series for N with
abelian factors and so N is solvable. On the other hand, by the correpsondence
theorem for normal subgroups we have
{eG/N } = N/N E Hd+1 /N E · · · E Hn /N = G/N
n−d
so defining Ki := Hd+i /N for i = 0, . . . , n − d it follows that (Ki )i=1
is a normal
series for G/N . Finally, note that by the Third Isomorphism Theorem we have
Ki+1
Hd+i+1 /N ∼ Hd+i+1
=
=
Ki
Hd+i /N
Hd+i
Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Page 3 of 6
and so each Ki+1 /Ki must be simple and abelian for i = 0, 1, . . . , n − d − 1. Thus
(Ki )n−d
i=1 is a composition series for G/N and G/N is a solvable group.
Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Page 4 of 6
Question 2. (Total of 10 points) Prove that no group of order 56 is simple.
Solution. a) Suppose G is a group of order 56 = 23 × 7. Then by Sylow III the
number of Sylow 7-subgroups is congruent to 1 mod 7 and divides 56 so there
must be either 1 or 8 such subgroups. If there is only one such subgroup, then
it must be normal (since for each p the set of Sylow p-subgroups is closed under
conjugation) and so we assume there are 8 such subgroups. Since they are cylic,
they can only intersect at the identity, and so we see there are 8 × 6 + 1 = 48
distinct elements of G of order 7.
On the other hand, the number of Sylow 2-subgroups is odd and divides 56, so there
must either be 1 or 7 such subgroups. Assume there are 7 of these subgroups, each
of which has cardinality 8. Choose two distinct Sylow 2-subgroups H1 , H2 . Then
|H1 ∩ H2 | ≤ 4 by Lagrange’s theorem, so that |H1 ∪ H2 | ≥ 16 − 4 = 12. But now
we have 48 + 12 = 60 elements of the group, which is a contradiction. Hence there
must only one Sylow 2-subgroup, which is normal.
Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Page 5 of 6
Question 3. (Total of 18 points)
√
√
Let 4 2 denote the positive fourth root of 2 and consider the extension Q( 4 2, i) : Q.
√
a) (4 points) Find [Q( 4 2, i) : Q].
√
b) (3 points) Find all the conjugates of 4 2 and i over Q.
√
c) (4 points) Let σ, τ ∈ Aut Q( 4 2, i) : Q be defined by
√
√
σ : 4 2 7→ i 4 2
σ : i 7→ i
√
√
4
4
τ : i 7→ −i.
τ : 2 7→ 2
√
Write down the elements of the subgroup of Aut Q( 4 2, i) : Q generated by σ and
√
τ (e.g. ι, σ, σ 2 , . . . ), describing each by its action on 4 2 and i.
√
d) (3 points) To which familiar group is Aut Q( 4 2, i) : Q isomorphic?
e) (4 points) Find the fixed fields of the subgroup {1, σ 2 , τ, σ 2 τ }.
√
√
Solution. a) Clearly i ∈
/ Q( 4 2) and is a root of x2 + 1; consequently, [Q( 4 2, i) :
√
√
Q( 4 2)] = 2. On the other hand, by Eisenstein we see that irr( 4 2, Q)(x) = x4 − 2
√
√
and so [Q( 4 2) : Q] = 4. By the Tower Law, it follows that [Q( 4 2, i) : Q].
√
√ √
√
√
b) The conjugates of 4 2 over Q are given by { 4 2, i 4 2, − 4 2, −i 4 2}. The conjugates
of i over Q are {i, −i}.
c) The automorphisms belonging to the subgroup generated by σ and τ are:
√
√
ι
ι : 4 2 7→ 4 2
ι : i 7→ i
√
√
4
4
σ
σ : 2 7→ i 2
σ: i →
7 i
√
√
4
4
2
2
2
σ :i→
7 i
σ
σ : 2 7→ − 2
√
√
4
4
3
3
3
σ
σ : 2 7→ −i 2
σ :i→
7 i
√
√
τ
τ : 4 2 7→ 4 2
τ : i 7→ −i
√
√
στ
στ : 4 2 7→ i 4 2
στ : i 7→ −i
√
√
4
4
2
2
σ τ
σ τ : 2 7→ − 2
σ 2 τ : i 7→ −i
√
√
σ3τ
σ 3 τ : 4 2 7→ −i 4 2
σ 3 τ : i 7→ −i
Note that all these automorphisms are distinct and they in fact are all the elements
√
of Aut Q( 4 2, i) : Q: since each element must be mapped to a conjugate over Q
√
by any ψ ∈ Aut Q( 4 2, i) : Q there are at most 8 choices of ψ.
√
d) Aut Q( 4 2, i) : Q ∼
= D4 , the dihedral group of symmetries of the square: τ corresponds to reflection across the vertical axis and σ to a π/2 anti-clockwise rotation.
√
e) Any element x ∈ Q( 4 2, i) can be expressed as
√
√
√
√
√
√
4
4
4
4
x = a1 + a2 2 + a3 2 + a4 ( 2)3 + a5 i + a6 i 2 + a7 i 2 + a8 i( 2)3
Math 256 (11) - Midterm Test 2
Spring Quarter 2016
Page 6 of 6
for some a1 , . . . , a8 ∈ Q. Now we see that
√
√
√
√
√
√
4
4
4
4
σ 2 (x) = a1 − a2 2 + a3 2 − a4 ( 2)3 + a5 i − a6 i 2 + a7 i 2 − a8 i( 2)3 ;
√
√
√
√
√
√
4
4
4
4
τ (x) = a1 + a2 2 + a3 2 + a4 ( 2)3 − a5 i − a6 i 2 − a7 i 2 − a8 i( 2)3 ;
√
√
√
√
√
√
4
4
4
4
σ 2 τ (x) = a1 − a2 2 + a3 2 − a4 ( 2)3 − a5 i + a6 i 2 − a7 i 2 + a8 i( 2)3 .
√
Consequently, the fixed field is Q( 2).