Download PRACTICE QUESTIONS FOR FINAL

Math 330 - Abstract Algebra I
Spring 2009
PRACTICE QUESTIONS FOR FINAL - WORKED
SOLUTIONS
These questions are intended to represent approximately how difficult and
long the Final Exam will be, and also to indicate some of the types of questions
that might arise. They should not be contrued as a complete list of the topics
that are examinable.
(1) Let
GL2 (R) =
a
c
b
d
| a, b, c, d ∈ R, ab − cd 6= 0 ,
which is a group with the usual operation of multiplication.
a b
Let H be the subset of matrices
, with a, b, d ∈ R and ad 6= 0.
0 d
(a) Prove that H is a subgroup of GL2 (R;
(b) What is the identity of H?
(c) Give an explicit formula for the inverse of
a
0
b
d
.
Solution:
(b) The identity element of GL2 (R) is
1 0
0 1
and this is also the identity element of H.
(c) The inverse of
is
1
a
0
a
0
b
c
b
− ad
1
d
as can be easily seen by multiplying these two matrices together.
(a) We just checked that H contains the identity element.
Also, the above formula shows that if A ∈ H then A−1 ∈ H. Finally, we
make the following calculation to see that H is closed under multiplication:
a1 b1
a2 b2
a1 a2 a1 b2 + b1 d2
=
0 d1
0 d2
0
d1 d2
Note that since a1 d1 6= 0 and a2 d2 6= 0 we have (a1 a2 )(d1 d2 ) = 0. Therefore, the product of two matrices in H is also in H, as required.
(2) Consider the following two elements of S7 :
x
α
1
3
2
4
3
7
4
1
5
2
6
5
7
6
x
β
1
2
2
3
3
1
4
4
5
5
6
7
7
6
and
(a) For each of the following elements of S7 , write them as a product of
disjoint cycles: α, β, αβ, β −1 αβ, α−1 β −1 αβ;
(b) For each the elements in part (a), calculate their order;
(c) For each of the elements in part (a), say whether they are even or
odd.
Solutions:
(a):
We have:
α
=
(1376524)
β
=
(123)(67)
αβ
=
(14)(275)
−1
=
(132)(67)
αβ
=
(1432675)
α−1
=
(1425673)
=
(1276354)
β
β
α
−1
−1 −1
β
αβ
(b): The orders are as follows:
α has order 7
β has order 6
αβ has order 6
β −1 αβ has order 7
α−1 β −1 αβ has order 7
(c): A seven cycle is a product of 6 2-cycles, so α, β −1 αβ and α−1 β −1 αβ
are all even.
A 3-cycle is a product of 2 2-cycles, so each of β and αβ are odd.
(3) Let G and H be groups, and let φ : G → H be a homomorphism which is
also onto. Let Z(G) = {g ∈ G | for all g0 ∈ G, g0 g = gg0 } be the center
of G, and Z(H) the center of H.
(a) Prove that φ(Z(G)) ⊆ Z(H).
(b) Give an example of groups G and H and a homomorphism φ : G → H
which is not surjective so that φ(Z(G)) 6⊆ Z(H). [Hint: Z(S3 ) = {1}]
Solution:
(a): Suppose that h ∈ H and g ∈ Z(G). We have to prove that hφ(g) =
φ(g)h. Well, φ is a surjective homomorphism, so there is some g0 ∈ G so
that φ(g0 ) = h. Therefore, we have
hφ(g)
= φ(g0 )φ(g)
= φ(g0 g)
= φ(gg0 )
= φ(g)φ(g0 )
= φ(g)h,
as required.
(b): Consider G = Z/3Z, and H = S3 . Define φ : G → H by φ(1 + 3Z) =
(123). Since G is abelian, Z(G) = G. However, as noted in the Hint,
Z(S3 ) = {1}. Now, φ(G) = he, (123), (132)i 6⊆ {1} = Z(S3 ).
(4) (Gallian, 14.18, p.269) Suppose that in the ring Z the ideal h35i is a proper
ideal of J, and J is a proper ideal of I. What are the possibilities for J?
What are the possibilities for I?
Solutions:
All ideals of Z are of the form mZ. We have mZ ⊆ nZ if and only if n | m.
Therefore, the ideals that contain h35i = 35Z are:
Z, 5Z, 7Z, and 35Z. The first three have 35Z as a proper ideal.
It is not hard to see that we must have I = Z, whereas the possibilities
for J are 5Z and 7Z.
√
(5) (Gallian,
√ 13.24, p.255) Let d be a positive integer. Prove that Q[ d] =
{a + b d | a, b ∈ Q} is a field.
Solution:
√
We consider two separate cases. If d has a square root in Q then Q[ 2] =
Q, which is certainly a field.
Therefore, suppose that d√does not have a square root in Q. It is straightforward to check that Q[ 2] is commutative:
√
√
√
(a1 + b1 d)(a2 + b2 d = (a1 a2 + db1 b2 ) + (a1 b2 + a2 b1 ) d
√
= (a2 a1 + db2 b1 ) + (a2 b1 + a1 b2 ) d
√
√
= (a2 + b2 d)(a1 + b1 d)
√
√
√
Also, 1 = 1 + 0 d ∈ Q[ d], so Q[ d] is a commutative ring with unity.
√
We therefore need to check that every
√ nonzero element of Q[ d] has a
multiplicative inverse. So, let a + b d be so that (a, b) 6= (0, 0). Then
a2 − b2 d 6= 0, since then either b2 = 0, in which case a and b are both
zero (which we’re assuming is not the case), or else (a/b)2 = d, so d has a
square root in Q, which we are also assuming is not the case.
√
√
Thus,
a2 − b2 d 6= 0 and so a − b d 6= 0 also (since a2 − b2 d = (a + b d)(a −
√
b d). Now, we have
√
(a + b d).
√
1
(a − b d) = 1,
a2 − b2 d
√
so the multiplicative inverse of a + b d is
a2
√
a
b
− 2
d.
2
2
−b d a −b d
a
−b
Finally, note that a2 − b2 d ∈ Q, so a2 −b
2 d , a2 −b2 d ∈ Q, which means that
√
√
every element of Q[ d] has a multiplicative inverse in Q[ d], so it is a
field, as required.