Math 434 (Abstract Algebra) Final Exam
Answers to Review Questions
1.
(a) c = (1 3 2) (4 5) , f = (1 2 4 5 3 6 7)
(b) There is more than one correct answer.
One possible answer is:
c = (1 2) (1 3) (4 5)
f = (1 7) (1 6) (1 3) (1 5) (1 4) (1 2)
(c) o(c) = 6 and o(f) = 7.
(d) c is an odd permutation, since it factors into 3
transpositions.
(e) f is an even permutation, since it factors into 6
transpositions.
2.
f 2 = (1, 6, 3) (2, 5) (4, 7) , f –1 = (1, 6, 3) (2, 7, 5, 4)
3.
(5, 6, 4, 3) or (3, 5, 6, 4)
4.
There is more than one correct answer.
One possible answer is (1, 2)(4, 5).
5. (a) The proof that qa is a permutation is exactly the
first of the three parts of the proof of Theorem 4.7
(Cayley’s Theorem); the proof that ma is a permutation is
very similar and is Exercise 4.2-9a, and you can show your
proof to the instructor if you want an opinion about its
correctness.
5.-continued
(b) The proof that Q = {qa | a  G} is a group with
respect to mapping composition is exactly the second of the
three parts of the proof of Theorem 4.7 (Cayley’s
Theorem); the proof that M = {ma | a  G} is a group with
respect to mapping composition is very similar and is
Exercise 4.2-9b, and you can show your proof to the
instructor if you want an opinion about its correctness.
(c) The proof that :GQ is an isomorphism for any
group G is exactly the third of the three parts of the proof
of Theorem 4.7 (Cayley’s Theorem); you should find that it
is only possible to prove that  is an isomorphism for
abelian groups G, and this is Exercise 4.2-9c, which you
can show to the instructor if you want an opinion about its
correctness.
(a) G , { e ,  , 2 3 , 4 , 5 } ,
{ e ,  }{ e ,  }{ e ,  }{ e ,  }{ e ,  } , { e }
(b) The only normal subgroups are
G, { e ,  , 2 3 , 4 , 5 }, and { e }.
(c) H = { e ,  , 2 3 , 4 , 5 }
G/H = {H, H} = {H, H} = {H, H} = {H, H} =
{H, H}
The multiplication table for G/H = {H, H} is as follows:
H H
H | H H
H | H H
(d) The set of left cosets will not satisfy all the
conditions required to be a group.
6.
7.
(a) Multiplication Table for G1:
[2]10
[4]10
[6]10
[8]10
------------------------------------------[2]10 |
[4]10
[8]10
[2]10
[6]10
|
[4]10 |
[8]10
[6]10
[4]10
[2]10
|
[6]10 |
[2]10
[4]10
[6]10
[8]10
|
[8]10 |
[6]10
[2]10
[8]10
[4]10
|
Multiplication Table for G2:
[4]20
[8]20
[12]20
[16]20
------------------------------------------[4]20 | [16]20
[12]20
[8]20
[4]20
|
[8]20 | [12]20
[4]20
[16]20
[8]20
|
[12]20 |
[8]20
[16]20
[4]20
[12]10
|
[16]20 |
[4]20
[8]20
[12]20
[16]20
|
(b) The multiplication tables in part (a) for each of G1
and G2 show that multiplication is closed for each set, that
there is an identity element ([6]10 for G1 and [16]20 for G2),
and that each element has an inverse (since the identity
element appears exactly once in each row/column).
Finally, we know that multiplication is associative in each
set by Theorem 2.29(b).
7.-continued
(c) Each of G1 and G2 is cyclic; that is, neither one is
like the Klein group. (From the multiplication tables it is
easy to see that [2]10 and [8]10 are generators of G1 , and that
[8]20 and [12]20 are generators of G2 .)
(d) G1 and G2 are isomorphic, since each one is a
cyclic group of order 4. An isomorphism can be defined by
mapping a generator of G1 to a generator of G2 after which
all the other mappings of elements are determined by the
“product” property.
8. (a) You can use Theorem 3.10 to construct a proof that
H1 is a group with ordinary addition, since H1 is a subset of
Z which we know is a group with ordinary addition.
(b) You can prove that H2 is not a group with ordinary
multiplication by showing that there is no multiplicative
identity; however, H2 is closed with respect to ordinary
multiplication.
(c) You can prove that H3 is a group with ordinary
addition, by showing that each of the four required
conditions is satisfied.
(d) You can prove that H4 is not a group with ordinary
multiplication either by giving an example to show that H4
is not closed or by showing that not every element has a
multiplicative inverse, such as 5 / 3; however, H4 does
satisfy the other two conditions necessary to be a group.
9. (a) To prove that R2 is a commutative ring, you will
find it useful to construct an addition table and a
multiplication table. From the addition table, you should
find it easy to demonstrate that each of the first five
conditions of Definition 5.1a (i.e., the first condition of
Definition 5.1b) is satisfied. From the multiplication table,
you should find it easy to demonstrate that each of the last
three conditions of Definition 5.1a (i.e., the last two
conditions of Definition 5.1b) are satisfied. Finally, from
the multiplication table, you should find it easy to
demonstrate that [9]12 is the unity.
(b) You will find it useful to construct a multiplication
table for R1 . From the multiplication table, you should find
it easy to demonstrate that R1 has no zero divisors.
(c) From the multiplication table for R2 , you should
find it easy to demonstrate that [6]12 is a zero divisor.
(d) [2]10 [4]10 [6]10 [8]10
(e) [3]12 [6]12 [9]12
10. Note: You should find that you can construct a proof
using Theorem 3.4 to show that each of M1 , M2 , M3 , and
M4 is a subring.
(a) You should be able to prove that M1 has no unity, and
therefore no element can have a multiplicative inverse.
(b) You should be able to prove that M2 has a unity which
is the 22 matrix with every entry equal to 1 / 2, and the
only matrices which have multiplicative inverses are the
unity and the 22 matrix with every entry equal to  1 / 2,
each of which has itself as its multiplicative inverse.
(c) You should be able to prove that M3 has a unity which
is the 22 matrix with every entry equal to 1 / 2, and every
element of M3 has a multiplicative inverse, specifically, the
arbitrary element
n
m
n

m
n
m
n  where n  0 has multiplicative inverse

m
m
 4n
m

 4n
m
4n 
m .

4n 
(d) You should be able to prove that M4 has a unity which
is the 22 matrix with every entry equal to 1 / 2, and every
element of M4 has a multiplicative inverse, specifically, the
arbitrary element
x
x

x
x 
1
 4x
where x  0 has multiplicative inverse  1

 4x
1
4x 
1 .

4x 
11. (a) Since G has order 12, the only possible orders of
subgroups are divisors of 12.
Subgroups of Order 12: Of course, the only possible
subgroup is G the group itself, and of course this subgroup
is normal.
Subgroups of Order 6: Subgroups of order 6 can only
contain elements whose order is a divisor of 6. These
subgroups can be found by looking for elements of order 6
and pairing an element of order 2 with an element of order
3. These subgroups are {e, a, a2, a3, a4, a5, a6},
{e, a2, a4, b, a2b, a4b}, {e, a2, a4, ab, a3b, a5b }.
All of these subgroups are normal.
Subgroups of Order 4: Subgroups of order 4 can only
contain elements whose order is a divisor of 4. These
subgroups can be found by looking for elements of order 4
or by pairing two different elements of order 2. These
subgroups are {e, a3, b, a3b}, {e, a3, ab, a4b},
{e, a3, a2b, a5b}.
None of these subgroups is normal.
Subgroups of Order 3: Subgroups of order 3 can only be
those generated by an element of order 3 (since 3 is prime).
The only such subgroup is {e, a2, a4}, and this subgroup is
normal.
Subgroups of Order 2: Subgroups of order 2 can only be
those generated by an element of order 2 (since 2 is prime).
These subgroups are {e, a3}, {e, b}, {e, ab}, {e, a2b},
{e, a3b}, {e, a4b}, {e, a5b}. The only one of these
subgroups which is normal is {e, a3}.
Subgroups of Order 1: Of course, the only possible
subgroup is the trivial group {e}, and of course this
subgroup is normal.
11.-continued
(b) G/H = {H, aH, bH, abH} (Note: this of course is
not the only way to represent the four elements of this
quotient group.) The corresponding multiplication table is
as follows:
H
aH
bH
abH
H | H
aH
bH
abH
aH | aH
H
abH
bH
bH | bH
abH
H
aH
abH | abH
bH
aH
H
12. To prove that x = (a, e), that is, to prove that x is the
greatest common divisor of a and e, you need to show that
x satisfies each of the three conditions of Definition 2.11.
You should find it trivially easy to show that the first
condition is satisfied.
To show that the second condition is satisfied, first use the
facts that x | d and d | a , and then use the facts that x | c and
c|e .
To show that the third condition is satisfied, suppose that
y | a and y | e . First say why y | b and why y | c , next say
why y | d , and finally say why y | x .