MAT-314 Midterm Exam 2 Review

Abstract Algebra, Saracino
Second Edition
Midterm Exam 2
(Review)
“A person who never made a
mistake never tried anything new.“
– Albert Einstein -

Lehman College, Department of Mathematics
Midterm Exam 2 Review (1 of 10)
1(a) Let 𝐺 be a group with identity element 𝑒. Define the
center, 𝑍(𝐺), of 𝐺 as the set:
Show that 𝑍(𝐺) is a subgroup of 𝐺.
1(b) Let 𝐺 be a group with identity element 𝑒 and let 𝑎
be a fixed element of 𝐺. Define the centralizer, 𝐶 𝐺(𝑎), of
𝑎 in 𝐺 as the set:
Show that 𝐶 𝐺 𝑎 is a subgroup of 𝐺.
𝑍 𝐺 = 𝑎 ∈ 𝐺 | 𝑎𝑔 = 𝑔𝑎 for all 𝑔 ∈ 𝐺
𝐶 𝐺 𝑎 = 𝑔 ∈ 𝐺 | 𝑎𝑔 = 𝑔𝑎

1(a) Solution. Now, 𝑍(𝐺) is defined as the set:
(i) By definition, 𝑍 𝐺 ⊆ 𝐺.
(ii) Since 𝑒𝑔 = 𝑔𝑒 for all 𝑔 ∈ 𝐺 then 𝑒 ∈ 𝑍(𝐺) so 𝑍(𝐺) ≠ ∅.
(iii) Let 𝑎, 𝑏 ∈ 𝑍(𝐺). Then, 𝑎𝑔 = 𝑔𝑎 and 𝑏𝑔 = 𝑔𝑏 for all
𝑔 ∈ 𝐺. Therefore:
It follows that 𝑎𝑏 𝑔 = 𝑔(𝑎𝑏) for all 𝑔 ∈ 𝐺, and 𝑎𝑏 ∈ 𝑍 𝐺 ,
Therefore, 𝑍 𝐺 is closed under the group operation in 𝐺.
(iv) Let 𝑎 ∈ 𝑍(𝐺). Then, 𝑎𝑔 = 𝑔𝑎 for all 𝑔 ∈ 𝐺:
Hence 𝑎−1
𝑔 = 𝑔𝑎−1
and 𝑍 𝐺 is closed under inverses.
𝑍 𝐺 = 𝑎 ∈ 𝐺 | 𝑎𝑔 = 𝑔𝑎 for all 𝑔 ∈ 𝐺
𝑎𝑏 𝑔 = 𝑎 𝑏𝑔 = 𝑎 𝑔𝑏 = 𝑎𝑔 𝑏 = 𝑔𝑎 𝑏 = 𝑔(𝑎𝑏)
𝑎−1
𝑔 = 𝑎−1
𝑔−1 −1
= 𝑔−1
𝑎 −1
= 𝑎𝑔−1 −1
= 𝑔−1 −1
𝑎−1
= 𝑔𝑎−1

1(b) Solution. Now, 𝐶 𝐺(𝑎) is defined as the set:
(i) By definition, 𝐶 𝐺(𝑎) ⊆ 𝐺.
(ii) Since 𝑎𝑒 = 𝑒𝑎, then 𝑒 ∈ 𝐶 𝐺(𝑎), and 𝐶 𝐺(𝑎) ≠ ∅.
(iii) Let 𝑥, 𝑦 ∈ 𝐶 𝐺 𝑎 . Then, 𝑎𝑥 = 𝑥𝑎 and 𝑎𝑦 = 𝑦𝑎.
Therefore:
It follows that 𝑥𝑦 ∈ 𝐶 𝐺 𝑎 , and 𝐶 𝐺 𝑎 is closed under the
group operation in 𝐺.
(iv) Let 𝑥 ∈ 𝐶 𝐺 𝑎 . Then, 𝑎𝑥 = 𝑥𝑎, and
Hence 𝑥−1
𝑎 = 𝑎𝑥−1
and 𝑍 𝐺 is closed under inverses.
𝑎 𝑥𝑦 = 𝑎𝑥 𝑦 = 𝑥𝑎 𝑦 = 𝑥 𝑎𝑦 = 𝑥 𝑦𝑎 = 𝑥𝑦 𝑎
𝑥−1
(𝑎𝑥)𝑥−1
= 𝑥−1
𝑥𝑎 𝑥−1
𝑥−1
𝑎 𝑥𝑥−1
= (𝑥−1
𝑥)(𝑎𝑥−1
)
𝐶 𝐺 𝑎 = 𝑔 ∈ 𝐺 | 𝑎𝑔 = 𝑔𝑎

2. Which of the following groups are abelian? Which
one is cyclic? Justify your answer.
3(a) Let 𝐺1 and 𝐺2 be two groups with identity elements
𝑒1 and 𝑒2, respectively. Let 𝐻 be a subgroup of 𝐺1 and let
𝐾 be a subgroup of 𝐺2. Show that 𝐻 × 𝐾 is a subgroup
of 𝐺1 × 𝐺2.
3(b) Let 𝐺1 and 𝐺2 be two groups. Show that 𝐺1 × 𝐺2 is
abelian if and only if 𝐺1 and 𝐺2 are abelian.
(i) ℤ3 × 𝑆3
(ii) ℤ8 × ℤ9
(iii) ℤ4 × ℤ6

2. Solution. (i) ℤ3× 𝑆3 is nonabelian since 𝑆3 is
nonabelian. Hence it is also noncyclic,.
(ii) ℤ8 × ℤ9 is abelian since both ℤ8 and ℤ9 are abelian.
Also, ℤ8 × ℤ9 is cyclic because gcd 8, 9 = 1 .
(ii) ℤ4 × ℤ6 is abelian since both ℤ4 and ℤ6 are abelian.
But ℤ4 × ℤ6 is not cyclic since gcd 4, 6 = 2 ≠ 1 .
3(a) Solution. We want to show that 𝐻 × 𝐾 is a
subgroup of 𝐺1 × 𝐺2.
(i) 𝐻 × 𝐾 ⊆ 𝐺1 × 𝐺2 by definition.
(ii) 𝐻 × 𝐾 ≠ ∅, since 𝑒1, 𝑒2 ∈ 𝐻 × 𝐾
(iii) Let (𝑎1, 𝑏1) and (𝑎2, 𝑏2) be elements of 𝐻 × 𝐾, then
𝑎1, 𝑏1 𝑎2, 𝑏2 = 𝑎1 𝑎2, 𝑏1 𝑏2 ∈ 𝐻 × 𝐾 since 𝑎1 𝑎2 ∈ 𝐻,

(3a). Solution. (iii) Let (𝑎1, 𝑏1) and (𝑎2, 𝑏2) be elements of
𝐻 × 𝐾, then 𝑎1, 𝑏1 𝑎2, 𝑏2 = 𝑎1 𝑎2, 𝑏1 𝑏2 ∈ 𝐻 × 𝐾 since
𝑎1 𝑎2 ∈ 𝐻, and 𝑏1 𝑏2 ∈ 𝐾 by closure under subgroups.
(iv) If 𝑎, 𝑏 ∈ 𝐻 × 𝐾, then 𝑎, 𝑏 −1
= 𝑎−1
, 𝑏−1
∈ 𝐻 × 𝐾
since 𝑎−1
∈ 𝐻 and 𝑏−1
∈ 𝐾 since subgroups are closed
under inverses.
(3b). Let 𝐺1 and 𝐺2 be two groups. Show that 𝐺1 × 𝐺2 is
abelian if and only if 𝐺1 and 𝐺2 are abelian.
Solution (⟸). Let 𝐺1 and 𝐺2 be abelian and let (𝑎1, 𝑏1)
and (𝑎2, 𝑏2) be elements of 𝐺1 × 𝐺2, then 𝑎1, 𝑎2 ∈ 𝐺1 and
𝑏1, 𝑏2 ∈ 𝐺2. Then
𝑎1, 𝑏1 𝑎2, 𝑏2 = 𝑎1 𝑎2, 𝑏1 𝑏2
= 𝑎2 𝑎1, 𝑏2 𝑏1 = 𝑎2, 𝑏2 𝑎1, 𝑏1

(3b). Solution (⟹). Let 𝐺1 × 𝐺2 be abelian and let
𝑎1, 𝑎2 ∈ 𝐺1 and 𝑏1, 𝑏2 ∈ 𝐺2 then (𝑎1, 𝑏1) and (𝑎2, 𝑏2) are
elements of 𝐺1 × 𝐺2, and
But
and
Therefore, 𝑎1 𝑎2 = 𝑎2 𝑎1 and 𝑏1 𝑏2 = 𝑏2 𝑏1. It follows that
𝐺1and 𝐺2 are both abelian groups.
𝑎1, 𝑏1 𝑎2, 𝑏2 = 𝑎1 𝑎2, 𝑏1 𝑏2
𝑎2, 𝑏2 𝑎1, 𝑏1 = 𝑎2 𝑎1, 𝑏2 𝑏1
𝑎1, 𝑏1 𝑎2, 𝑏2 = 𝑎2, 𝑏2 𝑎1, 𝑏1

4(a) Let 𝐺 be a group and let 𝑥 be a fixed element of 𝐺.
Define a function 𝑓𝑥: 𝐺 → 𝐺 by 𝑓𝑥 𝑎 = 𝑥𝑎𝑥−1
. Then
𝑥𝑎𝑥−1
is called the conjugate of 𝑎 by 𝑥. Show that the
function 𝑓𝑥 is one-to-one and onto 𝐺.
4(b) Let 𝐺 be a group and let 𝑎 be an element of 𝐺.
Define a function 𝑓: 𝐺 → 𝐺 by 𝑓 𝑎 = 𝑎−1
. Show that the
function 𝑓 is one-to-one and onto 𝐺.

4(a) Solution. Let 𝑎, 𝑏 ∈ 𝐺, then 𝑓𝑥 𝑎 = 𝑥𝑎𝑥−1
and
𝑓𝑥 𝑏 = 𝑥𝑏𝑥−1
. Suppose 𝑓𝑥 𝑎 = 𝑓𝑥(𝑏), then
and 𝑎 = 𝑏 by left and right cancellation. Therefore 𝑓 is
one-to-one. Let 𝑐 ∈ 𝐺 be an arbitrary element. Suppose
𝑐 = 𝑓𝑥(𝑎) for some 𝑎 ∈ 𝐺, then
and 𝑥−1
𝑐𝑥 = 𝑥−1
𝑥𝑎𝑥−1
𝑥 = 𝑥−1
𝑥 𝑎 𝑥−1
𝑥 = 𝑎.
Therefore, for all 𝑐 ∈ 𝐺, there exists 𝑎 = 𝑥−1
𝑐𝑥 ∈ 𝐺 such
that 𝑐 = 𝑓𝑥 𝑎 . It follows that 𝑓𝑥 is onto 𝐺.
4(b) Let 𝑎, 𝑏 ∈ 𝐺, then 𝑓 𝑎 = 𝑎−1
and 𝑓 𝑏 = 𝑏−1
.
Suppose 𝑓 𝑎 = 𝑓(𝑏), then 𝑎−1 = 𝑏−1 and 𝑎 = 𝑏.
𝑥𝑎𝑥−1
= 𝑥𝑏𝑥−1
𝑐 = 𝑥𝑎𝑥−1

4(b) Solution Suppose 𝑓 𝑎 = 𝑓(𝑏), then 𝑎−1
= 𝑏−1
and
𝑎 = 𝑏. Therefore 𝑓 is one-to-one. Let 𝑐 ∈ 𝐺 be an arbitrary
element. Suppose 𝑐 = 𝑓(𝑎) for some 𝑎 ∈ 𝐺, then
and 𝑐−1
= 𝑎. Hence, for all 𝑐 ∈ 𝐺, there exists 𝑎 = 𝑐−1
∈ 𝐺
such that 𝑐 = 𝑓 𝑎 . It follows that 𝑓 is onto 𝐺.
𝑐 = 𝑎−1

5. Let 𝛼, 𝛽 ∈ 𝑆4 be two permutations, with:
(i) Write 𝛼 as a product of disjoint cycles.
(ii) Write 𝛽 as a product of 2-cycles (transpositions).
(iii) Write the product 𝛼𝛽 in cycle notation.
(iv) Is 𝛽 an even permutation? Justify your answer.
(v) Is 𝛼−1
an even permutation? Justify your answer.
(vi) Determine the order of 𝛽 in 𝑆4
𝛼 =
1 2
3 1
3
4
4
2
𝛽 =
1 2
3 4
3
1
4
2

5. (Solution) Let 𝛼, 𝛽 ∈ 𝑆4 be two permutations, with:
(i) Since 𝛼 1 = 3; 𝛼 3 = 4; 𝛼 4 = 2 and 𝛼 2 = 1,
then 𝛼 = (1 3 4 2).
(ii) Since 𝛽 1 = 3; 𝛽 3 = 1 and 𝛽 2 = 4; 𝛽 2 = 2,
then 𝛽 = (1 3)(2 4).
(iii) 𝛼𝛽 =
1 2
3 1
3
4
4
2
∘
1 2
3 4
3
1
4
2
=
1 2
4 2
3
3
4
1
= (1 4)
(iv) Since 𝛽 = (1 3)(2 4), then 𝛽 is an even permutation.
(v) 𝛼−1 =
1 2
2 4
3
1
4
3
= 1 2 4 3 = (1 3)(1 4)(1 2).
Thus, 𝛼−1
is an odd permutation (product of 3 2-cycles).
𝛼 =
1 2
3 1
3
4
4
2
𝛽 =
1 2
3 4
3
1
4
2

5. (Solution) (vi) Determine the order of 𝛽 in 𝑆4
Since 1 3 and (2 4) are disjoint cycles then the order of
𝛽 = (1 3)(2 4) is given by o 𝛽 = lcm(o 1 3 , o 2 4 ).
Now, (1 3) and (2 4) are of order 2, since they are 2-
cycles. It follows that o 𝛽 = lcm 2, 2 = 2.
The above result could have been computed directly:
𝛽 =
1 2
3 4
3
1
4
2
= (1 3)(2 4)
𝛽1
= 𝛽 =
1 2
3 4
3
1
4
2
≠
1 2
1 2
3
3
4
4
= 𝑒
𝛽2
=
1 2
3 4
3
1
4
2
∘
1 2
3 4
3
1
4
2
=
1 2
1 2
3
3
4
4
= 𝑒

6(a). Prove that the symmetric group 𝑆 𝑛 is nonabelian
for 𝑛 ≥ 3.
Hint: use the permutations 𝛼 = (1 2) and 𝛽 = (1 3).
First, show that 𝛼, 𝛽 ∈ 𝑆 𝑛 for 𝑛 ≥ 3. Finally, show that
𝛼𝛽 ≠ 𝛽𝛼.
6(b). Prove that the alternating group 𝐴 𝑛, consisting of
all even permutations of the symmetric group 𝑆 𝑛, is
nonabelian for 𝑛 ≥ 4.
Hint: use the permutations 𝜎 = (1 2 3) and 𝜏 = (2 3 4).
First, show that 𝜎, 𝜏 ∈ 𝐴 𝑛 for 𝑛 ≥ 4. Finally, show that
𝜎𝜏 ≠ 𝜏𝜎.

6(a). Prove that the symmetric group 𝑆 𝑛 is nonabelian
for 𝑛 ≥ 3.
Let 𝛼 = (1 2) and 𝛽 = (1 3). For 𝑛 ≥ 3. Then
It follows that 𝛼, 𝛽 ∈ 𝑆 𝑛 for 𝑛 ≥ 3. Finally, for 𝛼𝛽 we have
𝛼𝛽 1 = 𝛼 𝛽 1 = 𝛼 3 = 3. But for 𝛽𝛼 we have
𝛽𝛼 1 = 𝛽 𝛼 1 = 𝛽 2 = 2. Therefore 𝛼𝛽 ≠ 𝛽𝛼, and 𝑆 𝑛
is nonabelian for 𝑛 ≥ 3.
In cycle notation: 𝛼𝛽 = 1 2 1 3 and 𝛽𝛼 = (1 3)(1 2).
Therefore 𝛼𝛽 1 = 3 and 𝛽𝛼 1 = 2. Hence 𝛼𝛽 ≠ 𝛽𝛼,
and 𝑆 𝑛 is nonabelian for 𝑛 ≥ 3.
𝛽 =
1 2
3 2
3
1
… 𝑛
… 𝑛
𝛼 =
1 2
2 1
3
3
… 𝑛
… 𝑛

6(b). Prove that the alternating group 𝐴 𝑛 is nonabelian
for 𝑛 ≥ 4.
Let 𝜎 = (1 2 3) and 𝛽 = (2 3 4). For 𝑛 ≥ 4. Then
It follows that 𝜎, 𝜏 ∈ 𝐴 𝑛 for 𝑛 ≥ 4, since they are 3-cycles
that decompose into 2 transpositions (even permutation).
Hence 𝜎𝜏 = 1 2 3 2 3 4 and 𝜏𝜎 = (2 3 4)(1 2 3). Then
and
Hence 𝜎𝜏 ≠ 𝜏𝜎, and 𝐴 𝑛 is nonabelian for 𝑛 ≥ 4.
𝜏 =
1 2
1 3
3 4
4 2
… 𝑛
… 𝑛
𝛼 =
1 2
2 3
3
1
4 … 𝑛
4 … 𝑛
𝜎𝜏(1) = 𝜎(𝜏(1)) = 𝜎(1) = 2
𝜏𝜎(1) = 𝜏(𝜎(1)) = 𝜏(2) = 3

7. Let 𝐺 be the group ℤ6 under addition modulo 6. That
is, ℤ6 = 0, 1, 2, 3, 4, 5 , and let 𝐻 = ⟨3⟩ be the cyclic
subgroup of 𝐺 generated by the element 3 ∈ 𝐺.
(i) List the elements of 𝐻 in set notation.
(ii) Construct all left cosets of 𝐻 in 𝐺.
(iii) Determine all distinct left cosets of 𝐻 in 𝐺.
8. Let 𝐺 be a group and let 𝐻 be a subgroup of 𝐺. Show
that if 𝑎, 𝑏 ∈ 𝐺, then 𝑎𝐻 = |𝑏𝐻|. That is, the cardinality
of every left coset of 𝐻 in 𝐺 is the same.
Hint: Define a function 𝑓: 𝑎𝐻 → 𝑏𝐻 by 𝑓 𝑎ℎ = 𝑏ℎ for all
ℎ ∈ 𝐻. First, show that 𝑓 is one-to-one, then show that 𝑓
is onto 𝐺.

7. Let 𝐺 be the group ℤ6 under addition modulo 6. That
is, ℤ6 = 0, 1, 2, 3, 4, 5 , and let 𝐻 = 3 . Then
(i) 𝐻 = 3 = 3, 3 + 3 = 0 = 0, 3 .
(ii) Construct all left cosets of 𝐻 in 𝐺:
(iii) The distinct left cosets of 𝐻 in 𝐺 are:
0 + 𝐻 = 0 + 0, 0 + 3 = 0, 3 = 𝐻
1 + 𝐻 = 1 + 0, 1 + 3 = 1, 4
2 + 𝐻 = 2 + 0, 2 + 3 = 2, 5
3 + 𝐻 = 3 + 0, 3 + 3 = 3, 0
4 + 𝐻 = 4 + 0, 4 + 3 = 4, 1
5 + 𝐻 = 5 + 0, 5 + 3 = 5, 2
0 + 𝐻 = 3 + 𝐻 = 0, 3 = 𝐻
1 + 𝐻 = 4 + 𝐻 = 1, 4
2 + 𝐻 = 5 + 𝐻 = 2, 5

Right Cosets (6 of 8)
8. Solution Let 𝐻 be a subgroup of 𝐺. Let 𝑎, 𝑏 ∈ 𝐺. To
show that 𝑎𝐻 = 𝑏𝐻 , we will show that there is a one-
to-one and onto function from 𝑎𝐻 to 𝑏𝐻. Define a
function 𝑓: 𝑎𝐻 → 𝑏𝐻 by 𝑓 𝑎ℎ = 𝑏ℎ for all 𝑎ℎ ∈ 𝑎𝐻.
To show that 𝑓 is one-to-one, let 𝑎ℎ1, 𝑎ℎ2 ∈ 𝑎𝐻 for some
ℎ1, ℎ2 ∈ 𝐻 . Then, 𝑓 𝑎ℎ1 = 𝑏ℎ1 and 𝑓 𝑎ℎ2 = 𝑏ℎ2. If
𝑓 𝑎ℎ1 = 𝑓 𝑎ℎ2 , then 𝑏ℎ1 = 𝑏ℎ2. By left-cancellation of
the element 𝑏, we have ℎ1 = ℎ2 and thus 𝑎ℎ1 = 𝑎ℎ2.
Hence 𝑓 is one-to-one.
To show 𝑓 is onto, let 𝑐 ∈ 𝑏𝐻 be arbitrary, then we have
𝑐 = 𝑏ℎ for some ℎ ∈ 𝐻. But 𝑏ℎ = 𝑓(𝑎ℎ) for 𝑎ℎ ∈ 𝑎𝐻. It
follows that 𝑓 is onto 𝑎𝐻.

9. Let 𝐺 be a group and let 𝐻 and 𝐾 be subgroups of 𝐺,
such that 𝐻 = 33 and 𝐾 = 55. Show that 𝐻 ∩ 𝐾 is
cyclic.
10. Justify your answer to the following questions:
(a) Is 𝐴4 a normal subgroup of 𝑆4.
(b) Is 𝐻 = 0, 3, 6, 9 a normal subgroup of ℤ12
(c) Is the subgroup 𝐻 = 1, −1 a normal subgroup of
the group 𝒬8 = 1, 𝑖, −𝑖, 𝑗, −𝑗, 𝑘, −𝑘 under multiplication
with 𝑖2 = 𝑗2 = 𝑘2 = 𝑖𝑗𝑘 = −1.

9. Solution. Now, since 𝐻 and 𝐾 are subgroups of 𝐺
then 𝐻 ∩ 𝐾 is a subgroup of 𝐻 and a subgroup of 𝐾,
respectively.
By Lagrange’s Theorem |𝐻 ∩ 𝐾| divides 𝐻 = 33.
Therefore 𝐻 ∩ 𝐾 = 1, 3, 11 or 33. Similarly, 𝐾 = 55.
Hence 𝐻 ∩ 𝐾 = 1, 5, 11 or 55. Thus 𝐻 ∩ 𝐾 = 1, or 11.
If 𝐻 ∩ 𝐾 = 1, then 𝐻 ∩ 𝐾 = 𝑒 = ⟨𝑒⟩ is cyclic. Else, if
𝐻 ∩ 𝐾 = 11, then 𝐻 ∩ 𝐾 is a group of prime order and
is therefore cyclic by a corollary of Lagrange’s Theorem.

10. Solutions.
(a) 𝐴4 a normal subgroup of 𝑆4 because 𝑆4 = 4! = 24
and 𝐴4 = 4!/2 = 12. By Lagrange’s Theorem, the
index [𝑆3: 𝐴4] of 𝐴4 in 𝑆3 is given by:
It follows that 𝐴4 is an index 2 subgroup of 𝑆4 and is
therefore normal.
(b) ℤ12 is an abelian group. Therefore any subgroup of
ℤ12 is normal, including 𝐻 = 0, 3, 6, 9 .
(c) 𝐻 = 1, −1 is a subgroup of the center of 𝒬8, since
both 1 and −1 commute with every element of 𝒬8. It
follows that 𝐻 is a normal subgroup of 𝒬8.
[𝑆3: 𝐴4] =
𝑆4
𝐴4
=
24
12
= 2

MAT-314 Midterm Exam 2 Review

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MAT-314 Midterm Exam 2 Review