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MA 541 – Modern Algebra I
Solutions to HW #1
1. Which of the following are groups under the given operation? Justify your
answers.
(a) The collection of odd integers under +.
• This is not a group, since it is not closed. Consider any element
m. Then m + m = 2m, which is even.
(b) The collection of even integers under +.
• This is a group.
• Consider any pair of elements m, n ∈ 2Z. Then m = 2x and
n = 2y for some pair of integers x and y. Therefore, m + n =
(2x) + (2y) = 2(x + y), which is even, so the operation is binary.
• Consider elements a, b, and c in 2Z. Then, by associativity of
addition in the integers, (a + b) + c = a + (b + c), so the operation
is associative.
• Consider any element a ∈ 2Z. Then 0 + a = a + 0 = a, so there
is an identity, namely 0.
• Consider any element a ∈ 2Z. Then a has an inverse, namely
−a, since a+(−a) = 0, which we showed above to be the identity
element.
(c) Z7 under multiplication.
• This is not a group, since the element 0 does not have an inverse.
For any a ∈ (Z7 , ·), 0 · a = 0 6= 1.
(d) Z7 − {0} under multiplication (that is, Z7 with zero removed).
• This is a group.
• Consider elements a, b, and c in Z7 − {0}. Then, by associativity
of multiplication in the integers, (a · b) · c = a · (b · c), so the
operation is associative.
• Consider any element a ∈ Z7 − {0}. Then 1 · a = a · 1 = a, so
there is an identity, namely 1.
• Consider any element a ∈ Z7 − {0}. Then a has an inverse. The
justification follows: 1 · 1 = 1 ≡ 1 (mod 7); 2 · 4 = 8 ≡ 1 (mod
7); 3 · 5 = 15 ≡ 1 (mod 7); 4 · 2 = 8 ≡ 1 (mod 7); 5 · 3 = 15 ≡ 1
(mod 7); 6 · 6 = 36 ≡ 1 (mod 7).
(e) The collection of positive real numbers under multiplication.
• This is a group.
• For any pair of elements a, b ∈ R+ , we have a · b > 0, since a > 0
and b > 0. So the operation is binary.
• Consider elements a, b, and c in R+ . By the associativity of
multiplication in the reals, (a · b) · c = a · (b · c), so the operation
is associative.
1
• Consider any element a ∈ R+ . Then 1 · a = a · 1 = a, so there is
an identity, namely 1.
• Consider any element a ∈ R+ . Since a > 0, we have 1/a > 0.
Since a · (1/a) = 1 each element of R+ has an inverse in R+ .
(f) The collection of positive real numbers under division.
• This is not a group, since the operation is not associative. For
example, (1/3)/2 = 1/6 6= 1/(3/2) = 2/3.
2. For b ∈ Zm , we say that b is a unit if there exists an element c ∈ Zm such
that b · c ≡ 1 (mod m). For example, 2 is a unit in Z5 since 2 · 3 ≡ 1 (mod
5).
(a) For each of the following m determine all of the units in Zm .
m = 3; 4; 5; 6; 7; 8; 9
•
•
•
•
•
•
•
m
m
m
m
m
m
m
=
=
=
=
=
=
=
3.
4.
5.
6.
7.
8.
9.
The
The
The
The
The
The
The
units
units
units
units
units
units
units
are
are
are
are
are
are
are
1 and 2.
1 and 3.
1, 2, 3, and 4.
1 and 5.
1, 2, 3, 4, 5, and 6.
1, 3, 5, and 7.
1, 2, 4, 5, 7, and 8.
(b) Many observations to make!
(c) In any of the above examples, do the non-zero elements of Zm form
a group?
• (Z3 − {0}, ·), (Z5 − {0}, ·), and (Z7 − {0}, ·) are all groups.
3. (a) For each a ∈ Z7 , compute a, a + a, a + a + a, etc. until the pattern
is clear. Any observations?
•
•
•
•
•
•
•
a = 0.
a = 1.
a = 2.
a = 3.
a = 4.
a = 5.
a = 6.
We
We
We
We
We
We
We
have
have
have
have
have
have
have
0,
1,
2,
3,
4,
5,
6,
0,
2,
4,
6,
1,
3,
5,
0,
3,
6,
2,
5,
1,
4,
and the pattern repeats.
4, 5, 6, 0, and the pattern
1, 3, 5, 0, and the pattern
5, 1, 4, 0, and the pattern
2, 6, 3, 0, and the pattern
6, 4, 2, 0, and the pattern
3, 2, 1, 0, and the pattern
repeats.
repeats.
repeats.
repeats.
repeats.
repeats.
(b) For each a ∈ Z8 , compute a, a + a, a + a + a, etc. until the pattern
is clear. Any observations?
• a = 0. We have 0, 0, 0, and the pattern repeats.
• a = 1. We have 1, 2, 3, 4, 5, 6, 7, and the pattern repeats.
• a = 2. We have 2, 4, 6, 0, and the pattern repeats.
2
•
•
•
•
•
a = 3.
a = 4.
a = 5.
a = 6.
a = 7.
We
We
We
We
We
have
have
have
have
have
3,
4,
5,
6,
7,
6,
0,
2,
4,
6,
1,
4,
7,
2,
5,
4,
0,
4,
0,
4,
7, 2, 5, 0, and the pattern repeats.
and the pattern repeats.
1, 6, 3, 0, and the pattern repeats.
and the pattern repeats.
3, 2, 1, 0, and the pattern repeats.
(c) For each a ∈ Z7 , compute a, a · a, a · a · a, etc. until the pattern is
clear. Any observations?
•
•
•
•
•
•
•
a = 0.
a = 1.
a = 2.
a = 3.
a = 4.
a = 5.
a = 6.
We
We
We
We
We
We
We
have
have
have
have
have
have
have
0,
1,
2,
3,
4,
5,
6,
0,
1,
4,
2,
2,
4,
1,
0,
1,
1,
6,
1,
6,
6,
and the pattern repeats.
and the pattern repeats.
and the pattern repeats.
4, 5, 1, and the pattern repeats.
and the pattern repeats.
2, 3, 1, and the pattern repeats.
1, and the pattern repeats.
(d) For each a ∈ Z8 , compute a, a · a, a · a · a, etc. until the pattern is
clear. Any observations?
•
•
•
•
•
•
•
•
a = 0.
a = 1.
a = 2.
a = 3.
a = 4.
a = 5.
a = 6.
a = 7.
We
We
We
We
We
We
We
We
have
have
have
have
have
have
have
have
0,
1,
2,
3,
4,
5,
6,
7,
0,
1,
4,
1,
0,
1,
4,
1,
0,
1,
0,
3,
0,
5,
0,
7,
and the pattern repeats.
and the pattern repeats.
0, 0, 0, and the pattern continues.
1, and the pattern repeats.
0, 0 and the pattern continues.
1, and the pattern repeats.
0, 0, 0, and the pattern continues.
1, and the pattern repeats.
(e) Any overall observations?
• Many observations!
3