F11MTH3175 GroupTheory
Quiz 1, Solutions
Name:
Please explain all your work.
1. Use Euclidian algorithm:
4pts
(a) Find the greatest common divisor gcd(660, 105) = d.
Answer:
s
1
0
1
-3
t
0
1
-6
19
660
105
30
15
0
6
3
2
Therefore gcd(660, 105) = 15 .
4pts
(b) Find a pair of integers s, t ∈ Z such that s · 660 + t · 105 = d.
Answer: From the above table: −3 · 660 + 19 · 105 = 15 .
2. Let p, q, r be three distinct primes. Let a = p5 q and b = q 3 r2 .
4pts
(a) Find the greatest common divisor gcd(a, b).
Answer: a = p5 q = p5 q 1 r0 and b = q 3 r2 = p0 q 3 r2 .
gcd(a, b) = pmin{5,0} q min{1,3} rmin{0,2} = p0 q 1 r0 = q
4pts
Therefore gcd(a, b) = q .
(b) Find the least common multiple lcm(a, b).
Answer: a = p5 q = p5 q 1 r0 and b = q 3 r2 = p0 q 3 r2 .
lcm(a, b) = pmax{5,0} q max{1,3} rmax{0,2} = p5 q 3 r2 = q
Therefore gcd(a, b) = p5 q 3 r2 .
3. Consider the group U (20).
2pts
3pts
(a) How many elements are there in U (20)?
Answer: |U (20)| = #{elements in U (20)} = ϕ(20) = ϕ(22 5) =
ϕ(22 )ϕ(5) = (2 − 1)2(5 − 1) = 2 · 4 = 8. Therefore U (20) has 8 elements .
(b) List all elements of U (20).
Answer: U (20) = {all integers in Z20 which have mutiplicative inverses in Z20 } =
{0 ≤ i ≤ 19 | i is relatively prime to 20} = {1, 3, 7, 9, 11, 13, 17, 19} .
1
F11MTH3175 GroupTheory
4pts
Quiz 1, Solutions
Name:
(c) Find the multiplicative inverse of 7 in U (20).
Answer: Find integer 1 ≤ t ≤ 19 such that t · 7 = 1 (mod20).
One way: Since the numbers are small, one can guess and see that 3·7 = 21 = 1(mod20).
Therefore multiplicative inverse of 7 in U (20) is 3 .
Another way: Since gcd(20, 7) = 1, there exist integers s, t ∈ Z such that s · 20 + t · 7 = 1.
Use Euclidean algorithm and find s, t.
s t
1 0
0 1
1 -2
-1 3
20
7
6
1
0
2
1
6
Therefore −1 · 20 + 3 · 7 = 1. Hence 3 · 7 = 1(mod20).
Therefore multiplicative inverse of 7 in U (20) is 3 .
2
F11MTH3175 GroupTheory
8pts
Quiz 1, Solutions
Name:
4. Let G be a group. Suppose (ab)−1 = a−1 b−1 for all elements a, b ∈ G. Prove that G is abelian.
Proof: Assumption is: (ab)−1 = a−1 b−1 for all elements a, b ∈ G.
Want to show ab = ba for all elements a, b ∈ G.
——————Let a, b ∈ G. Then ab ∈ G since G is closed under operation.
(ab)−1 = a−1 b−1 for all elements a, b ∈ G by assumption.
Multiply (ab)−1 = a−1 b−1 by ab on the left and get:
(ab)(ab)−1 = (ab)(a−1 b−1 )
e = (ab)(a−1 b−1 ) since (ab)(ab)−1 = e by the inverse property for the element ab ∈ G.
Multiply e = (ab)(a−1 b−1 ) by b on the right:
eb = ((ab)(a−1 b−1 )) b
b = [(ab)(a−1 b−1 )]b since e is the identity element in G.
b = (ab)[(a−1 b−1 )b] by associative law in G.
b = (ab)[a−1 (b−1 b)] by associative law again in G.
b = (ab)[a−1 e] by inverse property for b ∈ G.
b = (ab)[a−1 ] by identity property for e ∈ G. Write:
b = (ab)a−1
Multiply b = (ab)a−1 by a on the right side:
ba = ((ab)a−1 )a ba = (ab)(a−1 a) by associative law.
ba = (ab)(e) by inverse property for ∈ G.
ba = (ab) by identity property of e ∈ G.
Therefore ab = ba for all a, b ∈ G. (This actually uses symmetric property of =.)
Remark: There are several other ways of proving this.
8pts
5. Show that {1, 2, 3} under multiplication modulo 4 is not a group.
/ {1, 2, 3}, the set {1, 2, 3} is not closed under
One way: 2 · 2 = 4 = 0(mod4). Since 0 ∈
operation, hence {1, 2, 3} is not a group.
Second way: Make Cayley table:
·
1
2
3
1
1
2
3
2
2
0
2
3
3
2
1
Since new element 0 appeared in the table, the set is not closed under operation.
Third way: Since 1 ∈ {1, 2, 3} is the identity and there is no y ∈ {1, 2, 3}
such that y · 2 = 1(mod4), it follows that 2 does not have an inverse.
Fourth way: Since gcd(2, 4) = 2 6= 1, it follows that 2 does not have multiplicative inverse
(mod4).
3
F11MTH3175 GroupTheory
9pts
6. Find A−1
Quiz 1, Solutions
Name:
4 5
where A =
in Gl2 (Z11 ).
2 1
Answer: Since det(A) = 4 · 1 − 5 · 2 = 4 − 10 = −6 = 5(mod11), the matrix A is invertible,
i.e. A ∈ Gl2 (Z11 ).
d −b
a b
1
−1
. Therefore:
For a matrix A =
the inverse matrix is given by A = ad−bc
−c a
c d
1 −5
1 −5
1 −5
1
1
1
−1
= −6
=5
A = 4·1−5·2
−2 4
−2 4
−2 4
Find
1
5
∈ Z11 , i.e. find multiplicative inverse of 5 ∈ Z11 .
Since gcd(11, 5) = 1 you can use Euclidean algorithm to find s, t ∈ Z such that s · 11 + t · 5 = 1,
and get:
(−4) · 11 + 9 · 5 = 1. Therefore 9 · 5 = 1(mod11). Hence 51 = 9 ∈ Z11 .
−1
A
1 −5
9 −45
9 10
=9
=
=
−2 4
−18 36
4 3
Check: AA−1 = I and A−1 A = I.
4 5 9 10
56
AA =
=
2 1 4 3
22
9 10 4 5
56
−1
A A=
=
4 3 2 1
22
−1
55
1
=
23
0
55
1
=
23
0
0
=I
1
0
=I
1
4
−1
A
9 10
=
4 3