S11MTH3175GroupTheory
Quiz 1- Solutions
Name:
Please explain all your work.
1. Use Euclidean algorithm:
(a) Find g = gcd(660, 102). Euclidian algorithm:
660
1
0
1
−2
102
0
1
-6
13
660
102 6
48 2
6 8
0
gcd(660, 102) = 6
(b) Find a pair of integers α, β such that 660α + 102β = g.
Solution:
• α = −2, β = 13 and we have
• 660(−2) + 102(13) = 6 since −1320 + 1326 = 6.
(c) Find the general solution for the pairs of integers α, β such that 660α + 102β = g.
• If one pair of integers that give a solution is (α0 , β0 ) then the general solution will be:
(α + k 102
, β − k 660
)
6
6
• 660(−2 + k
102
660
) + 102(13 − k
) = 6 where k can be any integer.
6
6
1
S11MTH3175GroupTheory
Quiz 1- Solutions
Name:
2. (a) List all elements of U (18).
Solution:
•
•
•
•
Z18 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
U (18) = {x ∈ Z18 | x has a multiplicative inverse (mod 18)}
U (18) = {x ∈ Z18 | x is relatively prime to 18}
U (18) = {x ∈ Z18 | gcd(18, x) = 1}
• U (18) = {1, 5, 7, 11, 13, 17}
(b) Find multiplicative inverse of 7 in U (18).
Solution:
•
•
•
•
Multiplicative inverse of 7 in U (18) is an integer y ∈ U (18) such that 7y = 1 ∈ U (18)
Find y ∈ U (18) such that 18k + 7y = 1 for some integer k.
Since gcd(18, 7) = 1 there exist integers α, β ∈ Z such that 18α + 7β = 1
Euclidian algorithm:
18
1
0
1
-1
2
•
•
•
•
7
0
1
-2
3
−5
18
7
4
3
1
0
2
1
1
3
α = 2, β = −5 and 18(2) + 7(−5) = 1
Find y ∈ U (18) such that (−5)(mod18) = y
y = 13 since −5 = 18(−1) + 13
So 7 · 13 = 1 (mod 18).
• Therefore 7−1 = 13 ∈ U (18)
• Another way would be to compute each product with 7 and see:
7 · 1 = 7(mod18)
7 · 5 = 35 = 17(mod18)
7 · 7 = 49 = 13(mod18)
7 · 11 = 77 = 5(mod18)
7 · 13 = 91 = 1(mod18) implies 7−1 = 13 ∈ U (18)
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S11MTH3175GroupTheory
−1
3. Find A
Quiz 1- Solutions
Name:
2 5
where A =
in M2 (Z15 ).
2 1
Solution:
1
d
−b
a b
, which is defined only if ad − bc 6= 0.
• Let A =
. Then A−1 =
c d
ad − bc −c a
1
1
1 −1 5
2 5
1 −5
1 −5
−1
• If A =
, then A =
=
=
.
2 1
2 · 1 − 5 · 2 −2 2
−8 −2 2
8 2 −2
• All values must be computed in Z15 .
•
1
8
= 2 ∈ Z15 since (8)(2) = 16 = 1(mod15)
−1
• A
• A
−1
1 −1 5
−1 5
−2 10
13 10
=2
=
=
∈ M2 (Z15 ).
=
2 −2
4 −4
4 11
8 2 −2
13 10
=
∈ M2 (Z15 )
4 11
• It is a good idea to check your answer by computing:
2 5 13 10
46 75
1 0
−1
AA =
=
=
∈ M2 (Z15 )and
2 1 4 11
30 31
0 1
13 10 2 5
46 75
1 0
−1
A A=
=
=
∈ M2 (Z15 ).
4 11 2 1
30 31
0 1
3
S11MTH3175GroupTheory
Quiz 1- Solutions
Name:
4. Let G be a group. Suppose g 2 = e for all g ∈ G. Prove that G is Abelian.
Proof: Assume g 2 = e for all g ∈ G.
Want to show ab = ba for all a, b ∈ G .
(a) Let a, b ∈ G be any two elements in G.
(b) Then ab ∈ G, (since G is closed under binary operation).
(c) Therefore (ab)2 = e, (since this is true for all elements in G).
(d) (ab)2 = (ab)(ab), by definition of square of any element.
(e) (ab)(ab) = (ab)2 = e, from d) and c)
(f) (ab)(ab) = e, (transitivity of ”=”)
(g) a(b(ab)) = (ab)(ab), (associative law in group G)
(h) a(b(ab)) = e, (from (g), (f) and transitivity of ”=”).
(i) a(a(b(ab))) = ae, (multiply the above by a on the left)
(j) (aa)(b(ab)) = a, (associative law on the left side and property of identity e on the right
side)
(k) (e)(b(ab)) = a, using the assumption that g 2 = e for all g ∈ G (in particular this is true
for a ∈ G.
(l) (e)(b(ab)) = e(b(ab)) = (eb)(ab)) = b(ab) = (ba)b, (first just delete unnecessary parenthesis, second is associative law, third is property of identity e and the forth is associative
law again.
(m) (ba)b = a, (use (l) and (k) and transitivity of ”=”)
(n) (ba)b = a, (multiply this by b on the right side and get:
(o) ((ba)b)b = ab,
(p) ((ba)b)b = (ba)(bb) = (ba)b2 = (ba)e = ba (apply associative law & definition of b2 &
property that g 2 = e &property of identity e - exactly in this order for each ”=” ).
(q) ab = ba, follows from (o) and (p).
(r) Therefore G is Abelian (commutative).
• Remark: Usually you will not write every single step that I wrote above, but even if you
skip steps you should understand which steps you skipped and what are the logical steps
that you are using.
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S11MTH3175GroupTheory
Quiz 1- Solutions
Name:
5. Give an example of a group G and two elements a, b ∈ G such that ab 6= ba.
Solution:
• Group G
Let G = Gl2 (R)
= {all invertible 2x2 matrices with real number
entries}
x y
G = Gl2 (R) =
| x, y, z, w ∈ R, (xw − yz) 6= 0
z w
G is a group under matrix multiplication.
• Two elements a, b ∈ G
1 2
4 0
Let a =
and b =
.
0 3
0 5
det(a) = (1)(3) − (2)(0) = 3 6= 0 and 1, 2, 0, 3 ∈ R. therefore a ∈ Gl2 (R) = G
det(b) = (4)(5) − (0)(0) = 20 6= 0 and 4, 0, 0, 5 ∈ R. therefore b ∈ Gl2 (R) = G
• Check that ab 6= ba
1 2 4
ab =
0 3 0
4 0 1
ba =
0 5 0
0
4 10
=
5
0 15
2
4 8
=
3
0 15
Therefore ab 6= ba.
• There are many other examples, but in each case you have to:
– state clearly what is the group G (including binary operation),
– give two elements a, b and make sure that they are in the group,
– check that ab 6= ba.
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