Math 3110
Name:
Big Quiz #1
Answer Key
September 23rd , 2013
Be sure to show your work!
1. (20 points) Definition and Basics
(a) Suppose that G is a non-empty set equipped an operation. What 4 things do I need to check to see if G is a
group?
1: Closure: For every a, b ∈ G, we have ab ∈ G.
2: Associativity: For every a, b, c ∈ G, we have (ab)c = a(bc).
3: Identity: There exists some element e ∈ G such that for all x ∈ G we have ex = xe = x.
4: Inverses: For each g ∈ G there exists some g −1 ∈ G such that gg −1 = g −1 g = e.
What additional property needs to hold for G to be an Abelian group?
5: Commutativity: ab = ba for all a, b ∈ G.
(b) Let G = Q≤0 be the non-positive rational numbers. Prove G is not a group under addition.
Q≤0 under addition satisfies the closure axiom (non-positive plus non-positive is still non-positive), associativity
holds (addition is associative), and has an additive identity (0 is non-positive). It even satisfies the commutativity
axiom (so if it was a group it would be an Abelian group). However, Q≤0 fails to contain inverses. So to show
that it fails to be a group, we will provided a concrete counter-example.
Notice that −1 ∈ Q≤0 . However, its inverse −(−1) = +1 6∈ Q≤0 (since +1 6≤ 0). Thus −1 has no (additive)
inverse (in Q≤0 ) so Q≤0 is not a group (under addition).
(c) Let G = R (the real numbers). Prove G is not a group under subtraction.
The reals under subtraction satisfy the closure axiom, but pretty much everything else goes wrong. For example,
the identity axiom does not hold (so we cannot really talk about inverses).
Notice that x − 0 = x for all x ∈ R. So if R had a “subtractive” identity it would have to be 0. However,
0 − x = −x 6= x (unless x = 0), so although 0 is a right identity it is not a left identity. To be a group R (with
subtraction) must have an identity which works on both sides. Thus R is not a group.
Associativity also fails to hold. We could show R with subtraction is not a group by demonstrating subtraction
is not associative. This is a bit simpler than showing there is no identity.
Consider 1, 2, 3 ∈ R. (1 − 2) − 3 = −1 − 3 = −4 6= 2 = 1 − (−1) = 1 − (2 − 3). Thus subtraction is not associative
and so R under subtraction is not a group.
2. (20 points) Working mod 20.
(a) What is the inverse of 3 in the group Z20 ? What operation makes Z20 a group?
Z20 is an abelian group under addition mod 20. The inverse of 3 is −3 = 17 (the inverse with respect to addition
mod 20).
(b) Is 3 an element of U (20)? If not, why not? If so, why so & what is its inverse?
Yes, 3 ∈ U (20) since 3 and 20 are relatively prime (their GCD is 1). The multiplicative inverse of 3 could be
computed using the extended Euclidean algorithm, but this is overkill. We should be able to “guess” it’s inverse
by multiplying 3 by various integers which are relatively prime to 20. After trying a few numbers, we see that
3 · 7 = 7 · 3 = 21 = 1 mod 20. So 3−1 = 7.
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(c) List all of the distinct cyclic subgroups, hxi, of Z20 .
A cyclic group has a unique subgroup of order k for each k dividing its order. So we have unique subgroups of
orders: 1, 2, 4, 5, 10, and 20. Given k (a divisor of n), n/k generates the cyclic subgroup of order k in Zn . Thus
we have. . .
h0i = {0}, h10i = {0, 10}, h5i = {0, 5, 10, 15}, h4i = {0, 4, 8, 12, 16},
h1i = {0, 1, 2, . . . , 18, 19} = Z20 are the distinct cyclic subgroups of Z20 .
h2i = {0, 2, 4, . . . , 16, 18},
and
(d) What is the order of 15 in Z20 ?
Three ways to compute this: (1) The GCD of 15 and 20 is 5 so the order of 15 is 20/5 = 4. (2) Consider
h15i = {0, 15, 15 + 15, . . . } = {0, 15, 10, 5} so |15| = |h15i| = 4. (3) 15 6= 0, 15 + 15 = 10 6= 0, 15 + 15 + 15 = 5 6= 0,
15 + 15 + 15 + 15 = 0. So the fourth (additive) power of 15 is 0, thus |15| = 4.
(e) Draw the subgroup lattice of Z20 [Note: 20 = 22 · 5].
We know that there is exactly one subgroup per
divisor of 20. These subgroups are arranged according to divisibility, so to draw a subgroup lattice we should first draw a divisibility lattice for
the divisors of 20. Then, again, to get a subgroup
of order k we can use the generator 20/k.
3. (20 points) More Modular Arithmetic.
(a) List the elements of U (8). Then find their orders and the list the elements in cyclic subgroup generated by
that element. [Note: There may be more spaces than you need.]
The elements of U (8) are exactly the integers between 0 and 8 − 1 = 7 which are relatively prime to 8. We can
find the cyclic subgroup generated by x by computing 1, x, x2 , x3 , etc. The size of this subgroup is the order of
its generator.
(b) Is U (8) cyclic?
x=
1
3
5
7
|x| =
1
2
2
2
hxi =
h1i = {1}
h3i = {1, 3}
h5i = {1, 5}
h7i = {1, 7}
Yes
/
No
(Circle the correct answer.)
U (8) is not cyclic since it does not have a generator (i.e. no element of order 4).
(c) Let A =
1
3
2
. Is A ∈ GL2 (Z7 )? If so, find A−1 . If not, explain why not.
4
1 2
= 1(4) − 3(2) = −2 = 5 (mod 7). Notice that 5 ∈ U (7) since 5 and 7 are relatively prime. Since
3 4
5−1 exists, the matrix has an inverse. We find 5−1 by “guessing”. Since 5 · 3 = 1 (mod 7) we have 5−1 = 3.
−1
a b
d −b
−1
The 2 × 2 matrix inverse formula says:
= (ad − bc)
.
c d
−c a
−1
1 2
4 −2
4 5
4 5
5 1
= (1(4) − 3(2))−1
= 5−1
=3
=
3 4
−3 1
4 1
4 1
5 3
det
2
(d) Find 36−1 mod 151 using the extended Euclidean algorithm.
Dividing with remainder we get: 151 = 36(4) + 7, 36 = 7(5) + 1, and 7 = 1(7) + 0. The last non-zero remainder
(i.e. 1) is the GCD of 151 and 36.
Working backwards, we have 1 = (1)36 + (−5)7. Subbing in 151 − 36(4) = 7 we get 1 = (1)36 + (−5)[151 − 36(4)]
so that 1 = (21)36 + (−5)151.
Therefore, GCD(151, 36) = (21)36 + (−5)151 = 1. This means that 36(21) = 1 (mod 151), so 36−1 = 21.
4. (20 points) Recall that D4 = {1, x, x2 , x3 , y, xy, x2 y, x3 y} = hx, y | x4 = 1, y2 = 1, (xy)2 = 1i.
(a) Use the relations for D4 to derive the relation: yx = x−1 y.
We have xyxy = (xy)2 = 1. Thus multiplying by x−1 on the left we get x−1 xyxy = x−1 1 so yxy = x−1 . Now
multiply by y on the right and get yxyy = x−1 y. But y 2 = 1 so yx = yxy 2 = x−1 y.
(b) Fill in the Cayley table for D4 :
1
x
x2
x3
y
xy
x2 y
x3 y
1
1
x
x2
x3
y
xy
x2 y
x3 y
x
x
x2
x3
1
xy
x2 y
x3 y
y
x2
x2
x3
1
x
x2 y
x3 y
y
xy
x3
x3
1
x
x2
x3 y
y
xy
x2 y
y
y
x3 y
x2 y
xy
1
x3
x2
x
xy
xy
y
x3 y
x2 y
x
1
x3
x2
x2 y
x2 y
xy
y
x3 y
x2
x
1
x3
x3 y
x3 y
x3 y
xy
y
x3
x2
x
1
(c) Find the inverse and order of each element in D4 .
g=
1
x
x2
x3
y
xy
x2 y
x3 y
g −1 =
1
x3
x2
x
y
xy
x2 y
x3 y
|g| =
1
4
2
4
2
2
2
2
(d) Do the rotations form a subgroup of D4 ? If so, why? If not, why not?
Yes, the rotations form a subgroup. To check if a (non-empty) set is a subgroup of a finite group, we just need
to verify that closure holds. In the case of rotations, a rotation composed with a rotation is a rotation. Thus
closure holds and so it’s a subgroup.
5. (20 points) Proofs!
(a) Choose one of the following:
I. Prove that f : Z → Z defined by f (x) = 5x is 1-1 but not onto.
Suppose that f (m) = f (n). Then 5m = 5n and so m = n. Thus f is 1-1.
Consider 1 ∈ Z. Suppose that f (x) = 1. Then 5x = 1 and so x = 1/5 6∈ Z. Therefore, 1 is not in the range
of f and so f is not onto.
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II. Let G be a group and g ∈ G. Prove that |g| = |g −1 |
(i.e. g and its inverse have the same order).
k
By definition hgi = {g k | k ∈ Z} and hg −1 i = { g −1
| k ∈ Z} = {g ` | ` = −k, k ∈ Z} = {g ` | ` ∈ Z}
−1
since −Z = Z. Therefore, hgi = hg i. Since the cyclic subgroups generated by both g and g −1 are
equal and since the size of a cyclic subgroup is the order of the element generating it, we have that
|g| = |hgi| = |hg −1 i| = |g −1 |.
Alternatively we could note that g k = e iff (g k )−1 = e−1 iff g −k = e. So any power which sends g to the
identity also sends g −1 to the identity (and vice-versa). Thus g and g −1 have the same order.
(b) Choose one of the following:
(You must use a subgroup test in your proof.)
I. Prove that SL2 (Z) = A ∈ Z2×2 | det(A) = 1 is a subgroup of GL2 (Z).
Suppose A ∈ SL2 (Z). Then det(A) = 1 so A ∈ GL2 (Z) = {A ∈ Z2×2 | det(A) = ±1}. Therefore,
SL2 (Z) ⊆ GL2 (Z). Since det(I2 ) = 1, we have I2 ∈ SL2 (Z) and so SL2 (Z) is non-empty. We still need to
check closure and closure under inversion.
Suppose A, B ∈ SL2 (Z). Then det(AB) = det(A)det(B) = 1 · 1 = 1. Therefore, AB ∈ SL2 (Z).
Suppose A ∈ SL2 (Z). Then det(A−1 ) = det(A)−1 = 1−1 = 1. Therefore, A−1 ∈ SL2 (Z).
II. Prove that H = {n ∈ Z | n = 10x + 6y for some x, y ∈ Z} is a subgroup of Z.
Obviously H is a non-empty subset of Z, so we just need to check closure under addition and inverses (i.e.
negation).
Suppose a, b ∈ H. Then there exists x, y, u, v ∈ Z such that a = 10x + 6y and b = 10u + 6v. We have
that a + b = (10x + 6y) + (10u + 6v) = 10(x + u) + 6(y + v) ∈ H since x + u and y + v are integers. Also,
−a = −(10x + 6y) = 10(−x) + 6(−y) ∈ H since −x and −y are integers. Therefore, H is a subgroup of Z.
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