Math 110A
Exam 2 Solutions
1. Give an example for each of the following. You do not need to justify your response.
(a) A non-abelian group
Lots of possibilities here. Here’s one: the quaternions.
(b) A subset of S3 that is not a subgroup of S3.
Lots of possibilities here. Here’s one: {(12)}.
2. True or False. Please briefly justify your response.
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Z6 is a subgroup of Z12 .
Z6 is not a subset of Z12, because [a] in Z6 is different thatn [a] in Z12 .
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Let a, b, x, y ∈ Z. If d = ax + by, then d = (a, b).
The GCD is the SMALLEST positive integer that can be expressed as a linear
combination of a and b.
Here is a counterexample, to show the statement is false. Let a = 4 and b = 6. Then
10 = a(1) + b(1), but 10 is not the GCD of 4 and 6.
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If p is prime, then Zp is a group with respect to multiplication.
[0] does not have a multiplicative inverse.
3. (a) Find elements [a], [b] ∈ Z8 such that [a][b] = [0], but [a] 6= [0] and [b] 6= [0].
More than one possibility, here’s one: [2][4] = [0], but [2] 6= [0] and [4] 6= [0].
(b) What is the additive inverse of [2] in Z8 ?
The additive inverse of [2] is [−2], or equivalently [6].
(c) List all elements in Z8 that have a multiplicative inverse.
U (Z8 ) = {[1], [3], [5], [7]}
(d) What is the multiplicative inverse of [3] in Z8 ?
[3][3] = [9] = [1]
Thus the multiplicative inverse of [3] is [3].
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Math 110A
Exam 2 Solutions
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4. Define ∗ on Z by a ∗ b = a + b − 2. Prove that Z is a group with respect to ∗.
Proof. If x, y ∈ Z, then x + y − 2 ∈ Z, so ∗ is a binary operation on Z.
Let x, y, z ∈ Z.
x ∗ (y ∗ z) = x ∗ (y + z − 2) = x + (y + z − 2) − 2 = x + y + z − 4
(x ∗ y) ∗ z = (x + y − 2) ∗ z = (x + y − 2) + z − 2 = x + y + z − 4
Thus x ∗ (y ∗ z) = (x ∗ y) ∗ z, so ∗ is associative.
We claim that 2 is the identity. Let x ∈ Z.
x∗2 = x+2−2 = x
2∗x = 2+x−2 = x
Thus 2 is the identity.
Let x ∈ Z. We claim that 4 − x is the inverse of x.
x ∗ (4 − x) = x + (4 − x) − 2 = 2
(4 − x) ∗ x = (4 − x) + x − 2 = 2
Thus the inverse of x is 4 − x.
Therefore ∗ is an associative binary operation on Z and Z contains the identity and inverses. Therefore
(Z, ∗) is a group.
5. Let G be a group and let a be a fixed element in G. Let H = {x ∈ G | axa−1 = x}. Prove H is a
subgroup of G.
Proof. H is a subset of G by definition.
Let e be the identity in G. Then aea−1 = aa−1 = e. Thus aea−1 = e, and so e ∈ H. Hence H 6= ∅.
Let x, y ∈ H. Thus axa−1 = x and aya−1 = y. Therefore we have the following.
xy = (axa−1 )(aya−1) = axya−1
Thus axya−1 = xy, hence xy ∈ H.
Moreover, x−1 = (axa−1 )−1 = (a−1 )−1 x−1 a−1 = ax−1 a−1 . Thus ax−1 a−1 = x−1 , hence x−1 ∈ H.
Therefore by the subgroup test, H is a subgroup of G.
Math 110A
Exam 2 Solutions
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6. Let [a] ∈ Zn . Prove that if [a] = [1], then (a, n) = 1.
Proof.
[a] = [1] ⇒ a ≡ 1(mod n)
⇒ n|a − 1
⇒ a − 1 = nk for some k ∈ Z
⇒ 1 = a(1) + n(−k)
Thus we have expressed 1 as a linear combination of a and n. Since 1 is the smallest positive integer,
then (a, b) = 1.
7. Let a, b ∈ Z, not both zero and let d = (a, b). Prove that an integer c can be expressed as a linear
combination of a and b if and only if d|c.
Proof. Assume a, b ∈ Z not both zero and let d = (a, b).
(⇒) Assume c can be expressed as a linear combination of a and b. So there exists integers x and y
such that c = ax + by. By definition, d|a and d|b. Hence d divides any linear combination of a
and b. Hence d|ax + by. In other words, d|c.
(⇐) Assume d|c. Then there exists an integer k such that c = dk. Since d is the GCD of a and b
there exists integers m and n such that d = am + bn. Therefore we have the following.
c = dk
= (am + bn)k
= a(mk) + b(nk)
Since m, n, k ∈ Z, then mk ∈ Z and nk ∈ Z. Thus we have expressed c as a linear combination
of a and b.