Math 4281 Final Exam Practice Summer 2020

1. Prove that if I1 ⊆ I2 ⊆ · · · are ideals in a ring R, then I =

⋃∞

n=1 In is an ideal of R

Proof. Since 0 ∈ I1 ⊂ I, I 6= ∅.

Let a, b ∈ I. Then there exist n,m ∈ Z+ such that a ∈ In and b ∈ Im. WLOG, say

m ≥ n. Then a ∈ In ⊆ Im, so a, b ∈ Im, which is an ideal, so a− b ∈ Im ⊆ I.

Further, let r ∈ R. Then ra, ar ∈ In ⊆ I because In is an ideal.

Hence, I is an ideal by the Ideal Test.

2. Show that n element has order 2 in Sn if and only if its cycle decomposition if a product

of commuting 2-cycles.

Proof. Suppose first that σ ∈ Sn is a (disjoint) product of 2-cycles. But since disjoint

cycles commute, σ2 would be a product of squared 2-cycles. Since (a b)2 = 1 for any

a, b, we conclude that σ is of order 2.

Suppose now that σ ∈ Sn isn’t a product of 2-cycles, but instead has some other n-cycle

(n 6= 2) in its decomposition. Stil, the product of disjoint cycles commutes. But every n-

cycle has order n. Therefore σ2 would contain the square of an n-cycle, but that cannot

be the identity. Therefore, it still permutes some elements, and thus σ2 itself cannot be

the identity.

3. Let G = {z ∈ Z | zn = 1 for some n ∈ Z+}.

(a) Prove that G is a group under multiplication (called the groups of roots of unity in

C).

Proof. For (a), let x, y ∈ G. Then by definition, xn = 1 and ym = 1 for some

n,m ∈ Z+. Then

(xy)nm = xnmynm = (xn)m(ym)n = 1m1n = 1,

so G is closed under multiplication.

Since G ⊆ C, and C is a field, associativity holds in G.

Since 11 = 1 and 1x = x = x1 for all x ∈ G, G has an identity element.

Let x ∈ G. Then xn = 1 for some n ∈ Z+. Note that this implies that x 6= 0, so

x ∈ C is a unit and there exists x−1 ∈ C. Now to see that x−1 ∈ G, note that

(x−1)n = (xn)−1 = 1−1 = 1.

Thus by definition, G is a group under multiplication.

(b) Prove that G is not a group under addition.

Proof. note that 1 ∈ G, but 1 + 1 = 2 /∈ G. Thus G is not closed under addition,

so it cannot be a group under the operation.

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Math 4281 Final Exam Practice Summer 2020

4. Construct a field with 343 elements. Justify your work.

Proof. Note that the polynomial f(x) = x3 + x+ [1] ∈ Z7[x] has no roots in Z7, so it is

irreducible. Thus by Theorem 15.11, F = Z7[x]/〈f(x)〉 if a field, and the elements in F

can be uniquely written as ax2 + bx+ c with a, b, c ∈ Z7, so F has 73 = 343 elements.

5. Suppose G is a finite group and x, g ∈ G.

(a) Prove that |x| = |g−1xg|.

Proof. For (a), since G is finite, |x| = n < ∞ and |g−1xg| = m < ∞ for some

n,m ∈ Z+. let e denote the identity element of G.

Using xn = e, and multiplying both sides by g−1 on the left and g on the right to

get g−1xng = g−1eg = e. But then

e = g−1xng = g−1xexex · · ·xexg = g−1xgg−1xgg−1x · · ·xgg−1xg = (g−1xg)n.

Thus |g−1xg| = m ≤ n.

Now e = (g−1xg)m = g−1xgg−1xg · · · g−1xg = g−1xmg. Multiplying both sides on

the left by g and the right by g−1, we obtain xm = geg−1 = e, so n ≤ m.

Hence we have shown that n = m, or |x| = |g−1xg|.

(b) Use part (a) to deduce that for all a, b ∈ G, |ab| = |ba|.

Proof. let x = ab and g = a. Then

|ab| = |x| = |g−1xg| = |a−1aba| = |ba|.

6. Find all of the elements in the subgroup K = 〈(12)(34), (125)〉 ≤ S5.

Proof. We compute the powers of the given elements and all products to obtain

K = {I, (12)(34), (125), (152), (25)(34), (15)(34)}

7. (a) Let G = 〈g〉 be a cyclic group of order 32. Compute the order of g24 in G, justifying

your answer.

Proof. By Theorem 23.6.ii, |g24| = 32

gcd(24,32)

= 32

8

= 4.

(b) Let G′ be a group, and let a ∈ G′ be an element of order 21. Let H = 〈a〉 and

K = 〈a7〉. Find all of the distinct left cosets of K in H and provide their elements.

Proof. We compute the sets gK for g ∈ H = 〈a〉, so all g = an for n ∈ Z. The six

distinct left cosets of K in H are:

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Math 4281 Final Exam Practice Summer 2020

K = 〈a7〉 = {1, a7, a14},

aK = a〈a7〉 = {a, a8, a15},

a2K = a2〈a7〉 = {a2, a9, a16},

a3K = a3〈a7〉 = {a3, a10, a17},

a4K = a4〈a7〉 = {a4, a11, a18},

a5K = a5〈a7〉 = {a5, a12, a19},

and a6K = a6〈a7〉 = {a6, a13, a20}.

(c) Compute the distinct left cosets of 〈R2〉 in D6 and provide their elements.

Proof. The group D6 = {I, R,R2, R3, R4, R5, r, rR, rR2, rR3, rR4, rR5} is the group

of symmetries of a regular hexagon. The subgroup 〈R2〉 = {I, R2, R4}, so its left

cosets are:

〈R2〉 = {I, R2, R4},

R〈R2〉 = {R,R3, R5},

r〈R2〉 = {r, rR2, rR4},

and rR〈R2〉 = {rR, rR3, rR5}.

8. Find all odd prime integers p so that x+ [2] divides f(x) = x4 + x+ [1] ∈ Zp[x].

Proof. x+ [2] divides f(x) is −[2] is a root of f(x). Note that

f(−[2]) = (−[2])4 − [2] + [1] = [15]

which will be zero in Zp if and only if p = 3 or 5.

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