辅导案例-I1
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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