4720 quiz abstract algebra

QUIZ/HOMEWORK 10

MATHEMATICS 4720—–INTRODUCTION TO ABSTRACT ALGEBRA I

INSTRUCTOR: DR. ZHENBO QIN

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1. (i) Up to isomorphisms, classify all abelian groups of order 1800.(ii) Point out which one is cyclic.

Answer.(i) Note that 1800 = 23× 32× 52. So up to isomorphisms, all the abelian groups

of order 1800 are:Z2 × Z2 × Z2 × Z3 × Z3 × Z5 × Z5,

Z2 × Z2 × Z2 × Z3 × Z3 × Z52 ,

Z2 × Z2 × Z2 × Z32 × Z5 × Z5,

Z2 × Z2 × Z2 × Z32 × Z52 ,

Z2 × Z22 × Z3 × Z3 × Z5 × Z5,

Z2 × Z22 × Z3 × Z3 × Z52 ,

Z2 × Z22 × Z32 × Z5 × Z5,

Z2 × Z22 × Z32 × Z52 ,

Z23 × Z3 × Z3 × Z5 × Z5,

Z23 × Z3 × Z3 × Z52 ,

Z23 × Z32 × Z5 × Z5,

Z23 × Z32 × Z52 .

(ii) Z23 × Z32 × Z52 is cyclic.

2. Let n ≥ 3. Prove that the only element σ of Sn satisfying σγ = γσ for allγ ∈ Sn is σ = ι, the identity permutation.Answer.

Let σγ = γσ for all γ ∈ Sn. Assume that σ 6= ι. We want to draw a contradiction.Since σ 6= ι, σ(i) = j for some i 6= j in S = {1, 2, 3, . . . , n}. Without loss

of generality, we may assume that i = 1 and j = 2. So σ(1) = 2. Applyingσ(12) = (12)σ to 1 ∈ S, we obtain σ(2) = 1. Applying σ(23) = (23)σ to 1 ∈ S, wesee that 2 = 3 which is a contradiction.

3. Let n ≥ 2 and H ≤ Sn. Prove that either all permutations in H are even orexactly half of them are even.Answer.

Assume that not all permutations in H are even. Then there exists an oddpermutation σ ∈ H. Let A (respectively, B) be the subsets of H consisting of even(respectively, odd) permutations in H. Then,

H = A∐

B.

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Now, since σ is odd, all the permutations in σA are odd. So σA ⊂ B. Hence|σA| ≤ |B|. Since |A| = |σA|, we have |A| ≤ |B|. Similarly, |B| ≤ |A|. It followsthat |A| = |B|.

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QUIZ/HOMEWORK 9



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1. Prove that a direct product of abelian groups is abelian.Answer.

Let G = G1×G2 where both G1 and G2 are abelian. Then for any two elements(a1, a2), (b1, b2) ∈ G, we have (a1, a2)·(b1, b2) = (a1b1, a2b2) = (b1a1, b2a2) = (b1, b2)·(a1, a2). So G is abelian.

2. Prove that a group of order 65 must be isomorphic to Z65.Answer.

Let G be a group of order 65. Then, |G| = pq where p = 13 and q = 5 areprimes. Since q 6 |(p− 1), G must be cyclic. Hence, G ∼= Z65.

3. Show that a finite abelian group is not cyclic if and only if it contains a subgroupisomorphic to Zp × Zp for some prime p.Answer.

(i) Let G be a finite abelian group which is not cyclic. Then,

G ∼= Zpr11× . . .× Zprn

n(2.1)

where p1, . . . , pn are primes. Note that the direct product Zpr11× . . .×Zprn

nis cyclic

if and only if pr11 , . . . , prn

n are pair-wise coprime. Since G is not cyclic, pi = pj forsome i 6= j. For simplicity, let i = 1 and j = 2. Now take the subgroup Zp1 of Zp

r11

,

and the subgroup Zp2 of Zpr21

. Then Zp1 × Zp2 × {1} × . . .× {1} is a subgroup of

Zpr11× . . . × Zprn

n. Via the isomorphism (2.1), G has a subgroup H isomorphic to

the subgroup Zp1 ×Zp2 ×{1}× . . .×{1}, which in turn is isomorphic to the groupZp1 × Zp2 . Finally, putting p = p1 = p2, we see that H is isomorphic to Zp × Zp.

(ii) Conversely, suppose that G contains a subgroup isomorphic to Zp × Zp forsome prime p. Note that Zp × Zp is not cyclic. So G is not cyclic since everysubgroup of a cyclic group is cyclic.

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QUIZ/HOMEWORK 8



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1. Let G be an abelian group (possibly infinite). Let

T = {a ∈ G| am = e for some m ≥ 1 depending on a}.(i) Prove that T ≤ G;(ii) Prove that G/T has no element–other than its identity element-of finite

order.Answer.

(i) Since e1 = e, we have e ∈ T . So T is nonempty. Next, let a, b ∈ T . Then,am = e and bn = e for some integers m, n ≥ 1. Since G is abelian,

(ab−1)mn = amn(b−1)mn = (am)n(bn)−m = ene−m = e.

So ab−1 ∈ T . Hence T is a subgroup of G.(ii) If the element Ta ∈ G/T is of finite order, then (Ta)n = T for some n ≥ 1.

So T (an) = T , and an ∈ T . By the definition of T , there exists m ≥ 1 such that(an)m = e. So anm = e. By the definition of T again, we conclude that a ∈ T .Hence Ta = T , i.e., the element Ta ∈ G/T is the identity element T in G/T .

2. Let H ≤ G, and let N be a normal subgroup of G. If |H| = 2006 and |N | = 5,compute |H ∩N | and |HN |.Answer.

By the Second Isomorphism Theorem, H ∩ N and HN are subgroups of G;moreover,

H/(H ∩N) ∼= (HN)/N. (4.1)

Since H ∩ N are subgroups of both H and N , |H ∩ N | divides both |H| and|N |. So |H ∩ N | divides gcd(|H|, |N |). Since |H| = 2006 and |N | = 5, we havegcd(|H|, |N |) = 1. Hence |H ∩N | = 1.

By (4.1), |H/(H ∩N)| = |(HN)/N |. So we obtain

|H|/|H ∩N | = |HN |/|N |.It follows that |HN | = |N | · |H|/|H ∩N | = 5 · 2006/1 = 10130.

3. Let G = (R∗, ·) and N = {1,−1}. Prove that

G/N ∼= (R+, ·)where R+ denote the set of all positive real numbers.Answer.

Define φ : G → R+ by φ(r) = |r|. Then φ is a group homomorphism since

φ(r1r2) = |r1r2| = |r1| · |r2| = φ(r1) · φ(r2).

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It is clear that φ is surjective. By the First Isomorphism Theorem,

G/ker(φ) ∼= R+.

Now, ker(φ) = {r ∈ G|φ(r) = 1} = {r ∈ G| |r| = 1} = {1,−1} = N . Hence,

G/N ∼= R+.

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QUIZ/HOMEWORK 7



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1. Show that any group homomorphism φ : G → G′ where |G| is a prime must beeither the trivial homomorphism or a one-to-one map.Answer.

Let H = ker(φ). By Lagrange’s Theorem, |H| | |G|. Since |G| is a prime, either|H| = 1 or |H| = |G|. When |H| = 1, H = {e} where e denotes the identity ofG; by Corollary 3.1.18, φ is one-to-one. When |H| = |G|, we have H = G, i.e.,ker(φ) = G; so φ maps every element of G to the identity of G′, i.e., φ is the trivialhomomorphism.

2. Let G be an abelian group. Let H and K be finite cyclic subgroups with |H| = rand |K| = s. Prove that if r and s are relatively prime, then G contains a cyclicsubgroup of order rs.Answer.

Assume that H = 〈a〉 and K = 〈b〉. We claim that the cyclic subgroup 〈ab〉 isof order rs, i.e., ord(ab) = rs.

Indeed, let m = ord(ab). Since H = 〈a〉 and |H| = r, ord(a) = r. In particular,ar = e. Similarly, ord(b) = s and bs = e. Noticing that G is abelian, we obtain

(ab)rs = arsbrs = (ar)s(bs)r = eser = e.

So m ≤ rs since m = ord(ab).On the other hand, we have

e = er = ((ab)m)r = (ab)rm = armbrm = (ar)mbrm = embrm = brm.

So rm is a multiple of s since ord(b) = s. Since r and s are coprime, m is a multipleof r. Similarly, we can prove that m is a multiple of s. Since r and s are coprime,the least common multiple of r and s is rs. In particular, m ≥ rs.

Therefore, m = rs, i.e., ord(ab) = rs.

3. Let G be a group of order pq where p and q are primes. Prove that every propersubgroup of G is cyclic.Answer.

Let H be a proper subgroup of G. Since |H| divides |G|, |H| is either 1 or aprime (i.e., p or q). If |H| = 1, then H = {e} = 〈e〉 is cyclic. Next, assume that|H| is a prime. Let a ∈ H and a 6= e. Then, the order of the cyclic subgroup 〈a〉is either 1 or equal to |H|. In the later case, H = 〈a〉 is cyclic. In the former case,〈a〉 = {e} contradicting to a 6= e.

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4. Let p and q be prime numbers. Find the number of generators of the cyclicgroup Zpq. (Hint: consider separately p = q and p 6= q)Answer.

Recall that s ∈ Zpq is a generator if and only if gcd(s, pq) = 1.So when p = q, s 6= 0, p, 2p, . . . , (p − 1)p. Hence the number of generators is

p2 − p = p(p− 1).Next, assume that p < q. Then s 6= 0, p, 2p, . . . , (q − 1)p, q, 2q, . . . , (p − 1)q.

Hence the number of generators is pq − q − (p− 1) = (p− 1)(q − 1).

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QUIZ/HOMEWORK 6



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1. Let H be a subgroup of index 2 in a group G. Prove that xH = Hx for everyx ∈ G.Answer.

If x ∈ H, then xH = H = Hx. In the following, we assume x 6∈ H.Since x 6∈ H, xH 6= H. So H and xH are two different left cosets. Since the

index of H in G is 2, we must have

G = H∐

xH.

Similarly,

G = H∐

Hx.

It follows that xH = Hx.

2. Let G be a group, and a ∈ G be the unique element of order 2. Show thatax = xa for all x ∈ G.Answer.

Note that (xax−1)2 = xax−1 · xax−1 = xa2x−1 = xx−1 = e, and that xax−1 6= esince a 6= e. So xax−1 is also an element of order 2. By the uniqueness, xax−1 = a.Hence xa = ax for every x ∈ G.

3. Let G be an abelian group. Define the map f : G → G by

f(g) = g−1

for g ∈ G. Prove that f is an automorphism of G (i.e., you need to show that f isboth a homomorphism and a bijection).Answer.

Let g, h ∈ G. Then, f(gh) = (gh)−1 = h−1g−1. Since G is abelian, f(g)f(h) =g−1h−1 = h−1g−1. Hence, f(gh) = f(g)f(h). So f is a homomorphism. Next, iff(g) = f(h), then g−1 = h−1 and so g = h. This shows that f is injective. Finally,let g ∈ G, then f(g−1) = (g−1)−1 = g. So f is surjective. Therefore, f is bijectiveand hence an automorphism of G.

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QUIZ/HOMEWORK 5



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1. If G is a finite group of even order, show that there exists an element a 6= esuch that a2 = e.Answer.

Assume that a2 6= e whenever a 6= e. We want to draw a contradiction. Notethat a2 6= e if and only if a 6= a−1. So the elements in G can be listed as:

e, a1, a−11 , a2, a

−12 , . . . , ak, a

−1k

for some nonnegative integer k. In particular, |G| = 1+2k is odd. This contradictsto the assumption that |G| is even.

2. Let G be a group and a ∈ G. Assume that the order m = ord(a) is finite. Provethat as = at if and only if s ≡ t (mod m).Answer.

(i) If s ≡ t (mod m), then s− t = rm for some r ∈ Z. So s = t + rm. It followsthat as = at+rm = at(am)r = ater = at.

(ii) Now suppose as = at. Then, as−t = e. By the Division Algorithm, (s− t) =mq + r where q ∈ Z and 0 ≤ r < m. So e = as−t = amq+r = (am)qar = eqar = ar,i.e., ar = e. Since 0 ≤ r < m and m is the smallest positive integer such thatam = e, we must have r = 0. Hence (s − t) = mq, i.e., (s − t) is a multiple of m.Therefore, s ≡ t (mod m).

3. Let a, b be elements of a group G. Show that if the order of ab is n, then theorder of ba is also n (i.e., ab and ba have the same order).Answer.

Assume that the order of ba is m. We want to show that m = n.Since the order of ab is n, we have (ab)n = e, i.e., (ab) (ab) . . . (ab)︸︷︷︸

(n−1) copies

= e. Multi-

plying a−1 from left, we obtain b (ab) . . . (ab)︸︷︷︸(n−1) copies

= a−1. Now multiplying a from right,

we see that b (ab) . . . (ab)︸︷︷︸(n−1) copies

a = e, i.e., (ba)n = e. So we must have m ≤ n.

By symmetry, we also have n ≤ m. Therefore, m = n.

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QUIZ/HOMEWORK 4



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1. Let G be an abelian group with identity e. Fix a positive integer n. Prove thatall the elements x of G satisfying the equation xn = e form a subgroup H of G.Answer.

Since en = e, e ∈ H. So H is not empty. Next, let x, y ∈ H. Then xn = e andyn = e. Since G is abelian, we see that (xy−1)n = xn(y−1)n = xn(yn)−1 = ee−1 = e.So xy−1 ∈ H. Therefore, H is a subgroup of G.

2. Let H ≤ G. Define, for a, b ∈ G, a ∼ b if a−1b ∈ H. Prove that this defines anequivalence relation on G, and show that [a] = aH = {ah|h ∈ H}. The sets aHare called left cosets of H in G.Answer.

(i) Since a−1a = e ∈ H, a ∼ a. So ∼ is reflexive. If a ∼ b, then a−1b ∈ H. SinceH is a group, b−1a = (a−1b)−1 ∈ H. Thus, ∼ is symmetric. Next,assume that a ∼ band b ∼ c. Then a−1b, b−1c ∈ H. Since H is closed, a−1c = (a−1b)(b−1c) ∈ H. Soa ∼ c. Thus ∼ is transitive. It follows that ∼ is an equivalence relation on G.

(ii) Note that b ∈ [a] if and only if a ∼ b if and only if a−1b ∈ H if and only ifb ∈ aH. Hence, [a] = aH.

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QUIZ/HOMEWORK 3



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1. Let G be a group. Prove that if (ab)2 = a2b2 for all a, b ∈ G, then G is abelian.

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QUIZ/HOMEWORK 2



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1. Let G be the set of integers with the operation ∗ defined by:

a ∗ b = a− b.

Is G with the operation ∗ a group? If YES, write down your proof. If NOT, statea group axiom which fails for G.

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QUIZ/HOMEWORK 1



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1. Let a = −120 and b = 72. Express gcd(a, b) as ma + nb for some integers mand n. Show details.

ANSWER:

4720 quiz abstract algebra

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