1. EXERCISE SET 1

Prob 1: Prove for all n > 1 that Z/nZ is not group under multiplication ofresidue classes. Let (Z/nZ)∗ denote the residue classes of integers coprime to n.This is a group under multiplication.

Sol: For Z/nZ to be group under multiplication, multiplicative inverse shouldexist for all elements belonging to Z/nZ.

a.b = 1mod(n) (0 ≤ a, b ≤ n− 1)

ab− 1 = kn (k ∈ Z+)

ab+ k′n = 1 (k′ = −k)

That is multiplicative inverse exist for only those elements which are relativelyprime to n. Obvious counter example is 0 ∈ Z/nZ, but no multiplicative inverse.

Prob 2: Let G = {x ∈ R | 0 ≤ x < 1} and for x, y ∈ G, let x ∗ y = (x+ y − [x+y]) = {x + y}. Prove that ∗ is a binary operation on G and that G is an abeliangroup under ∗.

Sol: For ∗ to be a binary operation, x ∗ y should be in G for x, y ∈ G, which istrue as 0 ≤ {x+ y} < 1.

(1) ∗ is associative: Let x, y, z ∈ G, then

(x ∗ y) ∗ z = x+ y + z − [x+ y + z]

x ∗ (y ∗ z) = x+ y + z − [x+ y + z],

hence associative.(2) x ∗ e = e ∗ x = x, is satisfied for all x ∈ G, if we take e = 0.(3) x ∗ x−1 = x−1 ∗ x = e, is satisfied by x−1 = 1− x ∈ G, for x = 0 x−1 = 0(4) {x+ y} = {y + x} hence group is abelian.

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

• Prove that G is group under multiplication.Sol:

(1) Let z1, z2, z3 ∈ G(z1 ∗ z2) ∗ z3 = z1 ∗ z2 ∗ z3z1 ∗ (z2 ∗ z3) = z1 ∗ z2 ∗ z3

hence multiplication is associative.(2) z ∗ e = e ∗ z = z, this is satisfied for all z ∈ G when e = 1 ∈ G(3) z∗z−1 = z−1∗z = e this is satisfied by z−1 = zn−1 ∈ G as (zn−1)n = 1

Hence G is a group under multiplication.

• Prove that G is not group under addition.Sol: z + e = e+ z = z is only satisfied by e = 0, but 0 does not belong

to G, hence G is not a group under addition.

Prob 4: Find order of each element of the additive group Z/12Z.Sol: |0| = 1 |1| = 12 |2| = 6 |3| = 4 |4| = 3 |5| = 12

|6| = 2 |7| = 12 |8| = 3 |9| = 4 |10| = 6 |11| = 121

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Prob 5: Let x be an element of G. Prove that, if |x| = n for some positiveinteger n then x−1 = xn−1.

Sol: |x| = n implies xn = 1 as x ∈ G then x−1 ∈ G such that x∗x−1 = x−1 ∗x =1. As xn = 1, multiplying both sides by x−1 we will get x−1 = xn−1.

Prob 6: Let G be a finite group and let x be an element of G of order n. Provethat if n is odd, then x = (x2)k

′for some integer k′ ≥ 1.

Sol: As n is odd, n = 2k + 1 where k ∈ Z+ ∪ {0}

xn = x2k+1 = 1

multiply both sides by x, we get (x2)k+1 = x, replacing k + 1 by k′ where k′ ∈ Z+

and hence x = (x2)k′

for some integer k′ ≥ 1.

Prob 7: Prove that if x2 = 1 for all x ∈ G then G is abelian.Sol: Let

xy 6= yx

xxyy 6= xyxy

x2y2 6= (xy)2

1 6= (xy)2

which is contradiction as, if x, y ∈ G then xy ∈ G and hence (xy)2 = 1 thereforexy = yx, and hence G is abelian.

Prob 8: Prove that A×B is abelian iff A and B are abelian.Sol: A×B = (x, y) for x ∈ A, y ∈ B let,

p, q ∈ A and r, s ∈ B(p, r) ∗ (q, s) = (pq, rs)(q, s) ∗ (p, r) = (qp, sr)

so (p, r) ∗ (q, s) = (q, s) ∗ (p, r) implies pq = qp and rs = sr for all p, q ∈ A andr, s ∈ B, that is A and B should be abelian.

Prob 9: Show that order of (a,b) is lcm of |a| and |b|.Sol: let, |a| = n and |b| = m, order of (a,b) should be divisible by both n

and m, but lcm(n,m) is smallest number to which n and m divides, hence order of(a,b)=lcm(|a|,|b|).

Prob 10: Prove that (a1a2 · · · an)−1 = a−1n a−1n−1 · · · a−11 for all a1, a2, · · · , an ∈ G

Sol: for a1, a2, · · · , an ∈ G, (a1a2 · · · an) ∈ G and hence it’s inverse also existin G i.e. (a1a2 · · · an)(a1a2 · · · an)−1 = 1 now multiplying successively by a−11 , a−12 ,till a−1n , we will get

(a1a2 · · · an)−1 = a−1n a−1n−1 · · · a−11 .

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2. EXERCISE SET 2

Prob 1: Let x and y be the elements of order 2 in any group G. Prove that ift = xy then tx = xt−1.

Sol: |x| = 2 and |y| = 2 implies x = x−1 and y = y−1

t = xy

t−1 = (xy)−1 = y−1x−1 = yx

tx = xyx

xt−1 = xyx

tx = xt−1

For Problem 2 and 3: Let X2n =⟨x, y|xn = y2 = 1, xy = yx2

⟩.

Prob 2: Show that if n = 3k, then X2n has order 6.Sol:

x = xy2 = (xy)y = (yx2)y = (yx)(xy) = (yx)(yx2) = y(xy)x2 = yyx2x2 = y2x4 = x4

i.e. x = x4 so we get x3 = 1, hence order of x2n is 6,

X2n = {1, x, x2, y, yx, yx2}.

Prob 3: Show that if gcd(3, n) = 1, then x satisfies the additional relation:x = 1. In this case deduce that X2n has order 2.

Sol:gcd(3, n) = 1

3a+ nb = 1

x1 = x3a+nb = (x3)a(xn)b = 1

so, X2n has order 2, X2n = {1, y}.

For Problem 4-8: Let Y =⟨u, v|u4 = v3 = 1, uv = v2u2

⟩Prob 4: Show that v2 = v−1.

Sol:v3 = 1

v−1vv2 = v−1

v2 = v−1

Prob 5: Show that v commutes with u3.Sol:

v2u3v = (v2u2)(uv) = (uv)(uv) = uv3u2 = u3

as, v2u3v = u3, multiplying both sides by v we get,u3v = vu3.

Prob 6: show that v commutes with u.Sol:

u4 = 1

u9 = u

(u3)3 = u

4

(v2u3v)3 = u

(v2u9v) = u

u9v = vu

uv = vu

Prob 7: Show that uv = 1.Sol:

uv = vu

u = vuv2

u = vv2u2v

u = u2v

u3u = u3u2v

uv = 1

Prob 8: Show that u = 1, deduce that v = 1, and conclude that Y = 1.Sol:

u4v3 = 1

u3uvv2 = 1

u3v2 = 1

u2uvv = 1

u2v = 1

u = 1

as, uv = 1 and u = 1 then v = 1, and hence Y=1.

Prob 9: Prove that if σ is them-cycle (a1a2 · · · am), then for all i ∈ {1, 2, · · · ,m},σi(ak) = ak+i, where k+i is replaced by it’s least positive residue mod(m). Deducethat |σ| = m.

Sol: Using process of induction, σ1(ak) = ak+1 is true. Let σm(ak) = ak+mthen for i = m + 1, σm+1(ak) = σ1σm(ak) = σ1(ak+m) = ak+m+1 i.e.equation istrue for m+ 1 whenever it is true for m, hence this equation is true for all i ∈ Z+ .Whenever k+ i ≥ m, is replaced by x where k+ i ≡ (x)mod(m) and 0 ≤ x ≤ m−1.As σi(ak) = ak+i, let ak = ak+i for all 0 ≤ k ≤ m − 1, then (k)mod(m) =(k + i)mod(m) which implies m|i, smallest i divisible by m is m hence |σ| = m.

Prob 10: Let σ be the m-cycle (1 2 3 · · · m). Show that σi is also a m-cycle iffi is relatively prime to m.

Sol 1: Suppose σi is m-cycle then

σi = (1, 1 + i, · · · , 1 + (m− 1)i)

and so {1, 1 + i, 1 + 2i, . . . , 1 + (m− 1)i} must be distinct mod(m). Let gcd(i,m) 6=1 then there exist prime p, such that i = pk1 and m = pk2 where (k1, k2, p ∈Z+), (0 < k1, k2 < m)

mk1 = ik2

1 +mk1 = 1 + ik2

1mod(m) = (1 + ik2)mod(m)

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which is contradiction, hence gcd(i,m) = 1 for σi be the m-cycle.Suppose gcd(i,m) = 1, then {1, 1 + i, . . . , 1 + (m+ 1)i} are distinct mod(m). If

not then

1 + li = 1 + ji (0 ≤ l < j ≤ m− 1), (0 < i ≤ m− 1)

(l − j)i = 0mod(m)

(l − j)i = km k ∈ Z+

Since gcd(i,m) = 1 there are integers k1 and k2 such that k1i+k2m = 1. Mutiplyingboth sides by l − j we get,

k1(l − j)i+ k2(l − j)m = l − jwhich implies m|l − j, but this is not possible since 0 < l − j < m. Therefore if iand m are relatively prime then σi is an m-cycle.

3. EXERCISE SET 3

Prob 1: Show that if n is not prime then Z/nZ is not a field.Sol: For Z/nZ to be a field each element should have multiplicative inverse.

ab = 1 (0 ≤ a, b ≤ n− 1)

ab+ kn = 1 (k ∈ Z)

which implies inverse of only those elements exist which are relatively prime to n,hence n has to be a prime number.

Prob 2: Let p be a prime. Prove that order of GL2(Fp) is p4 − p3 − p2 + p.Sol: |Fp| = p, GL2(Fp) is set of all 2× 2 matrices having entries from Fp whose

determinant is not zero. Let A =(A1 A2

)to be a column matrix for 2 × 2

matrices. For A ∈ GL2(Fp), there are p2 − 1 choices for A1 (A1 6= 0), determinantof 2× 2 is zero iff one column is a multiple of another i.e. A2 = kA2 where k ∈ Fp,det(A) = 0 iff A2 = {0A1, 1A1, · · · , (p−1)A1} so there are are only (p2−p) choicesfor A2, hence there are (p2 − 1)(p2 − p) choices for A ∈ GL2(Fp). Hence order ofGL2(Fp) is p4 − p3 − p2 + p.

Prob 3: Show that GLn(F) is non-abelian for any n ≥ 2 and any F.Sol: For GLn(F) to be abelian, for all A,B ∈ GLn(F) AB = BA, which is not

true for all n ≥ 2 and any F. Counter example is

(1 23 4

) (2 31 4

)=

(4 1110 25

),

where as

(2 31 4

) (1 23 4

)=

(11 1613 18

).

Prob 4: Compute order of each element in Q8.Sol: |1| = 1, | − 1| = 2, |i| = | − i| = |j| = | − j| = |k| = | − k| = 4.

Prob 5: Let ϕ : G→ H be a Homomorphism, prove that ϕ(xn) = ϕ(x)n for alln ∈ Z+ .

Sol: Using process of induction, for n = 1 ϕ(x1) = ϕ(x)1, let ϕ(xk) = ϕ(x)k forn = k+1, ϕ(xk+1) = ϕ(xkx1) = ϕ(xk)ϕx1 = ϕ(x)kϕ(x)1 = ϕ(x)k+1 which implies,equation is true for n = k + 1 whenever it is true for n = k hence equation is truefor all n ∈ Z+.

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Prob 6: Let G and H be groups and let ϕ : G→ H be a homomorphism. Provethat kernel of ϕ is subgroup of G.

Sol: Kernel of ϕ is nonempty as 1G ∈ ker(ϕ) as ϕ(1G1G) = ϕ(1G) ⇒ ϕ(1G) =1H . Let g1, g2 ∈ ker(ϕ), then g1g2 ∈ ker(ϕ) as ϕ(g1g2) = ϕ(g1)ϕ(g2) = 1H , hencekernel of ϕ is closed under multiplication. g−11 ∈ ker(ϕ) as,

ϕ(g−11 g1) = ϕ(1G)

ϕ(g−11 )ϕ(g1) = 1H

ϕ(g−11 ) = 1H

hence kernel of ϕ is closed under inverse, therefore it is a subgroup of G.

Prob 7: Define a map π : R2 → R by π((x, y)) = x. Prove that π is ahomomorphism and find kernel of π .

Sol: Let (x, y), (v, w) ∈ R2 π((x, y)(v, w)) = π((x + v, y + w)) = x + v andπ((x, y))π((v, w)) = x+ v, hence π is a homomorphism. Kernel of π is (0, y) wherey ∈ R.

Prob 8: Let G be any group. Prove that map from G to itself defined by g 7→ g2

is homomorphism iff G is abelian.Sol: Let π : G → G such that π(g) = g2 for all g ∈ G. Let a, b ∈ G, then

π(ab) = (ab)2 and π(a)π(b) = a2b2, let’s assume (ab)2 = a2b2 multiplying both sidesby a−1 and b−1 we get, a−1(ab)2b−1 = a−1a2b2b−1 ba = ab i.e. if π a homomorphismthen G is abelian. Now, take G to be abelian i.e. ab = ba multiplying both sides bya and b, we get a2b2 = (ab)2 which implies π(a)π(b) = π(ab) i.e. when G is abelianπ is homomorphism. Hence π is homomorphism iff G is abelian.

Prob 9: Prove that for each fixed nonzero k ∈ Q the map from Q to itselfdefined by q 7→ kq is an automorphism.

Sol: Define π : Q→ Q such that π(q) = kq, π is a homomorphism as

π(a+ b) = k(a+ b) = ka+ kb = π(a)π(b) (a, b ∈ Q)

π is injective as π(m) = π(n) implies m = n. π is surjective as for each q ∈ Q thereexist t ∈ Q such that k−1t = q where k−1 ∈ Q − {0}, hence π is bijective whichimplies it is an isomorphism Q onto Q hence by definition it is automorphism.

Prob 10: Let A be an abelian group and fix some k ∈ Z. Prove that themap a 7→ ak is a homomorphism. If k = −1 prove that this homomorphism is anisomorphism.

Sol : Define π : A → A such that π(a) = ak. Let a, b ∈ A π(ab) = (ab)k andπ(a)π(b) = akbk as A is abelian (ab)k = akbk hence π is homomorphism. π is abijection for k = −1 because each element has unique inverse and as A is a groupinverse exist for each element. Hence π is an isomorphism.

Prob 11: Show that the additive group R acts on the x, y plane R × R byr.(x, y) = (x+ ry, y).

Sol : Let a, b ∈ R, a.(b.(x, y)) = a.(x+by, y) = (x+by+ay, y) = (x+(a+b)y, y) =(a+ b).(x, y), and 1.(x, y) = (x+ 0y, y) = (x, y) hence by definition of group actionthe additive group R acts on the x, y plane R× R.

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Prob 12: Let G be a group acting on a set A and fix some a ∈ A. Show thatP = {g ∈ G| ga = a} is subgroup of G.

Sol : P is nonempty as 1 ∈ P .Let m,n ∈ P , (mn)a = m(n(a)) = m(a) = a,which implies mn ∈ P , i.e. P is closed under multiplication, m−1 ∈ P as

a = 1a = (m−1m)a = m−1(m(a)) = m−1(a)

which implies p is closed under inverses. Hence P is subgroup.

Prob 13: Let G be any group and let A = G. Show that the maps defined byg.a = gag−1 for allg, a ∈ G do satisfy the axioms of a group action.

Sol : Let m,n ∈ G m.(n.a) = m.(nan−1) = mnan−1m−1 = (mn)a(mn)−1 =(mn).a and 1.a = 1a1−1 = a hence both axioms are satisfied.

For Prob 14-15: Let G be a group and let G act on itself by left conjugation,so each g ∈ G maps G to G by x 7→ gxg−1 for fixed g ∈ G.

Prob 14: Prove that conjugation by g is an Automorphism.Sol : Define π : G → G such that π(x) = gxg−1. Let x, y ∈ G, π(xy) =

gxyg−1 = (gxg−1)(gyg−1) = π(x)π(y), hence homomorphism. π is injective as,π(x) = π(y) implies x = y. π is surjective as, for every x ∈ G there exist y suchthat g−1xg = y. Hence π is a bijection and conjugation by g is an automorphism.

Prob 15: Show that x and gxg−1 have the same order.Sol : Let |x| = n and (gxg−1)m = 1, which implies xm = 1 so, n|m smallest m

to which n divides except zero is n, hence |gxg−1| = n.

Prob 16: Let H be a group acting on set A. Prove that the relation ∼ on Adefined by a ∼ b iff a = hb for some h ∈ H is an equivalence relation.

Sol : a ∼ a as a = 1a where 1 ∈ H. Let a ∼ b then b ∼ a, as a ∼ b implies a = hbmultiplying both sides by h−1 we get, h−1a = b as h−1 ∈ H hence symmetric. Leta ∼ b and b ∼ c then a ∼ c, as a ∼ b implies a = h1b and b ∼ c implies b = h2cnow substituting we get, a = (h1h2)c as as (h1h2) ∈ H hence transitive. Thereforerelation ∼ is an equivalence relation.

For Prob 17-19: Let A be a nonempty set and let k be a positive integer withk ≤ |A|. The symmetric group SA acts on set B consisting of all subset of A ofcardinality k by σ.{a1, . . . , ak} = {σ(a1), . . . , σ(ak)}.

Prob 17: Prove that this is a group action.Sol : Let τ, σ ∈ SA then τ.(σ.{a1, . . . , ak}) = τ.{σ(a1), . . . , σ(ak)} = {τσ(a1), . . . , τσ(ak)} =

(τσ).{(a1), . . . , (ak)}, and 1.{a1, . . . , ak} = {1a1, . . . , 1ak} = {a1, . . . , ak}. Henceby definition this is a group action.

Prob 18: For which values k the action of SA on k-element subsets is faithfulwhere |A| = n.

Sol : Let G be a group. G acting on a set B is equivalent to giving a grouphomomorphism φ : G→ Aut(B), as follows. Consider the two sets

S1 = {f : G×B → B|f is a group action, that is, f satisfiesf(g, f(h, b)) = f(gh, b) and f(e, b) = b}

8

and

S2 = {φ : G→ Aut(B)|φ is a group homomorphism}.Define Φ : S1 → S2 by (Φ(f)(g))(b) = f(g, b). It is obvious that Φ(f)(g) ∈Aut(B) since f is an action. We need to check that Φ(f) is a group homomorphismfrom G → Aut(B). Let g1, g2 ∈ G then we need to check that Φ(f)(g1g2) =Φ(f)(g1) ◦ Φ(f)(g2). Both these are elements in Aut(B) and so it suffices to checkthey are equal when we apply them on b ∈ B. (Φ(f)(g1g2))(b) = f(g1g2, b) and(Φ(f)(g1) ◦ Φ(f)(g2))(b) = (Φ(f)(g1))(f(g2, b)) = f(g1, f(g2, b)) since f is a groupaction f(g1g2, b) = f(g1, f(g2, b)) which implies Φ(f) is a group homomorphism iff is a group action.

Define Ψ : S2 → S1 given as (Ψ(φ))(g, b) = φ(g)(b). We need to check that Ψ(φ)is a group action from G×B → B. Let g1, g2 ∈ G and denote by 1 the identity ele-ment of G. Then we need to check that (Ψ(φ))(g1g2, b) = (Ψ(φ))(g1, (Ψ(φ))(g2, b))and (Ψ(φ))(1, b) = b. (Ψ(φ))(g1g2, b) = φ(g1g2)(b) and (Ψ(φ))(g1, (Ψ(φ))(g2, b)) =(Ψ(φ))(g1, φ(g2)(b)) = φ(g1) ◦ φ(g2)(b), φ(g1g2) = φ(g1) ◦ φ(g2) as φ is a group ho-momorphism. (Ψ(φ))(1, b) = φ(1)(b) but as φ is a group homomorphism φ(1) goesto identity in Aut(B) but IdAut(B)(b) = b hence (Ψ(φ))(1, b) = b. As, whenever φis a group homomorphism Ψ(φ) satisfies both the properties of group action, henceΨ(φ) is a group action if φ is a group homomorphism.

Now we need to check that Φ◦Ψ = IdS2 , that is, we want to check (Φ◦Ψ)φ = φfor φ ∈ S2. As φ ∈ S2, equality can be checked by applying both sides on anelement g ∈ G. (Φ ◦Ψ)(φ)(g) = φ(g) now, both these are elements of Aut(B) so itis suffice to check that they are equal when applied on b ∈ B. (Φ ◦ Ψ)(φ)(g)(b) =(Φ((Ψ(φ))(g)))(b) = (Ψ(φ))(g, b) = φ(g)(b) hence, Φ ◦ Ψ = IdS2

. Also, we needto check that Ψ ◦ Φ = IdS1

i.e. we want to check (Ψ ◦ Φ)f = f both of these areelements of S1 as f ∈ S1, so it is suffice to check that they are equal when appliedon (g, b) ∈ G × B. (Ψ ◦ Φ)(f(g, b)) = (Ψ(Φ(f)))(g, b) = (Φ(f)(g))(b) = f(g, b)hence, Ψ ◦ Φ = IdS1

.Group action f is said to be faithful when Φ(f) is injective i.e. ((Φ(f))(g1))(B) =

((Φ(f))(g2))(B) implies g1 = g2, for all f ∈ S1 where g1, g2 ∈ G. For the givenproblem G = SA and elements of G are distinct permutations of A. Let B bea k-element subset of A. Let σx, σy ∈ SA be two distinct permutations of A.Since these permutations are distinct there exist a ∈ A such that σx(a) 6= σy(a).We now can choose a k-element subset B of A where k is less than n, such thata ∈ B but σ−1y σx(a) /∈ B. Since σx(B) contains σx(a) but σy(B) does not contain

σyσ−1y σx(a) = σx(a) hence, as σx 6= σy implies σx(B) 6= σy(B) hence action is

faithful. For k = n there will be only one element in B and hence only one elementin Aut(B) i.e. IdAut(B) and hence kernel of group homomorphism Φ(f) will bewhole of SA, hence action will not be faithful. Therefore action is faithful for0 ≤ k ≤ n− 1.

Prob 19: For which values of k the action of SA on ordered k-tuple is faithfulwhere |A| = n.

Sol : Group action f is said to be faithful when Φ(f) is injective i.e. ((Φ(f))(g1))(B) =((Φ(f))(g2))(B) implies g1 = g2 for all f ∈ S1 where g1, g2 ∈ G. For the givenproblem G = SA and elements of G are distinct permutations of A, and B is anordered k-tuple of A. Let σx, σy ∈ SA be two distinct permutations of A. Sincethese permutations are distinct there exist a ∈ A such that σx(a) 6= σy(a). We

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now can choose an ordered k-tuple B where k is less than n, such that a ∈ Bbut σ−1y σx(a) /∈ B. Since σx(B) contains σx(a) but σy(B) does not contain

σyσ−1y σx(a) = σx(a) hence, as σx 6= σy implies σx(B) 6= σy(B) hence action is

faithful. For k = n as σx(a) 6= σy(a) hence σx(B) 6= σy(B) since a ∈ B, becauseunlike k-element subset, ordered k-tuple has n elements in B. Hence action isfaithful for 0 ≤ k ≤ n.

Prob 20: Let A and B be groups. Prove that A×B ∼= B ×A.Sol : Define π : A×B → B ×A such that π(x, y) = (y, x) for x ∈ A,y ∈ B. Let

a, b ∈ A, c, d ∈ B then π((a, b)(c, d)) = π((ac, bd)) = (bd, ac) and π((a, b)π((c, d)) =(b, a)(d, c) = (bd, ac), hence π is homomorphism. π is injective as π(x, y) = π(a, b)implies x = a and y = b. π is surjective as for every element (y, x) ∈ B × A thereexist element (x, y) ∈ A×B. Hence π is bijection which implies it is isomorphism.

Prob 21: Show that {1, s} can not kernel of group homomorphism from D6 toany other group.

Sol : Let φ is a group homomorphism from G to H, and let k = ker(φ),then for x ∈ G, Kx = xK where Kx = {kx|k ∈ K} and xK = {xk|k ∈ K}.D6 = {1, r, r2, r3, s, sr, sr2}, {1, s} can not be kernel because of following counterexample. r.{1, s} = {r, rs} whereas {1, s}.r = {r, sr} as rs 6= sr in D6 becausers = sr−1 and r−1 = r2 in D6. Hence {1,s} can not be kernel.

4. EXERCISE SET 4

Prob 1: Find all subgroups of Z45, giving generator for each.Sol: If m|45 then Z45 has subgroup of order n/m given by < m >. The positive

divisors of 45 are 1, 3, 5, 9, 15, 45 then the corresponding subgroups are < 1 >,<3 >,< 5 >,< 9 >,< 15 >,< 45 >.

Prob 2: If x is an element of the finite group G and |x| = |G|, prove thatG =< x >.

Sol: Let |x| = n then for k < n where k ∈ Z+∪{0} all xk are different. If x ∈ Gthen all the powers of x should be elements of G as G is a group, and as |G| = |x|then all the elements of G are the those generated by x then by the definition ofgenerator of a group, G =< x >.

Prob 3: Find all generators of Z/48Z.Sol: m generates Z/48Z if gcd(m, 48) = 1. Therefore generators of Z/48Z are

{1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}.

Prob 4: Find all generators of Z/202Z.Sol: m generates Z/202Z if gcd(m, 202) = 1. Therefore generators of Z/48Z are

all the odd numbers till 202 except 101.

Prob 5: Find number of generators for Z/49000Z .Sol: Prime factorization of 49000 is 23 × 53 × 72. Then number of generators of

49000 is equal to 49000−(no. of divisors of 2, 5, 7 till 49000−no. of divisors of 10, 14, 35 till 49000+

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no. of divisors of 70 till 49000) that is equal to (49000−24500−9800−7000−4900−3500− 1400 + 700) = 16800.

Prob 6: In Z/48Z write out all elements of < a > for every a.Sol:

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47 = {Z/48Z}

2, 10, 14, 22, 26, 34, 38, 46 = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48}

3, 9, 15, 21, 33, 39, 45 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48}

4, 20, 28, 44 = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}

6, 18, 30, 42 = {6, 12, 18, 24, 30, 36, 42, 48}

8, 40 = {8, 16, 24, 32, 40, 48}

12, 36 = {12, 24, 36, 48}

16, 27 = {16, 32, 48}

24 = {24, 48}

48 = {48}.

Prob 7:List all subgroups of Z48.Sol: < 1 >,< 2 >,< 4 >,< 6 >,< 8 >,< 12 >,< 16 >,< 24 >,< 48 > are the

subgroups of Z48.

Prob 8: Let Z48 =< x >. For which integers a does map to ϕa defined byϕa : 1→ xa extend to an isomorphism from Z/48Z onto Z48.

Sol: All integers a such that (a, 48) = 1, that is, a = 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47

Prob 9: Let Z36 =< x >. For which integers a does the map φa defined byφa : 1→ xa extend to a well defined homomorphism from Z/48Z into Z36. Can φaever be a surjective homomorphism.

Sol: As order of 1 is 48 in Z/48Z, hence 1 must map to an element of orderdividing 48 i.e. a can be equal to {1, 2, 3, 4, 6, 8, 12} but since a can not be equal to36 hence this mapping can not be surjective homomorphism.

Prob 10: What is order of 30 in Z/54Z? Write out all of the elements and theirorders in < 30 >.

Sol : |30| = 5, < 30 >= {0, 30, 6, 36, 12, 42, 18, 48, 24} and their orders are{1,9,9,3,9,9,3,9,9} respectively.

Prob 11: Find all cyclic subgroups of D8.Sol : Following are the cyclic subgroups of D8: < 1 >,< s >,< sr >,< sr2 >

,< sr3 >,< r >,< r2 >.

Prob 12: Prove that following groups are not cyclic.

11

• Z2 × Z2.Sol : As Z2 = {0, 1}, hence Z2 × Z2 = ({0, 0}, {0, 1}, {1, 0}, {1, 1}) i.e|Z2×Z2| = 4, therefore if this group is cyclic there should exist an elementof order 4, but since there is no element of order 4 in given group, hencenot cyclic.

• Z2 × Z.Sol : Z2 × Z is either of the form {0, x} or of the form {1,x} where x ∈ Z.{0, x} can not be the generator as it can not generate {1, x}, hence it mustbe of the form {1,x} if x 6= ±1 then it can not generate {1,1}, and if x = ±1then they can not generate {0, 1} hence not cyclic.

• Z× Z.Sol : If {a, b} is generator then it’s obvious that −1 ≤ a, b ≤ 1. None ofthem except {1, 1} and {−1,−1} can generate {1, 1} but then {1, 2} cannot be generated by them, hence not cyclic.

Prob 13: Prove that the following pairs of groups are not isomorphic.

• Z× Z2,Z.Sol : There is a element {0, 1} in Z×Z2 of order 2, but there is not a singleelement of order 2 in Z.

• Q× Z2,Q.Sol : There is a element {0, 1} in Q × Z2 of order 2, but there is not asingle element of order 2 in Q.

Prob 14: Prove that Q×Q is not cyclic.Sol : Let {a, b} be the generator, then obviously a, b ∈ Q i.e a/2, b/2 ∈ Q but

{a, b} can not generate {a/2, b/2} hence not cyclic.

Prob 15: Assume |x| = n and |y| = m. Suppose that x and y commute:xy = yx. Prove that |xy| divides the least common multiple of m and n. Need thisbe true if x and y do not commute? Give an example of commuting elements x,ysuch that the order of xy is not equal to least common multiple of |x| and |y|.

Sol : Let k = lcm(|x|, |y|) so m|k and n|k. Then (xy)k = xkyk = 1 hence kdivides order of xy. No, we have a counter example. Let x = (1, 2) and y = (2, 3)then |x| = |y| = 2 but xy = (1, 2, 3) has order 3. Take G = Z/2Z and takex, y = (1, 1) as both have order 2, but xy = (0, 0) has order 1, here order of xy isnot equal to least common multiple of |x| and |y|.

Prob 16: Show that if H is any group and h is an element of H with hn = 1,then there is a homomorphism from Zn =< x > to H such that x→ h.

Sol : Since hn = 1, there is a m ∈ N such that | < h > | = m so m|n. Defineπ : Zn → H such that π(xk) = hk then π(xk)π(xl) = hk+l = π(xk+l), hence π is ahomomorphism where k, l ≤ m, if k, l ≥ m then they will be replaced by x, y, suchthat k ≡ x(modm) and l ≡ y(modm).

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Prob 17: Show that if H is any group with h ∈ H, then there is homomorphismϕ : Z→ H such that 1→ h.

Sol : Since Z is cyclic, it’s image under any homomorphism will be the cyclicgroup generated by the image of 1. Since < h > is isomorphic to some cyclic groupZn or Z, and hence map from Z to cyclic group which maps generator to generatoris a homomorphism.

Prob 18: Let p be a prime and let n be a positive integer. Show that if x is anelement of the group G such that xp

n

= 1 then |x| = pm for some m ≤ n.Sol : As x ∈ G and xp

n

= 1. Let |x| = a then a|pn. Let a =∏nk=1(pαk

k ) and let

b = pn − a then xb = 1 and hence a|b that is,∏nk=1(pαk

k )|pn which is equivalent to∏nk=1,k 6=t(p

αk

k )|pn−αt where pt = p this is possible only when α1, . . . , αn = 0 thatis a = pm where m ≤ n.

Prob 19: Let p be an odd prime and let n be a positive integer. Use binomial

theorem to show that (1 + p)pn−1 ≡ 1(mod pn) but (1 + p)p

n−2 6= 1(mod pn).Sol : Let ordp(n) denote the highest order of p that divides n. Writing

(pm

k

)=pm∏k−1j=1 (pm − j)k∏k−1j=1 j

we can see that ordp(j) = ordp(pm − j) as whenever j = kpα where k ∈ Z+ and

(k, p) = 1 then, (pm − j) = (pm − kpα) = pα(pm−α − k) as ordp(pm−α − k) =

0 hence ordp(j) = ordp(pm − j) = α. Hence ordp

(pm

k

)= m − ordp(k). Now,

(1 + p)pn−1

=∑pn−1

k=0 (pk(pn−1

k

)) as ordp(p

k(pn−1

k

)) = k + n− 1− ordp(k) for k ≥ 1

term k−1−ordp(k) ≥ 0 and hence except for k = 0 all term of binomial expansion

will be divisible by pn and hence (1 + p)pn−1 ≡ 1(mod pn).

(1 + p)pn−2

=∑pn−2

k=0 (pk(pn−2

k

)) as ordp(p

k(pn−2

k

)) = k + n − 2 − ordp(k) for

k ≥ 2 term k− 2− ordp(k) ≥ 2 and hence except for k = 0, 1 all terms of binomial

expansion will be divisible by pn and hence (1 + p)pn−2 ≡ 1 + pn−1(mod pn), which

implies (1 + p)pn−2 6= 1(mod pn).

Prob 20: Let n be an integer ≥ 3. Use the Binomial Theorem to show that

(1 + 22)2n−2 ≡ 1(mod 2n) but (1 + 22)2

n−3 6= 1(mod 2n).

Sol : (1+22)2n−2

=∑2n−2

k=0 (22k(2n−2

k

)) as ord2(22k

(2n−2

k

)) = 2k+n−2−ord2(k)

for k ≥ 1 term (2k− 2− ord2(k)) ≥ 0 and hence except k = 0 all terms of binomial

expansion will be divisible by 2n and hence (1 + 22)2n−2 ≡ 1(mod 2n).

(1 + 22)2n−3

=∑2n−3

k=0 (22k(2n−3

k

)) as ord2(22k

(2n−3

k

)) = 2k + n− 3− ord2(k) for

k ≥ 2 term (2k − 2− ord2(k)) ≥ 0 and hence except k = 0, 1 all terms of binomial

expansion will be divisible by 2n and hence (1+22)2n−3 ≡ (1+2n−1)(mod 2n) which

implies (1 + 22)2n−3 6= 1(mod 2n).

Prob 21: Show that (Z/2nZ)× is not cyclic for any n ≥ 3.Sol : Let H =< x > be a cyclic group, if |H| = n < ∞, then for each positive

integer a dividing n there is a unique subgroup of H of order a. This subgroup is

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cyclic subgroup < xd >, where d = na · · · ( chap 2 Theorem 7, Abstract Algebra by

Dummit and Foote).For n ≥ 3 there 3 terms 2n−1, 2n−1+1, 2n−1−1 ∈ (Z/2nZ)× and all are distinct

hence (Z/2nZ)× is not cyclic group.

Prob 22: Let G be a finite group and let x ∈ G.

• Prove that if g ∈ NG(< x >) then gxg−1 = xa for some a ∈ Z.Sol : NG(< x >) = {g ∈ G|g < x > g−1 =< x >} as < x > is cyclic

group generated by x. Then by definition of NG(< x >), gxmg−1 = xn form,n ∈ Z putting m = 1 and n = a we get gxg−1 = xa.

• Prove conversly that if gxg−1 = xa for some a ∈ Z then g < x > g−1 =<x >.

Sol : As gxg−1 = xa which implies gxkg−1 = xak where k ∈ Z, asxak ∈< x > hence gxkg−1 ∈< x > which implies g < x > g−1 ≤<x >. Let |x| = n then gxig−1 for 0 ≤ i ≤ n − 1 are all distinct, becausegxag−1 = gxbg−1 implies xa = xb that is |g < x > g−1| =< x >= n andhence g < x > g−1 =< x >.

Prob 23: Let G be a cyclic group order n and let k be an integer relativelyprime to n. Prove that map x→ xk is surjectice.

Sol: Let x be a generator of cyclic group G. Since (k, |x|) = 1 then < xk >= Gsince map is from G → G hence and both x and xk generates G hence map issurjective.

Prob 24: Let Zn be a cyclic group of order n and for each integer let σa : Zn →Zn by σa(x) = xa for all x ∈ Zn.

• Prove that σa is an automorphism of Zn iff (a, n) = 1.Sol: If (a, n) = 1 then < x >=< xa > where |x| = n hence there

is a surjective homomorphism as |x| = |xa| = n and since Zn is finitesurjective homomorphism implies injective. Hence (a, n) = 1 implies σa isan automorphism. Now, if σa is an automorphism then σa(x) = xa is alsoa generator which implies (a, n) = 1.

• Prove that σa = σb iff a ≡ b(modn).Sol: If σa = σb this should also hold true when applied on x ∈ G,

σa(x) = σb(x) which implies xa = xb hence a ≡ b(modn). If a ≡ b(modn)then a = kn+ b then xa = xkn+b = xb which immplies σa(x) = σb(x) thatis σa = σb.

• Prove that every automorphism of Zn is equal to σa for some integer a.Sol: Let σ be an automorphism of Zn. Then σ(x) = xa for some a, since

σ(x) ∈ Zn since x is an generator then as xa is also a generator and henceσ = σa.

• Prove that σa ◦ σb = σab. Deduce that map a → σa is an isomorphism of(Z/nZ)× onto the Aut(Zn).

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Sol: σa ◦ σb(x) = σa(xb) = xab = σab(x) hence σa ◦ σb = σab. Mapa → σa is surjective as by part (c) every automorphism is of the form σawhere (a, n) = 1 and every such a belongs in (Z/nZ)×. This map is alsoinjective as by part (b) σa = σb implies a = b where a, b ∈ (Z/nZ)×. Hencemap is an isomorphism.