Exercises on Groups

1. Prove that all the subgroups of (ℤ,+) are the subsets of the form nℤ = {nz | z ∈ ℤ}. Derive that all subgroup of ℤ are cyclic.

2. Prove that the intersection of arbitrary subgroups of a group G is a subgroup of G.

3. Prove that the union of two subgroups of a group G is a subgroup of G if and only if one of the two subgroups is contained in the other. Verify this property in the case of (ℤ,+).

4. Let A and B two subgroups of a group G. Let AB := {ab | a ∈ A, b ∈ B}. Prove that AB is a subgroup if and only if AB = BA (set equality).

5. Prove that the set T of matrices of the form [a, b; -b a] where a,b ∈ ℝ are not both null at the same time, is a subgroup of (GL_n(ℝ), ⋅).

6. Prove that, if a and b are elements of a group G such that ab = ba, then the subgroup generated X = {a,b} is Abelian.

7. Let G an Abelian group and n a positive integer. Let S the subset of G formed from all the n-th powers of the elements of G, that is S := {gⁿ | g ∈ G}. Prove that S is a subgroup of G.

8. Prove that the set U(ℤ_n) of invertible elements of ℤ_n is a group with respest to multiplication.

9. Prove that given a group (G, ⋅), given two arbitrary elements of a,b ∈ G each of the following equations

   ax = b,   ya = b

   admits a unique solution.

10. Prove that in a group (G, ⋅), the following cancellation laws hold true:

    ax = ay,   ⇒   x = y

    xa = ya,   ⇒   x = y

11. Let (G, ⋅) a group. Let s a given element of G. In g is defined another operation *:

    a * b := a ⋅ s ⋅ b   ∀a,b ∈ G

    Verify whether (G, *) is a group.

12. Given the set S = ℝ \ {−1} of all real numbers different from −1. Determine if S is a group

    • with respect to ordinary multiplication of real numbers.

    • with respect to the following operation: x * y := x + y + xy, ∀x,y ∈ S.

13. Let * an operation on ℕ defined as follows: a * b is obtained by doing adding up the elements a + b represented in base 10 and adding up modulo 10 the correspondent digits (9 ⊕ 4 = 3). In practice a ⊕ b is obtained doing the usual addition of a and b, with no carryover. For example

    • 1492 ⊕ 1987 = 2379

    • 772 ⊕ 1343 = 1015 (3+2 gives 5 in the one's place of the result. 4+7 gives 1 in the 10's place (11 = 1 mod 10). 3+7 gives 0 in the 100's place. 0+1 gives 1 in the 1000's place)

    Determine if (ℕ, ⊕) is a group. Repeat the same exercise, replacing base 10 with an arbitrary base n.

14. Prove that non Abelian groups are never cyclic.

15. Prove that if g is an element of a group G such that gⁿ = e for some n ∈ ℕ, then the period of g is a divisor of n.

16. Prove that an element a̅ ∈ ℤ_n has order n/d, where d = gcd(a,n).

17. Determine in ℤ₁₀₀₀₀ all the elements of order 4 and 8.

18. Let G = ⟨g | gⁿ = e⟩ a cyclic group with n elements. Prove that ⟨g^k⟩ has n/d elements, where d = gcd(n,k). Deduce that all and the only generators of G are of the form g^k with (n,k) = 1.

19. Let G = {(a c; 0 b)}| a,b, c ∈ ℝ, a,b ≠ 0}. Which are the orders of the elements of G. Which are the orders of G = {(a c; 0 b)}| a,b, c ∈ ℂ, a,b ≠ 0}.

Solutions

1. Let H ≤ ℤ, H ≠ 0. Then H will contain for sure a positive integer h. Let n ∈ H be the least of such positive integers. We have clearly ⟨n⟩ = nℤ ⊆ H. To prove the other inclusion it suffices to divide any a ∈ H by n: a = nq + r, 0 ≤ r < n and since both a and nq are in H we have r ∈ H hence to avoid a contradiction of the minimality of n, r = 0

2. x ∈ ⋂_i H_i ⇔ x ∈ H_i, ∀i. Hence ∀x,y ∈ ⋂_i H_i we have xy⁻¹ ∈ H_i, which is the condition to satisfy to be a subgroup.

3. Let H and K two subgroups of G. If either H ⊆ K or K ⊆ H clearly H ∪ K coincides with either H or K. Conversely, if H ⊈ K and K ⊈ K, there exists h ∈ H, h ≠ K and a k ∈ K, k ≠ H. Then hk cannot belong neither to H nor to K, hence hk ≠ H ∪ K, thus H ∪ K is not a subgroup. In the case of (ℤ, +) if H = 3ℤ and K =2ℤ, 3 ∈ H, 3 ≠ K, 2 ∈ K, 2 ≠ K. 5 should be in the union, but 5 ∈ H and 5 ≠ K.

4. Let AB = BA then ∀ab, a'b' ∈ AB we have ab(a'b')⁻¹ = abb'⁻¹a'⁻¹ = ab̅a̅ ∈ AB, where we set b̅ = bb'⁻¹ ∈ B and a̅ = a⁻¹ ∈ A. Therefore, there exists some a₁ ∈ A and some b₁ ∈ B such that b̅a̅ = a₁b₁, so we have that ab̅a̅ = aa₁b₁ ∈ AB.

   Conversely let AB a subgroup. We have to prove that ab ∈ AB implies ab ∈ BA and the converse. If ab ∈ AB, then also (ab)⁻¹ ∈ AB, that is (ab)⁻¹ = a₁b₁ for some a₁ ∈ A and b₁ ∈ B. Then ab = (a₁b₁)⁻¹ = b₁⁻¹a₁⁻¹ ∈ BA since (a₁b₁) (b₁^-1a₁^-1) = e, thus AB ⊆ BA; same argument for the other inclusion.

5. The product of elements of T is still an element of T: [a² −b² 2ab; -2ab a² −b²]. For a = 1, b = 0 we have I. The inverse of a matrix of the form [a, b; −b a] is

   [a/(a² + b²) −b/(a² + b²); b/(a² + b²) a/(a² + b²)]

6. Since ab = ba, we have ⟨a,b⟩ = {aⁱb^j | i,h ∈ ℤ}. From ab = ba it can be easily seen that a^hb^k = b^ka^h. Thus

   (aⁱb^j)(a^hb^k) = a^i+hb^j+k = a^h+ib^k+j = (a^hb^k)(aⁱb^j).

   that is two elements of ⟨a,b⟩ commute, thus ⟨a,b⟩ is Abelian.

7. Let aⁿ, bⁿ two elements in S, with a,b ∈ G. Then aⁿ(bⁿ)⁻¹ = (since the group is Abelian) = (ab⁻¹)ⁿ.

   To to see why it's requires that a and b commute with each other, it works. Let c mean b^-1. Check that (bⁿ)^-1 = cⁿ, and then aⁿ cⁿ = aaa...accc...c = (ac)(ac)(ac)...(ac) = (ac)ⁿ which is back in S.

8. Let a̅,b̅ ∈ ℤ_n. If they are invertible then also their product is ab⁻¹ = a̅⁻¹ ⋅ b̅⁻¹. The class 1̅ is the identity. If a̅ ∈ ℤ_n, then a̅ is invertible. Then its inverse a̅⁻¹ ∈ U(ℤ_n) since (a̅⁻¹)⁻¹ = a̅.

   More generally, with a similar argument, given a ring with unity R, the subset U(R) of all invertible elements of R is a group.

9. By multiplying both sides of the first equation on the left by a⁻¹: x = a⁻¹b (unique solution); Similarly for the other equation which only solution is: y = ba⁻¹.

10. By multiplying both sides of the first equation on the left by a⁻¹ and of both sides ofthe second equation on the right by a⁻¹.

11. The operation * is associative: (a*b) * c = (a ⋅ s ⋅ b) * c = (a ⋅ s ⋅ b) ⋅ s ⋅ c = (since ⋅ is associative) = a ⋅ s ⋅ (b ⋅ s ⋅ c) = a * (b * c). To find the nueter element e, since a * e = e * a = a, that is a ⋅ s ⋅ e = e ⋅ s ⋅ a = a, thus e = s⁻¹. The inverse of a is an element y such that a * y = e hence a ⋅ s ⋅ y = e = s⁻¹ thus y = s⁻¹a⁻¹s⁻¹.

12. It is not a group with respect of the operation of multiplication since closure is not verified a ⋅ (−1/a) = −1.

    (S, *) is a group since closure is verified: x + y + xy = −1 ⇐⇒ x + y + xy + 1 = 0 ⇐⇒ x(1 + y) + (1 + y) = 0 ⇐⇒ (x + 1) (y + 1) = 0 ⇐⇒ x + 1 = 0 or y + 1, that cannot happen since x,y ∈ S.

    Associativity: (x * y)*z = (x + y + xy) * z = x + y + xy + z + xz + yz + xyz. And x * (x * z) = ... = x + y + z +yz + xy + xz + xyz = (x * y) *z.

    Identity element: We must find an e ∈ S such that

    e * x = x = x * e &emps; ∀x ∈ S

    it suffices e * x, since * is commutative.

    e * x = e + x + ex = x   ∀x ∈ S   ⇒   e + ex = 0  ∀x ∈ S.

    Thus e(1 + x) = 0 for all x ∈ S. Now, 1 + x ≠ 0 if and only if x ∈ S, hence the last relation implies e = 0, which the identity.

    Inverse: x * x' = 0 implies x + x' + xx' = 0, from which x + x'(1 + x) = 0 that is x' = −x/(1 + x). Again 1 + x ≠ 0, sice x ∈ S. Thus every element of S is invertible, and (S, *) is a group.

13. Closure: The result of ⊕ is still a natural number.

    Associativity of ⊕: it is guaranteed by the associativity of addition modulo 10.

    Identity: the neutral element is the zero.

    Inverse: To find the inverse we should ask 905 ⊕ ___ = 0? The answer is 105; Thus the inverse of the number n₁n₂ ... n_h is the number n'₁n'₂ ... n'_h.

14. A cyclic group is generated by a single element say, a, with all its elements being powers of a. Since aⁱa^j = a^jaⁱ = a^i+j for any integers i,j, we see that every cyclic gorup is Abelian.

15. Let m the order of g. We divide n by m and we get n = mq + r, 0 ≤ r < m, from which

    gⁿ = g^mqg^r = (g^m)^qg^r = eg^r

    hence the remainder must be null r = 0 and n = mq.

16. Let a̅ ≠ 0 and k its order. By definition k is the least positive integer such that ka̅ = 0. Then it must be ka is a mulitple of n (hence ka is a multiple of a). Then ka is a common multiple of a and n; since it must be the least, it is the lcm(a, n), thus ka = an/(a ,n), which yields to k = n/(a, n).

17. Based on previous exercise we have to dermine the classes a̅ such that 2⁴5⁴/(a, 2⁴5⁴) = 4. It must result (a, 2⁴5⁴) = 2²5⁴. The classes of period 4 are thus: (2500,10000) = (7500,10000) = 2500, [2500] and [7500].

    With a similar procedure the elements having order 8 are [1250], [3750], [6250] and [8750].

18. The proof is the same employed in Proposition 5.7.5. The n-th roots of units are a cyclic group of order n. To say that ⟨g^k⟩ has n/d elements is the same as saying that the period of g^k is n/d.   ■

19. The only elements with finite period are I = (1 0; 0 1) (period 1), (1 c; 0 -1) period 2 ∀c ∈ R, (-1 0;0 -1) period 2. Indeed we have

    (a c; 0 b)ⁿ = (aⁿ c(aⁿ⁻¹ + aⁿ⁻² b + ... + bⁿ⁻¹); 0 b²)

    hence all the elements of G with a ≠ ± 1, b ≠ ±1 have infinite period. If a = b =1, the element (1 c; 0 1) satisfies the relation

    (1 c; 0 b)ⁿ = (1 nc; 0 1)

    hence if c ≠ 0 the period is infinite, if c = 0 we have the identirty.

    For G' the elements of infinite period are the elements: (a 0; a b) with a,b the n^th roots of unity.

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