Exercises on direct products

1. What group ℤ₅ x ℤ₈ is isomorphic to?

2. Determine whether the group ℤ₁₆ is isomorphic to the direct product ℤ₄ x ℤ₄.

3. Check whether (ℚ, +) is the direct product of two of its proper subgroups.

4. Prove that a group with order 15 is necessarily cyclic.

5. Prove that U(ℤ₇) is a direct product. Check whether it is cyclic or not.

6. Calculate the possible orders of the elements of ℤ x ℤ₆. Is it a cyclic group? And what about ℤ x ℤ?.

7. Which of the following groups having the same order are isomorphic?

   ℤ₁₆,   ℤ₈ x ℤ₂,   ℤ₄ x ℤ₄,   ℤ₄ x ℤ₂ x ℤ₂,   ℤ₂ x ℤ₂ x ℤ₂ x ℤ₂

8. Calculate the orders of the elements ℤ₂ x D₅.

9. Check whethet S₄ is a direct product of its non-trivial subgroups.

10. Let G = H x K the direct product of two groups H and K. Let Z(H) the center of H and Z(K) the center of K. Check whether the following identity holds true

    Z(H x K) = Z(H) x Z(K)

    with Z(H x K) the center of H x K. If the identity is true, prove it.

11. Prove that the group E₂ of all planar isometries is the semidirect product of the orthogonal group O₂(ℝ) and of the group T of translations.

Solutions

1. By Corollary 7.17.15 ℤ₅ x ℤ₈ is cyclic since (8,5) = 1, hence isomorphic to ℤ₄₀. ■

2. ℤ₁₆ not isomorphic to ℤ₄ x ℤ₄ which is not cyclic by Corollary 7.17.15: it ha no element of order 16.  ■

3. Any two nontrivial subgroups of Q intersect nontrivially. Take p/q ∈ H₁ and r/s ∈ H₂. By closure unders addition p/q + p/q + ... = p ∈ H₁ (q times addition). Similarly r ∈ H₂. Hence p ⋅ r ∈ H₁ ∩ H₂.  ■

4. A group G of order 15 = 5 ⋅ 3, has a Sylow 3-subgroup and a Sylow 5-subgroup. They are both unique and have obviously non trivial intersection since the orders of each are different. Let a a generator of H and b one of K, then the element ab has order 15. Thus G is cyclic. ■

5. U(ℤ₇) is the set of invertible elements of ℤ₇. In ℤ₇ every nonzero element is invertible, hence U(ℤ₇) = {[1], [2], [3], [4], [5], [6]. It is cyclic.

6. (0, [0]₆) is the identity element with order 1; (z, [a]₆) is always inf, if z ≠0; (0, [5]₆) has order 6 since 6 ⋅ [5]₆ = [30]₆ = [0]₆. (0,[2]₆) and (0, [4]₆) have order 3; (0,[3]₆) has order 2.

   ℤ x ℤ is not cyclic since there is no element (n,m) ∈ ℤ x ℤ which generate the whole group (if you some over (1,1) you don't get (0,0)). ■

7. Except ℤ₁₆ which is cyclic the other groups resulting by direct product are not cyclic by Proposition 7.17.14, so not isomorphic to ℤ₁₆. For the other groups you can look at order of different elements, how many elements in ℤ₄ x ℤ₄ have order 4? How many elements in ℤ₄ x ℤ₂ x ℤ₂ have order 4?

   ℤ₈ x ℤ₂

   	ℤ2
   		[0]2	[1]2
   ℤ8
   	[0]8	(0,0) order = 1	(0,1) 2
   	[1]8	(1,0) 1	(1,1) 8
   	[2]8	(2,0) 4	(2,1) 4
   	[3]8	(3,0) 4	(3,1) 4
   	[4]8	(4,0) 2	(4,1) 2
   	[5]8	(5,0) 8	(5,1) 8
   	[6]8	(6,0) 3	(6,1) 4
   	[7]8	(7,0) 8	(7,1) 8

   ℤ₄ x ℤ₄

   	ℤ4
   		[0]4	[1]4	[2]2	[3]4
   ℤ8
   	[0]4	(0,0) order = 1	(0,1) 4	(0,2) 2	(0,3) 4
   	[1]4	(1,0) 4	(1,1) 4	(1,2) 4	(1,3) 4
   	[2]4	(2,0) 2	(2,1) 4	(2,2) 2	(2,1) 4
   	[3]4	(3,0) 4	(3,1) 4	(3,2) 4	(3,3) 4

   ℤ₂ x ℤ₂ x ℤ₂ x ℤ₂:

   o((0,0,0,0)) = 1   o((1,0,0,0)) = 2   o((0,1,0,0)) = 2,   o((0,0,1,0)) = 2,   o((0,0,0,1)) = 2
   o((1,1,0,0)) = 2   o((0,1,1,0)) = 2   o((0,0,1,1)) = 2,   o((1,0,0,1)) = 2,
   o((1,1,1,0)) = 2   o((0,1,1,1)) = 2   o((1,0,1,1)) = 2,   o((1,1,1,1)) = 2,
   o((1,0,1,0)) = 2,  o((0,1,0,1)) = 2   o((1,0,1,0)) = 2

   ℤ₄ x ℤ₂ x ℤ₂:

   o((0,0,0)) = 1,   o((0,1,0)) = 2,   o((0,0,1)) = 2   o((0,1,1)) = 2
   o((1,0,0)) = 4,   o((1,0,1)) = 4,  o((1,1,0)) = 4,   o((1,1,1)) = 4
   o((2,0,0)) = 2,   o((2,0,1)) = 2,  o((2,1,0)) = 2,   o((2,1,1)) = 2
   o((3,0,0)) = 4,   o((3,0,1)) = 4,   o((3,1,0)) = 4,   o((3,1,1)) = 4,
   

   Hence none of the groups above are isomorphic.  ■

8. D₅ is the group of symmetries of a regular pentagon. It has ten elements: e, four counterclockwise rotations r_i (72° = 360°/5, 144°, 216°, 288°) and the five reflections l_i about perpendiculars from a vertex to the opposite side.

   ℤ₂ x D₅

   	ℤ2
   		[0]2	[1]2
   D5
   	e	([0]2, e) order = 1	([1]2, e) 4
   	r1	([0]2, r1) order = 5	([1]2,r1) 10
   	r2	([0]2, r2) 5	([1]2, r2) 10
   	r3	([0]2, r3) 5	([1]2, r3) 10
   	r4	([0]2, r4) 5	([1]2, r4) 10
   	l1	([0]2, l1) 2	([1]2, l1) 2
   	l2	([0]2, l2) 2	([1]2, l2) 2
   	l3	([0]2, l3) 2	([1]2, l3) 2
   	l4	([0]2, l4) 2	([1]2, l4) 2
   	l5	([0]2, l5) 2	([1]2, l5) 2

   so the possible orders in D₅ are 1, 2, 5, 10.  ■

9. The two normal subgroups of S₄ are A₄ = {id, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)} and K = {id, (12)(34), (1,3),(2,4), (2,3) (1,4)} which have a non-trivial intersection hence by Theorem 7.17.4 is not possible to express S₄ as direct product.  ■

10. It is true. (x, y) ∈ Z(H x H) ⇐⇒ (x, y) (h, k) = (h, k) (x, y) for all (h, k) ∈ H x K ⇐⇒ (xh, yk) = (hx, ky) for all h ∈ H, k ∈ K ⇐⇒ xh = hx, yk = ky for all h ∈ H, k ∈ K ⇐⇒ x ∈ Z(H), y ∈ Z(K) ⇐⇒ (x, y) ∈ Z(H) x Z(K) ■.

11. We have already studied the isometries which form a group under composition of function, the so called Euclidean group E₂. Euclidean group is built from the translational group T and the orthogonal group O₂(R).

    By checking that the conditions of Theorem 7.17.18 are satisfied i.e:

    • T is a normal subgroup of E₂,

    • E₂ = TO₂,

    • T ∩ O₂ = {I}.

    We have that E₂ is the semidirect product T x_φ O, where φ is a certain homorphism.

    We see that the composition of a rotation, a translation, and the inverse rotation is another translation i.e. T is a normal subgroup. Suppose f ∈ O, τ in T, and τ(0) = v. Then for each x ∈ ℝ² we have

    ftf⁻¹(x) = f (v + f⁻¹(x))
    = f(v) + f(f⁻¹(x))  because f in linear
    = f(v) + x

    Therefore the conhugate ftf⁻¹ is translation by the vector f(v), hence T is a normal subgroup of E₂.

    The intersection of T and O is just the identity transformation because every non-trivial translation moves the origin, whereas every element of O keeps the origin fixed.

    Now we show that E₂ = TO = OT, that is a general element E₂ is either a rotation about the origin followed by a translation, or a rotation followed by a traslation. Let f : ℝ² ⟶ ℝ² be an arbitrary isometry. Define g : ℝ² ⟶ ℝ² by g(x) = f(x) − f(0). Then g fixes the origin, so g ∈ O₂. Clearly, f = tg, where τ(x) = x + f(0). By taking the inverse of f we have f⁻¹ = τ⁻¹ o⁻¹, so E₂ = TO₂ as well.

    In matrix notation, the general direct isometry of the Euclidean plane is given by

    $$\left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)$$

    where x and y are Cartesian coordinates; it represents a rotation (counter-clockwise) through an angle θ followed by a translation by distance a in the x-direction and a distance b in the y-direction. The same can also be obtained by a suitable translation followed by a rotation.

    Thus E₂ = T x_φ O is a semidirect product where φ is the homomorphism φ: O ⟶ Aut(T) induced by conjugation

    φ: O₂(R) ⟶ Aut(T)
    φ_o(t) = oto⁻¹ ■

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