Group Theory: Math30038, Sheet 6

Solutions GCS

1. Consider the group D of rigid symmetries of a regular n-gon (which may be turned over). Prove that this group has order 2n, is non-abelian, and can be generated by two elements each of order 2. Show that D has a cyclic subgroup of index 2. Solution Each axis of symmetry of a regular n-gon either passes through a vertex and the midpoint of the opposite side (if n is odd) or a pair of opposite vertices or mid-points of sides (if n is even), Either way, these give rise to n elements of order 2 in D which are relections in these axes. Label the vertices with the numbers from 1 to n in clockwise order. Suppose that n = 2r + 1 is odd. In this event let x = (2, n)(3, n − 1) ⋯ (r + 1, r + 2) and y = (1, 2)(n, 3) ⋯ (r + 1, r + 3). Now z = xy = (1, 2, 3, …, n) and yx = z^{−1} ≠ z since n ≥ 3 for non-triviality. Thus D is non-abelian. In the degenerate case n = 2 the group is in fact abelian. Now ⟨z⟩ is the cyclic group of order n consisting of rotations, and x⟨z⟩ is the set of n reflections mentioned earlier. This exhausts D and D = ⟨x, y⟩. Next suppose that n = 2r is even. This time let

x = (1, 2)(n, 3) ⋯ (r + 2, r + 3)

and

y = (1, 3)(n, 4) ⋯ (r + 1, r + 3)

so xy = (1, 2, 3, …, n) and the same story unfolds.

2. Consider the group D of rigid symmetries of the integers: so D is the group of all bijections θ from ℤ to ℤ which preserve distance. Thus θ must have the property that if x, y ∈ ℤ, then |x − y| = |(x)θ − (y)θ|. Prove that this group has infinite order, is non-abelian, and can be generated by two elements each of order 2. Show that D has a cyclic subgroup of index 2. Solution Let

x = (1, −1)(2, −2)(3, −3) ⋯

and

y = (0, 1)(−1, 2)(−2, 3) · · · .

Now xy is the map which sends z to z +1 for every integer z. Now ⟨xy⟩ is a cyclic subgroup of D. Any element t of D which is not in ⟨xy⟩ must send 0 to c and 1 to c − 1. Now t(xy)^{−c+1} sends 0 to 1 and 1 to 0. The definition of D ensures that t(xy)^{−c+1} = y and so t = y(xy)^{c−1} ∈ y⟨xy⟩. Thus the index of ⟨xy⟩ in D is at most 2. However, y ∉ ⟨xy⟩ since y reverses the direction of the integers. Thus |D : ⟨xy⟩| = 2.

3. Let D = ⟨x, y⟩ where o(x) = o(y) = 2 and x ≠ y. Let z = xy and put H = ⟨z⟩.

(a) Prove that x^{−1}zx = y^{−1}zy = z^{−1}. Solution x^{−1}zx = x^{−1}xyx = yx = y^{−1}x^{−1} = z^{−1}. Also y^{−1}zy = y^{−1}xy^{2} = y^{−1}x^{−1} = z^{−1}.

(b) Prove that x, y ∉ H.

Solution If x ∈ H, then x^{−1}zx = z^{−1} but also x^{−1}zx = z since H is cyclic and therefore abelian. Thus z^{2} = 1. A cyclic group can contain at most one subgroup of order 2, and so x = z = xy so y = 1 which contradicts o(y) = 2. Therefore x ∉ H. If y ∈ H then x = zy ∈ H which we know is not the case, so y ∉ H.

(c) Prove that |G : H| = 2. Solution Since x^{−1} = x and y^{−1} = y, and both x and y square to 1, any element of D must be either 1, or xyx ⋯ xyx or yxy ⋯ yxy or a power of z. Now xyx ⋯ xyx ∈ x⟨z⟩ and yxy ⋯ yxy = xxyxy ⋯ yxy ∈ x⟨z⟩. Thus the index of H is at most 2. Since x ∉ H it follows that |G : H| = 2.

(d) Let n = o(z) ∈ ℕ∪{∞}. For each possible value of n let G be called D_{n}. Show that the multiplication in D_{n} is completely determined (i.e. the number n nails down the group). Solution We have shown that in each group D there is a cyclic subgroup H = ⟨z⟩ of index 2, and that xH ≠ H. Therefore every element of D_{n} is uniquely expressible at x^{ε}h where ε ∈ {0, 1} and h ∈ H. Note that z^{x} = z^{−1} so h^{x} = h^{−1} for all h ∈ H. Multiplication is as follows: suppose that h, k ∈ H = ⟨z⟩. We have

\begin{aligned}h · k & = & hk \\ h · xkk & = & xh^{x}k = xh^{−1}k \\ xh · k & = & xhk \\ xh · xk & = & h^{x}k = h^{−1}k \end{aligned}

Thus the only issue is how multiplication happens in H, but H = ⟨s⟩ is cyclic and its multiplication is determined by the single number n = o(s).

(e) For each n ∈ ℕ ∪ {∞}, determine the centre of D_{n}. Solution There is no group D_{1}. The group D_{2} is of order 4 and so is abelian. Let us suppose that n ≥ 3. Now ⟨xy⟩ contains at most one element of order 2. If n is odd or infinite, there is no such element, but if n = 2r is even, there is such an element w = z^{r} and since w^{x} = w^{y} = w, w is central. If t = xz^{m} is in D_{n} but not in H = ⟨xy⟩ for some integer m, then t^{y} = yxyyz^{m}y = xxyxyyz^{m}y = xz^{2}z^{−m}. Now t^{y} = 1 if and only if z^{2m} = z^{2} so o(z) divides 2m − 2. However t^{x} = t if and only if xz^{m} = xz^{−m} and so o(z) divides 2m. Thus if t is central then o(z) must divide both 2m − 2 and 2m and so must divide 2. Thus for n > 2 the centre of D_{n} is the trivial subgroup unless n is finite and even, in which case the centre of D_{n} is {1, z^{n/2}}, a cyclic subgroup of order 2.

(f) Determine the conjugacy classes of D_{8}. Solution I meant to ask about D_{4}, so we will do both cases. Let G be the group under discussion. It is easy to verify that the centralizer of x has order 4, and is {1, z^{2}, x, xz^{2}}. Therefore the conjugacy class of x in D_{4} has size 2. Note that x^{y} = yxy = xxyxy = xz^{2}. Thus the conjugacy class of x is {x, xz^{2}}. Now y = xz and the centralizer of xz at least contains 1, xz, z^{2} and xz^{3} so the conjugacy class of y has size at most 2. However y^{x} = xyx = zx = xz^{−1} = xz^{3}. Thus the conjugacy class of y (= xz) is {xz, xz^{3}}. The centre of G is {1, z^{2}} and so each of the elements of this set is in a conjugacy class of size 1. Finally, the elements z and z^{3} are non-central and therefore are in conjugacy classes of size greater than 1. The remaining conjugacy class is therefore {z, z^{3}}. In summary, the conjugacy classes of D_{4} are

{1}, {z^{2}}, {z, z^{3}}, {x, xz^{2}}

and

{xz, xz^{3}}.

Let us work with D_{8}. The centralizer of x contains 1, x, z^{4} and xz^{4} so the conjugacy class of x has size at most 16/4 = 4. Now x^{y} = yxy = xz^{2}, and x^{yx} = xz^{6} and x^{yxy} = xz^{2}z^{2} = xz^{4}. Thus the conjugacy class of x is {x, xz^{2}, xz^{4}, xz^{6}}. Similar calculations reveal that the conjugacy class of y = xz is {xz, xz^{3}, xz^{5}, xz^{7}}. The elements of H are all centralized by H and so are either central, or are in conjugacy classes of size 2. Each element of H is conjugate to its inverse. Therefore the conjugacy classes of D_{8} are

{x, xz^{2}, xz^{4}, xz^{6}}, {xz, xz^{3}, xz^{5}, xz^{7}}, {1}, {z^{4}}, {z, z^{7}}, {z^{2}, z^{6}}

and

{z^{3}, z^{5}}.

(g) Do you recognize D_{6}? Solution I meant to ask about D_{3}, a non-abelian group of order 6, which is a copy of S_{3}. Notice that S_{3} is generated by two distinct elements of order 2 (as one should expect).

4. Suppose that G is a non-abelian finite group of order 2p where p is a prime number. Prove that G is generated by two elements order 2. Solution The fact that G is non-abelian forces the prime p to be odd. The group G contains an element x of order 2 and an element t of order p by Cauchy’s theorem. Now G = ⟨x, t⟩ is non-abelian so x and t do not commute. Now y = x^{t} ≠ 1 and y^{2} = t^{−1}xxt = 1 so y has order 2. Let T = ⟨x, y⟩ so T has even order more than 2, and dividing 2p. Therefore T = G and by previous theory, G is a copy of D_{p}.

5. We define a subgroup Q of S_{8} by letting i = (1, 2, 3, 4)(5, 6, 7, 8), j = (1, 5, 3, 7)(2, 8, 4, 6) (and NOT (2, 6, 4, 8) as earlier stated) and put Q = ⟨i, j⟩. Let k = ij and z = i^{2}. This group was the basis of William Rowan Hamilton’s generalization of the complex numbers called the Quaternions.

(a) Show that i^{2} = j^{2} = k^{2} = z and z^{2} = 1.

Solution This is routine.

(b) Show that ij = k, jk = i and ki = j. Solution This is routine.

(c) Show that ji = zk, kj = zi and ik = zj. Solution This is routine.

(d) Show that z is in the centre of Q. Solution z = i^{2} = j^{2} so z commutes with both i and j, so z is central.

(e) Show that Q = ⟨i⟩ ∪ z⟨i⟩.

Solution This is nonsense. z ∈ ⟨i⟩ and therefore ⟨i⟩ ∪ z⟨i⟩ = ⟨i⟩. However, for any element u ∉ ⟨i⟩ it will be the case that Q = ⟨i⟩ ∪ u⟨i⟩. The choices for u are j, zj, k and zk. Let us choose u to be k, and demonstrate this decomposition in this case. Any element of Q is a word in i and j. Now j = ik^{−1} = ik^{3} so every element of Q is a word in i and k (with positive exponents). Now any occurrence of ik can be replaced by kiz Using this repeatedly the ks can migrate to the left. When this is done use z = i^{2} to eliminate z. If the power of k involved is even, then use k^{2n} = i^{2n} to eliminate k. If the power of k involved is odd, all but one k can be eliminated. Thus Q = ⟨i⟩ ∪ k⟨i⟩.

(f) Show that |Q| = 8. Solution This follows immediately from the previous part.

(g) Show that Q and D_{8} (in Question 1) are both non-abelian groups of order 8, but they contain different numbers of elements of order 4. . Solution I meant to write ‘Show that Q and D_{4} (in Question 1) are both non-abelian groups of order 8,’ I fear that D_{8} has order 16. In D_{4} there are 5 involutions; there are 4 reflections and one rotation through π. In Q however, only z has order 2. One may verify tha t the remaining six non-identity elements of Q all have order 4. Now Q is non-abelian since ij ≠ ji. D_{4} is non-abelian because xy ≠ yx.

(h) Determine the conjugacy classes of Q. Solution They are

{1}, {z}, {i, zi}, {j, zj} and {k, zk}.

There are any number of ways to demonstrate this.

(i) On which bridge are the quaternions inscribed? Solution The Sir William Rowan Hamilton Bridge, Dublin.

6. Let G denote the set of invertible n by n matrices with complex entries. This is a group under multiplication of matrices. Give a transversal for the conjugacy classes of G. Hint: the course MA20012 does this (and not much else). Solution Jordan normal forms with “λ” non-zero.

7. Show that if N is a normal subgroup of G, then N must be a union of conjugacy classes of G including the conjugacy class of the identity element. Deduce that the only normal subgroups of A_{5} are 1 and A_{5}, but that A_{4} has a normal subgroup M which is neither 1 nor A_{4}. Solution Since g^{−1}Ng = N for all g ∈ G, it follows that if n ∈ B, then the conjugacy class of n is a subset of N. The conjugacy classes of A_{5} are the conjugates of id, (1, 2, 3), (1, 2)(3, 4), (1, 2, 3, 4, 5) and (1, 2, 3, 5, 4). These conjugacy classes have sizes 1, 20, 15, 12 and 12 respectively. The only sums these sizes which include 1 and which divide 60 are 1 and 60, so A_{5} is a simple group. An analogous calculation for A_{4} yields that the conjugacy classes of A_{4} have sizes 1, 3, 4 and 4. In this case 1 + 3 is a divisor of 12, so it is possible that there is a normal subgroup of size 4. It so happens that

{id, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}

is a (normal) subgroup of A_{4} of this order.