Mathematics 243, section 3 – Algebraic Structures Solutions for Problem Set 8 November 16, 2012

‘A’ Section

1. Consider the 2 × 2 matrices I_{2}, S, X, Y, D, T defined by

I_{2} = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) , S = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})
X = (\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}) , Y = (\begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix})
D = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) , T = (\begin{matrix}0 & −1 \\ −1 & 0 \end{matrix})

Let G = {I_{2}, S, S^{2}, S^{3}, X, Y, D, T }. Construct the operation table for matrix multiplication on this set and verify that G is a group under this operation.

Solution: The table looks like this (taking the products in order row × column in all cases):

\begin{matrix}· & I_{2} & S & S^{2} & S^{3} & X & Y & D & T \\ I_{2} & I_{2} & S & S^{2} & S^{3} & X & Y & D & T \\ S & S & S^{2} & S^{3} & I_{2} & D & T & Y & X \\ S^{2} & S^{2} & S^{3} & I_{2} & S & Y & X & T & D \\ S^{3} & S^{3} & I_{2} & S & S^{2} & T & D & X & Y \\ X & X & T & Y & D & I_{2} & S^{2} & S^{3} & S \\ Y & Y & D & X & T & S^{2} & I_{2} & S & S^{3} \\ D & D & X & T & Y & S & S^{3} & I_{2} & S^{2} \\ T & T & Y & D & X & S^{3} & S & S^{2} & I_{2} \end{matrix}

The table shows that G is closed under matrix multiplication, so this is a binary operation. Matrix multiplication is always associative as we saw before, so this operation is associative. The element I_{2} is an identity. Finally, each element has an inverse for matrix multiplication. I, S^{2}, X, Y, T, D are their own inverses, while S^{−1} = S^{3} and (S^{3})^{−1} = S.

2. a. Let G = ℤ_{12} under the operation of addition modulo 12. Determine the cyclic subgroups ⟨[a]⟩ for all [a] ∈ ℤ_{12}. Solution: We have

⟨[0]⟩ = {[0]}
⟨[1]⟩ = ℤ_{12} = ⟨[5]⟩ = ⟨[7]⟩ = ⟨[11]⟩
⟨[2]⟩ = {[0], [2], [4], [6], [8], [10]} = ⟨[10]⟩
⟨[3]⟩ = {[0], [3], [6], [9]} = ⟨[9]⟩
⟨[4]⟩ = {[0], [4], [8]} = ⟨[8]⟩
⟨[6]⟩ = {[0], [6]}

b. Conjecture a general formula for the number of elements in ⟨[a]⟩ in terms of the integers a and 12. Check out your conjecture on the corresponding list of cyclic subgroups of ℤ_{20} constructed in class on November 14.

Solution: The pattern that fits all of these examples is that the number of elements in ⟨[a]⟩ ⊆ ℤ_{n} is equal to n/ gcd(a, n). For instance, with n = 12 and a = 10, we have gcd(10, 12) = 2, and ⟨[10]⟩ has 12/2 = 6 elements. Moreover, all the a with the same gcd(a, n) apparently generate the same cyclic subgroup. Comment: These patterns do hold in general and we will prove them shortly.

‘B’ Section

1. Let

G = \{(\begin{matrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{matrix}) | a, b, c ∈ ℝ\}

Show that G is a group under the operation of matrix addition.

Solution: We know that M_{3×3}(ℝ) is a group under matrix addition. So a sneaky method here is to apply the “shortcut method” of Theorem 3.10 in the text to show that G is a subgroup of M_{3×3}(ℝ). G is definitely nonempty since it contains elements for all triples a, b, c of real numbers. Next, if

X = (\begin{matrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{matrix})  and  Y = (\begin{matrix}0 & a′ & b′ \\ 0 & 0 & c′ \\ 0 & 0 & 0 \end{matrix})

are in G, then

X − Y = (\begin{matrix}0 & a − a′ & b − b′ \\ 0 & 0 & c − c′ \\ 0 & 0 & 0 \end{matrix}) ∈ G

as well. Therefore G is a subgroup of M_{3×3}(ℝ), hence a group.

2. Let

G = \{(\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix}) | a, b, c ∈ ℝ\}

Is G is a group under the operation of matrix multiplication. If so, say why; if not say which properties in the definition fail.

Solution: We know that GL_{3}(ℝ) is a group under matrix multiplication. So a sneaky method here is to apply the “shortcut method” of Theorem 3.10 in the text to show that G is a subgroup of GL_{3}(ℝ). G is definitely nonempty since it contains elements for all triples a, b, c of real numbers. Next, if

X = (\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix})  and  Y = (\begin{matrix}1 & a′ & b′ \\ 0 & 1 & c′ \\ 0 & 0 & 1 \end{matrix})

are in G, then

Y^{−1} = (\begin{matrix}1 & −a′ & a′c′ − b′ \\ 0 & 1 & −c′ \\ 0 & 0 & 1 \end{matrix}) .

Hence

XY^{−1} = (\begin{matrix}1 & a − a′ & a′c′ − b′ − ac′ + b \\ 0 & 1 & c − c′ \\ 0 & 0 & 1 \end{matrix}) ∈ G

as well. Therefore G is a subgroup of GL_{3}(ℝ), hence a group.

3. Let G be a group, and consider the relation R on G defined by xRy ↔ y = axa^{−1} for some a ∈ G. (If xRy is true, the a that works in the equation will depend on x, y.) R is called the conjugacy relation on G.

a. Show that conjugacy is an equivalence relation on G.

Solution: For all x ∈ G, we have xRx since x = exe^{−1}. Therefore R is reflexive. If xRy, then y = axa^{−1}. If we multiply both sides of this equality by a^{−1} on the left and a on the right, we get x = a^{−1}ya = a^{−1}y(a^{−1})^{−1}. It follows that yRx is also true, so R is symmetric. Finally, if xRy and yRz, then y = axa^{−1} and z = byb^{−1} for some a, b ∈ G. Substituting, we get z = b(axa^{−1})b^{−1} = (ba)x(ba)^{−1} by associativity and the reverse order law for inverses. Therefore R is transitive.

b. Show that G is an abelian group if and only if the equivalence classes for the conjugacy relation satisfy [x] = {x} for all x ∈ G.

Solution: We have [x] = {x} if and only if x = axa^{−1} for all a ∈ G, or equivalently if and only if xa = ax. This is true for all classes [x] if and only if xa = ax for all a and all x in G. That is the same as saying the operation in G is commutative, or that G is an abelian group.

c. More generally, show that if [x] = {x} for some element x ∈ G, then xa = ax for all a ∈ G and conversely. The set of such elements x is called the center of G: The center is the subset of G defined by:

Z(G) = {x ∈ G | xa = ax for all a in G}.

Solution: One direction is essentially the same as the argument for part of the proof of part b, except we are no longer assuming that [x] = {x} holds for all x ∈ G, just for some particular x ∈ G. This proof is “reversible,” so it can be phrased like this:

[x] = {x} ⇔ x = axa^{−1} for all a ∈ G

⇔ xa = axa^{−1}a for all a ∈ G

⇔ xa = ax(a^{−1}a) for all a ∈ G, by associativity ⇔ xa = axe for all a ∈ G, by definition of inverse ⇔ xa = ax for all a ∈ G, by definition of identity element.

Hence [x] = {x} if and only if x ∈ Z(G), the center of G. d. Show that the center of G (as defined in part c) is a subgroup of G.

Solution: We again use the criterion of Theorem 3.10: Z(G) is not empty since it always contains the identity element e ∈ G – recall ex = x = xe for all x ∈ G. Next, let x, y ∈ Z(G), then xa = ax and ya = ay for all a ∈ G. But then y^{−1}yay^{−1} = y^{−1}ayy^{−1} as well so ay^{−1} = y^{−1}a for all a ∈ G. Hence by associativity and this last observation, for all a ∈ G:

(xy^{−1})a = x(y^{−1}a) = x(ay^{−1}) = (xa)y^{−1} = (ax)y^{−1} = a(xy^{−1}).

This shows xy^{−1} ∈ Z(G), so Z(G) is a subgroup of G. e. Let x ∈ G be a fixed element and define C_{x} = {a ∈ G | x = axa^{−1}}. Show that C_{x} is a

subgroup of G (C_{x} is called the centralizer of x). Solution: The idea of the proof is similar to that of the proof of part d. First, C_{x} is not empty since C_{x} contains at least the identity element. Let a, b ∈ C_{x}. Then x = axa^{−1} = bxb^{−1}. It also follows by multiplying both sides of the equality x = bxb^{−1} by b^{−1} on the left and b on the right that b^{−1}xb = x. But then by the reverse order law and associativity,

(ab^{−1})x(ab^{−1})^{−1} = ab^{−1}xba^{−1} = a(b^{−1}xb)a^{−1} = axa^{−1} = x.

It follows that ab^{−1} ∈ C_{x}, so C_{x} is a subgroup by Theorem 3.10. f. Let G = S(A) be the group of permutations of A = {a, b, c}. Using the names for the

elements of this group we introduced in Problem Set 3, find all of the equivalence classes for the conjugacy relation on G (there are three of them), determine the centralizer of each element of G, and the center of G. How are the sizes of the equivalence class of x and the number of elements of the centralizer of x related in each case? Solution: Recall that S(A) = {I_{A}, R_{a}, R_{b}, R_{c}, C_{1}, C_{2}}. Computing we find that there are exactly three conjugacy classes:

[I_{A}] = {I_{A}}
[R_{a}] = {R_{a}, R_{b}, R_{c}} = [R_{b}] = [R_{a}]
[C_{1}] = {C_{1}, C_{2}}

To start, since I_{A} is the identity element of this group, which commutes with every element, it follows from part c above, that [I_{A}] = {I_{A}}. The rest can be read off from the operation table you derived in Problem Set 3. For example,

R_{b} = R_{c} ∘ R_{a} ∘ R_{c}^{−1},

so R_{b} ∈ [R_{a}]. Similarly

R_{c} = R_{b} ∘ R_{a} ∘ R_{b}^{−1},

so R_{c} ∈ [R_{a}]. On the other hand conjugating C_{1} or C_{2} by any element of S(A) always yields either C_{1} or C_{2}, so those elements form another conjugacy class. The centralizers are as follows:

C_{I_{A}} = S(A)
C_{C_{1}} = C_{C_{2}} = {I_{A}, C_{1}, C_{2}}
C_{R_{a}} = {I_{A}, R_{a}}
C_{R_{b}} = {I_{A}, R_{b}}
C_{R_{c}} = {I_{A}, R_{c}}.

(For instance, from the group table for S(A), we see I, C_{1}, C_{2} all commute with C_{1}, but R_{a} ∘ C_{1} = R_{b} ≠ R_{c} = C_{1} ∘ R_{a}, R_{b} ∘ C_{1} = R_{c} ≠ R_{a} = C_{1} ∘ R_{b}, and R_{c} ∘ C_{1} = R_{a} ≠ R_{b} = C_{1} ∘ R_{c}. These computations show that none of R_{a}, R_{b}, R_{c} are in the centralizer of C_{1}.) In each case, the product of the size of the conjugacy class times the order of the centralizer equals 6 = |S(A)|. (Equivalently, the size of the conjugacy class of x is \frac{6}{|C_{x}|} in each case.)

4. Let H and K be subgroups of a group G.

a. Show that H ∩ K is also subgroup of G.

Solution: Both H and K contain e, so H ∩K ≠ ∅. Next, let x, y ∈ H ∩K, then xy^{−1} ∈ H since H is a subgroup. Similarly xy^{−1} ∈ K since K is a subgroup. Thus xy^{−1} ∈ H ∩ K. It follows that H ∩ K is a subgroup of G by Theorem 3.10.

b. Find an example where H ∪K is a subgroup of G and one where H ∪K is not a subgroup of G.

Solution: Consider the subgroups H = {I_{A}, R_{a}} and K = {I_{A}, R_{b}} of S(A) from question 3f above. H ∪K is not a subgroup of S(A) because H ∪K is not closed under composition R_{a}∘R_{b} = C_{1} ∉ H ∪K. On the other hand if H = K, then H ∪K = H = K is a subgroup.