Mathematics 243, section 1 – Algebraic Structures Solutions for Exam 3 December 8, 2003

I. A) To show h_{a} is one-to-one, assume that h_{a}(x) = h_{a}(y) for some x, y ∈ G. Then:

xa = ya ⇒ (xa)a^{−1} = (ya)a^{−1} ⇒ x = y

Hence h_{a} is one-to-one. To see that h_{a} is onto, let y ∈ G be a general element and consider x = ya^{−1}. Then we have

h_{a}(x) = xa = (ya)a^{−1} = y

Hence every element of G is in the image of h_{a}, so h_{a} is onto.

B) The result from part A says that in the group operation table for G, the column for a ∈ G contains each element of G exactly once.

II. A) Comment: Too many people were thinking in a very sloppy way in working on this question. The problem does not ask for the generators of Z_{24}, it asks for the generators of the cyclic subgroup H = ⟨[15]⟩ ⊂ Z_{24}.

Best solution: Recall the general fact in the proofs to know for this exam: In a finite cyclic group G = ⟨a⟩ of order n, ⟨a^{k}⟩ = ⟨x^{d}⟩ where d = gcd(n, k). Here Z_{24} is cyclic (generator [1] under addition) n = 24, and gcd(15, 24) = 3. So the same subgroup is generated by any [k] with gcd(k, 24) = 3. This gives

k = [3], [9], [15], or [21].

Also correct, but more “brute force” approach: Compute the elements of

H = ⟨[15]⟩ = {[0], [15], [6], [21], [12], [3], [18], [9]}.

Then check one-by-one to see which of them generate all of H. Only [3], [9], [15], [21] do. For example,

⟨[18]⟩ = {[0], [18], [12], [6]}

only has order 4, so [18] is not a generator of H. But

⟨[3] = {[0], [3], [6], [9], [12], [15], [18], [21]} = H,

so [3] is also one of the generators of H.

B) By direct computation,

σ = (1532)(24368)(12479) = (14795368)(2) = (14795368)

which has order 8, since it is an 8-cycle.

III. A) Since ϕ is a group homomorphism,

ϕ(e_{G}) = ϕ(e_{G} · e_{G}) = ϕ(e_{G}) · ϕ(e_{G})

hence, after multiplying both sides by ϕ(e_{G})^{−1} in H, we have

\begin{aligned}& ϕ(e_{G})^{−1}ϕ(e_{G}) = ϕ(e_{G})^{−1}ϕ(e_{G})ϕ(e_{G}) \\ ⇒ & e_{H} = ϕ(e_{G}) \end{aligned}

B) Recall the definition:

ker(ϕ) = {x ∈ G | ϕ(x) = e_{H}}.

First, ker(ϕ) is nonempty since by part A, e_{G} ∈ ker(ϕ). Next, if x, y ∈ ker(ϕ), then

ϕ(xy) = ϕ(x)ϕ(y) = e_{H}e_{H} = e_{H}

Hence xy ∈ ker(ϕ). Finally, if x ∈ ker(ϕ), then by a general property of homomorphisms that we proved in class,

ϕ(x^{−1}) = (ϕ(x))^{−1} = e_{H}^{−1} = e_{H}

Hence x^{−1} ∈ ker(ϕ). We have shown the three properties needed to see that ker(ϕ) is a subgroup of G.

IV. A) Let A = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) and B = ( \begin{matrix}c & d \\ −d & c \end{matrix} ) be two general matrices in G – it is not

sufficient to consider only the product A^{2}. Then

AB = ( \begin{matrix}a & b \\ −b & a \end{matrix} ) ( \begin{matrix}c & d \\ −d & c \end{matrix} ) = ( \begin{matrix}ac − bd & ad + bc \\ −(ad + bc) & ac − bd \end{matrix} )

Since the product AB has the same form (that is, the diagonal entries are equal and the off-diagonal entries are negatives of each other), the product will be in G as long as we know that at least one of these is nonzero. Note that the det(A) = a^{2} + b^{2} ≠ 0 and det(B) = c^{2} + d^{2} ≠ 0. Hence det(AB) = det(A) det(B) = (a^{2} + b^{2})(c^{2} + d^{2}) = (ac − bd)^{2} + (ad + bc)^{2} ≠ 0. It follows that at least one of ac − bd and ad + bc is nonzero, and G is closed under matrix products.

B) The general formula for matrix inverses for 2 × 2 matrices shows if A = ( \begin{matrix}a & b \\ −b & a \end{matrix} ),

then

A^{−1} = ( \begin{matrix}\frac{a}{a^{2}+b^{2}} & \frac{−b}{a^{2}+b^{2}} \\ \frac{b}{a^{2}+b^{2}} & \frac{a}{a^{2}+b^{2}} \end{matrix} )

Again, the two diagonal entries are the same, and the off-diagonal entries are negatives. So G is closed under matrix inverses.

Note: It follows from the two parts of this problem that G is a subgroup of GL_{2}(R), since G is clearly not empty.

Extra Credit. To do this in any reasonable way, you have to remember the 2 × 2 rotation

matrices we saw on one of the problem sets: Let A = ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ). If θ = 2π/7, then

A will have order 7:

A^{7} = ( \begin{matrix}cos(7θ) & − sin(7θ) \\ sin(7θ) & cos(7θ) \end{matrix} ) = ( \begin{matrix}cos(2π) & − sin(2π) \\ sin(2π) & cos(2π) \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .