Solution Set 8: Part A

Benjamin Bakker Mathematics 25a Prof. Berger

November 14, 2003

A.1. In each of the following exercises, you are given a set G and a composition law (x, y) ↦ x⊗y. Determine whether (G, ⊗) is a group, and if it is, find the unit element, give a formula for the inverse of an element, and determine whether G is abelian.

(1) G = ℝ and x ⊗ y = x + y + xy.

Solution. If G were a group, then 0 must be the identity, since 0⊗x = x⊗0 = x+0+0·x = x. But −1 cannot have an inverse since

(−1) ⊗ x = −1 + x + (−1)x = −1 ≠ 0

Thus, G is not a group. □

(2) G = (ℝ \ {0}) × ℝ and (x, u) ⊗ (y, v) = (xy, xv + u).

Solution. G is a group. Multiplication is associative since

(x, u) ⊗ [(y, v) ⊗ (z, w)] = (x, u) ⊗ (yz, yw + v) = (xyz, xyw + xv + u)
[(x, u) ⊗ (y, v)] ⊗ (z, w) = (xy, xv + u) ⊗ (z, w) = (xyz, xyw + xv + u)

The unit is (1, 0), for

(1, 0) ⊗ (x, u) = (x, u + 0) = (x, u) = (x, x · 0 + u) = (x, u) ⊗ (1, 0)

For a given (x, y), the inverse is (x, y)^{−1} = (1/x, −y/x) (x ≠ 0 ), since

(x, y) ⊗ (1/x, −y/x) = (1, −y + y) = (1, 0) = (1, y/x − y/x) = (1/x, −y/x) ⊗ (x, y)

The group is nonabelian since

(1, 2) ⊗ (2, 1) = (2, 3) ≠ (2, 5) = (2, 1) ⊗ (1, 2)

A much easier way to show G is a group would have been to identify

(x, u) ↔ (\begin{matrix}x & u \\ 0 & 1 \end{matrix})

and then noting that ⊗ is just matrix multiplication, that the identity has the above form, and that the inverse of any matrix of the above form has the above form. □

(3) G = {(x, y) ∈ ℝ^{2}|x^{2} + y^{2} = 1} and (x_{1}, y_{1}) ⊗ (x_{2}, y_{2}) = (x_{1}x_{2} − y_{1}y_{2}, x_{1}y_{2} + x_{2}y_{1}).

Solution. The multiplication here is identical to the multiplication law of complex numbers (x, y) = x + yi, and G is just the unit circle in the complex plane, S^{1} = {z ∈ ℂ||z| = 1}.

I did the last one out in gory detail, so here I’ll just use the trick that I mentioned at the end of (2). If we identify

(x, y) ↔ (\begin{matrix}x & −y \\ y & x \end{matrix})  (1)

then ⊗ is matrix multiplication, since

(\begin{matrix}x_{1} & −y_{1} \\ y_{1} & x_{1} \end{matrix}) (\begin{matrix}x_{2} & −y_{2} \\ y_{2} & x_{2} \end{matrix}) = (\begin{matrix}x_{1}x_{2} − y_{1}y_{2} & −x_{1}y_{2} − x_{2}y_{1} \\ x_{1}y_{2} + x_{2}y_{1} & x_{1}x_{2} − y_{1}y_{2} \end{matrix})  (2)

We can then consider G as the set of all matrices of the form (1) with determinant 1 (this is the condition that x^{2} + y^{2} = 1). G is closed under multiplication, for by (2) the product of the matrices corresponding to (x_{1}, y_{1}), (x_{2}, y_{2}) has the form of (1), and the determinant is the product of the determinants of the factors—namely, 1. The identity matrix is in G, corresponding to (1, 0), and every matrix (x, y) has determinant 1 and therefore is invertible; explicitly, the inverse is

(\begin{matrix}x & −y \\ y & x \end{matrix})^{−1} = (\begin{matrix}x & y \\ −y & x \end{matrix}) ⇒ (x, y)^{−1} = (x, −y)

which is clearly in G. G is in fact abelian, since

(\begin{matrix}x_{2} & −y_{2} \\ y_{2} & x_{2} \end{matrix}) (\begin{matrix}x_{1} & −y_{1} \\ y_{1} & x_{1} \end{matrix}) = (\begin{matrix}x_{1}x_{2} − y_{1}y_{2} & −x_{1}y_{2} − x_{2}y_{1} \\ x_{1}y_{2} + x_{2}y_{1} & x_{1}x_{2} − y_{1}y_{2} \end{matrix})

A.2. Let G be a finite group and let a, b ∈ G be such that (ab)^{m} = e for some m ≥ 1. Prove that (ba)^{m} = e.

Solution.

e = beb^{−1} = b(ab)^{m}b^{−1} = b [\underbrace{(ab)(ab) ⋯ (ab)(ab)}_{m times}] b^{−1} = (ba)^{m}

A.3. Let G be a group such that for every a ∈ G, we have a^{2} = e. Prove that G is abelian.

Solution. Note first that in an arbitrary group G, if an element x ∈ G has two inverse y, y′, then because inverses are required to be both-sided,

y = ye = y(xy′) = (yx)y′ = ey′ = y′

and therefore the inverse is unique.

For any a, b ∈ G we have (ab)(ba) = ab^{2}a = a^{2} = e, so (ab)^{−1} = ba. But by hypothesis, any element is its own inverse, so ab = ba.

A.4.

(1) Find all groups G which have no non-trivial automorphisms.

Solution. Conjugation by an element x ∈ G, f : a ↦ xax^{−1}, is an automorphism since: f(a)f(b) = (xax^{−1})(xbx^{−1}) = xabx^{−1} = f(ab); a ∈ ker f ⇒ xax^{−1} = e ⇒ a = e; for any a ∈ G, f(x^{−1}ax) = a. For any x ∈ G, conjugation by x is trivial, so for all a ∈ G, xax^{−1} = a ⇒ xa = ax, and G must be abelian. We showed in lecture that for any abelian group, x ↦ x^{−1} is an automorphism; if it is trivial, then for every x ∈ G, x = x^{−1} ⇒ x^{2} = e.

Suppose there are two non-unit elements a, b ∈ G. Let H = (e, a, b, ab), and pick representatives x_{i} of the nontrivial cosets. Let K be the subgroup generated by the x_{i}. Note that if x_{i} ≠ x_{j} then we can’t have x_{i}x_{j} ∈ H, or else x_{i}H = x_{j}H. Since x_{i}^{2} = e, it follows that H ∩ K = (e). The map H × K → G, (h, k) ↦ hk is an isomorphism; it is surjective by construction, and injective by the above (H ∩ K = (e), so if hk = e ⇒ h^{−1} = k ∈ H ∩ K). The map f : H → H, that switches a, b is a nontrivial automorphism as can be readily checked by computation, so we have an induced nontrivial automorphism g : H ×K → H ×K, g(h, k) = (f(h), k). Therefore, there cannot be two non-unit elements in G.

Only (e) and ℤ/2ℤ can have no nontrivial automorphisms. Since an automorphism must fix e, both of these groups do, in fact, have no nontrivial automorphisms.

(2) Find all groups G which have no non-trivial subgroups.

Solution. Certainly, the identity group (e) has no nontrivial subgroups. If G ≠ (e), then there is some x ∈ G not equal to e. The subgroup generated by x (i.e., the set of powers of x, negative, positive, and zero) cannot be (e) and therefore must be the whole group, so G is cyclic—ℤ/nℤ ≅ G for some n ≠ 1 (the isomorphism is given by the map ¯1 ↦ x). ℤ has nontrivial subgroup 2ℤ, so n ≠ 0. If n is not prime, then there is some p ≠ 1, n dividing n, in which case the subgroup of ℤ/nℤ generated by p is nontrivial. If n is prime, however, then for any nonzero x ∈ ℤ/nℤ, the subgroup generated by x must have order d greater than 1 and dividing n, and is therefore the whole group. Thus, the only groups (up to isomorphism) with no nontrivial subgroups are (e) and ℤ/pℤ for p prime.

A.5. Find all continuous homomorphisms f : (ℝ, +) → (ℝ, +).

Solution. For any homomorphism f : G_{1} → G_{2}, and any x ∈ G_{1}, f(x) = f(e_{1}x) = f(e_{1})f(x) ⇒ f(e_{1}) = e_{2}. Thus, for homomorphisms of ℝ into ℝ, f(0) = 0. f = 0 is one possible continuous homomorphism.

Suppose f(a) ≠ 0 for some a ∈ ℝ (note that a ≠ 0). Then for any n ∈ ℤ,

f(na) = f(\underbrace{a + ⋯ + a}_{n times}) = nf(a)

Similarly, for any nonzero m ∈ ℤ, f(a) = f(m(a/m)) = mf(a/m), so f(a/m) = f(a)/m. More generally, for any n/m ∈ ℚ, f(na/m) = nf(a)/m.

We claim that f(x) = f(a)x/a. By the above, this is true for the rational multiples aℚ of a. aℚ is dense in ℝ, so for every x ∈ ℝ, there is a sequence x_{n} ∈ aℚ converging to x. f is continuous, so

f(x) = lim f(x_{n}) = lim f(a)x_{n}/a = \frac{f(a)}{a} lim x_{n} = f(a)x/a

as claimed. In particular, if f(1) = f(a)/a = b, then f(x) = f(a)x/a = bx, and thus f is simply multiplication by f(1). This clearly is, in fact, a continuous homomorphism, so the only continuous homomorphism are multiplication by a real number (including 0).