Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 an Automorphism of a Group G Is an Isomorphism G → G. I. Prove That

Teddy Einstein Math 4320

HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G → G. i. Prove that Aut G is a group under composition.

Proof. Let f, g ∈ Aut G. Then f○g is a bijective homomorphism G → G because compositions of homomorphisms are homomorphisms and compositions of bijections are bijections. Hence f ○ g ∈ Aut G. If idG ∶ G → G is deﬁned by g ↦ g, it is trivially an automorphism. For f ∈ Aut G, f ○ idG = f = idG ○ f. Thus the identity map on G is the identity in Aut G. Recall that if G, H are groups and q ∶ G → H is an isomorphism, then its inverse function −1 f ∶ H → G is an isomorphism as well. Thus if f ∶ G → G is an automorphism, its inverse −1 function f ∶ G → G is also an automorphism. Hence Aut G is closed under inverses. Function composition is associative, so we have associativity. Thus Aut G satisﬁes the axioms of a group.

−1 ii. Prove that γ ∶ G → Aut G deﬁned by g ↦ γg where γg ∶ h ↦ g hg is a homomorphism.

Proof. Let g, k ∈ G. Observe that for h ∈ G:

−1 −1 −1 −1 −1 γ(gh)(k) = (gh) k(gh) = h g kgh = h (g kg)h = γh(γg(k)) = [γ(h) ○ γ(k)](h)

so γh is a homomorphism of groups.

iii. Prove that ker γ = Z(G).

Proof. Observe that

−1 ker γ = {g ∈ G ∶ γg = idG} = {g ∈ G ∶ g xg = x ∀x ∈ G} = {g ∈ G ∶ xg = gx ∀x ∈ G} = Z(G),

yielding the desired result.

iv. Prove Im γ ⊲ Aut G.

Proof. Let g ∈ G so that γg is an arbitrary element of Im γ. We need to prove that given −1 f ∈ Aut G, f ○ γg ○ f ∈ Im γ. Now we see that for k ∈ G:

−1 −1 −1 −1 −1 −1 −1 −1 −1 9 −1 f ○ γg ○ f(k) = f (g f(k)g) = f (g ) ⋅ f ○ f(k) ⋅ f (g) = (f (g)) ck ⋅ f (g) = γf −1(g)(k)

−1 −1 using the fact that f is a homomorphism. Hence f ○ γg ○ f = γf −1(g) ∈ Im γ, so Im γ ⊲ Aut G. 1 2

Problem 2: 2.113 (20 pts) −1 −1 Let G be a group with x, y ∈ G. Deﬁne the commutator [x, y] = x y xy and the commutator ′ −1 −1 1 subgroup of G = ⟨x y xy ∶ x, y ∈ G⟩ ≤ G. ′ i. Prove that G ⊲ G.

Proof. Let x, y ∈ G and h ∈ G. Observe that:

−1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 h x y xyh = h x y hh xyh = (h x hh y h)(h xhh yh) = [h xh, h yh]

−1 −1 −1 −1 because for any k ∈ G, (h kh) = h k h. Hence the conjugate of any commutator is contained in G′. ′ ′ Now let α ∈ G be a product of a commutators αi ∈ G so that:

α = α1α2 . . . αn

−1 ′ Observe that for g ∈ G, g αig ∈ G by the preceding. Now we see that: n −1 −1 −1 −1 −1 −1 −1 −1 g αg = g α1α2 . . . αng = g α1g gα2g gα3g g . . . gg αng = M g αig i=1

−1 ′ −1 −1 ′ where each g αig ∈ G , so g αg is a product of commutators. Thus g αg ∈ G . Hence ′ G ⊲ G.

′ ii. Prove that G~G is abelian.

−1 −1 ′ ′ −1 −1 ′ Proof. Let x, y ∈ G. Observe that the coset x y xyG = G because x y xy ∈ G . Hence ′ ′ ′ ′ ′ ′ ′ xG yG = xyG = yxG = yG xG , so G~G is abelian.

′ iii. If ϕ ∶ G → A is a homomorphism, where A is an abelian group, prove that G ≤ ker ϕ. ′ Conversely, if G ≤ ker ϕ, prove that Im ϕ is abelian.

Proof. Suppose x, y ∈ G and let ϕ ∶ G → A be a homomorphism where A is abelian. Observe −1 −1 −1 −1 that ϕ(x y xy) = ϕ(x) ϕ(y) ϕ(x)ϕ(y). Since A is abelian, we can rearrange to ﬁnd:

−1 −1 −1 −1 ϕ(x y xy) = ϕ(x) ϕ(x)ϕ(y) ϕ(y) = 1A

−1 −1 ′ so x y xy ∈ ker ϕ. Thus every commutator lies in ker ϕ which is a subgroup of G, so G , the smallest subgroup of G containing all commutators of G, must be a subgroup of ker ϕ. ′ Now conversely, suppose G ≤ ker ϕ where ϕ ∶ G → H is a group homomorphism. By the ﬁrst isomorphism theorem, Im ϕ ≅ G~ ker ϕ. Thus it suﬃces to show G~ ker ϕ is abelian. For x, y ∈ G: −1 −1 x y xy ker ϕ = ker ϕ

1Note that the commutator subgroup contains ALL (ﬁnite) PRODUCTS of commutators which may not be commutators themselves. 3

−1 −1 ′ because x y xy ∈ ker ϕ since G ≤ ker ϕ. Hence we have by properties of quotient groups:

xy ker ϕ = yx ker ϕ ⇔ x ker ϕy ker ϕ = y ker ϕx ker ϕ

which shows that G~ ker ϕ is abelian. Hence Im ϕ is abelian.

′ iv. If G ≤ H ≤ G, prove that H ⊲ G.

′ ′ Proof. By proposition 2.123, H ⊲ G if and only if H~G is a normal subgroup of G~G . ′ 2 Observe that G~G is abelian and every subgroup of an abelian group is normal. Thus ′ ′ H~G is normal in G~G which is abelian by the previous part. We then conclude that H is normal in G.

Problem 3: Supplemental Let (G, ∗) and (H, ⊗) be groups and let f ∶ H → Aut G be a homomorphism. Deﬁne a binary operation ⊕ on the set G × H by:

(g1, h1) ⊕ (g2, h2) = (g1 ∗ f(h1)(g2), h1 ⊗ h2)

Prove that (G × H, ⊕) is a group.

′ ′ ′ Proof. Let g, g ∈ G and h, h ∈ H. We see that f(h)(g ) ∈ G because f(h) ∈ Aut G. Thus by closure ′ of G, we have g ∗ f(h)(g ) ∈ G. Similarly, h1 ⊗ h2 ∈ H, so:

′ ′ ′ (g, h) ⊕ (g , h ) = (g ∗ f(h)(g ), h1 ⊗ h2) ∈ G × H so (G × H) is closed under ⊕. Observe that:

(1G, 1H )(g, h) = (1G ∗ f(1H )(g), 1H ⊗ h) = (1G ∗ g, 1H ⊗ h) = (g, h)

(g, h)(1G, 1H ) = (g ∗ f(h)(1G), h ⊗ 1H ) = (g ∗ 1G, h ⊗ 1H ) = (g, h) making (1G, 1H ) the identity in (G × H, ⊕). Note that f(1H ) = idG because f is a homomorphism (which means identity maps to identity), and f(h)(1G) = 1G because f(h) is a homomorphism which means it takes identity to identity as well.

For inverses, if we want (g, h) ⊕ (a, b) = (1G, 1H ), we need (g ∗ f(h)a, h ⊗ b) = (1G, 1H ). Thus we −1 −1 need b = h and f(h)(a) = g . Of course, such an a ∈ G exists because f is an automorphism and is surjective. Thus setting a, b with these parameters, we obtain an inverse for (g, h). ′′ ′′ Finally, we need to check associativity of the operation. Let g ∈ G and h ∈ H. Then check:

′ ′ ′′ ′′ ′ ′ ′′ ′′ (g, h) ⊕ ((g , h ) ⊕ (g , h )) = ((g, h) ⊕ (g , h )) ⊕ (g , h ) this is annoying but straightforward and is hence omitted.

2This is easy, if A is an abelian group, for all a, b ∈ A, a−1ba = b, so every subgroup is ﬁxed under conjugation. 4

Problem 4: Supplemental

Denote the group constructed in the previous exercise by G ×f H. For all n ≥ 3 ﬁnd fn such that

D2n ≅ In ×fn I2.

n 2 −1 Proof. Recall that D2n = ⟨a, bSa = b = 1, ab = ba ⟩. With this presentation, every element of D2n k m 3 can be written in the form a b where 0 ≤ k < n, 0 ≤ b < 2. Naively, we would want (k, m) ∈ In × I2 k m to correspond to a , b where ⋅ denotes the equivalence class of the specified integer ⋅ either mod 2 or mod n as appropriate. k m k′ m′ k+k′ m+m′ When multiplying together a b a b , if m = 0, then we get a b . Otherwise if m = 1, we k −k′ m′ k−k′ m+m′ get a a bb = a b . In other words, if we want to realize D2n in the desired form, we will want to apply some kind of inversion when the second coordinate of the first term is 1. Specifically, suppose we have an f ∶ I2 → Aut In giving the desired transformation and let x, y, z, w ∈ Z. We want (x, y)(z, w) = (x+f(y)(z), y+w). In order for the corespondence we want, we need x+f(y)(z) = x + z ′ 4 if y = 0 and x − z if y = 1. Hence define f ∶ Z → Aut In by:

′ t f (t) = h ↦ (−1) h

′ ′ thus f (t) is either the identity or inversion, so f (t) is always an automorphism of In. Check ′ ′ t+t′ t t′ ′ ′ ′ ′ f (t + t ) = (h ↦ (−1) h) = (h ↦ (−1) (−1) h) = (f (t) ○ f (t )). Hence f is a homomorphism ′ whose kernel is precisely 2Z, then even numbers. Since I2 = Z~2Z, f (t) induces a well defined homomorphism f ∶ I2 → Aut In such that f(0) is identity and f(1) is inversion. k m Now set our group G = In ×f I2. Define a map ϕ ∶ D2n → G defined by ϕ ∶ a b ↦ (k, m). We need ′ ′ ′ ′ to check that ϕ is a homomorphism. Let k, m, k , m be integers with 0 ≤ k, k < n and m, m ∈ {0, 1}. Suppose first that m is even:

k m k′ m′ ϕ(a b )ϕ(a b ) = (k, 0)(k′, m′) = (k + f(0)(k′), 0 + m′) = (k + k′, m + m′)

k+k′ m+m′ = (k + k′, m + m′) = ϕ(a b ) On the other hand if m is odd:

k m k′ m′ ϕ(a b )ϕ(a b ) = (k, 1)(k′, m′) = (k + f(1)(k′), 1 + m′) = (k − k′, 1 + m′)

k−k′ 1+m′ k 1 k′ m′ k m k′ m′ = ϕ(a b ) = ϕ(a b a b ) = ϕ(a b a b ) Thus ϕ is a homomorphism. m k m k If ϕ(a b ) = (0, 0), then with 0 ≤ k < n, 0 ≤ m < 2, then m = 0, k = 0, so a b = e, so ϕ is injective. m k Trivially, (m, k) = ϕ(a b ), so ϕ is surjective. Thus ϕ is an isomorphism of groups.

3In order to see this, observe that whenever a b appears to the left of a power of a, the relation ab = ba−1 allows it to be moved to the right of any power of a. 4 Note that we are writing In as an additive group here with operation +. 5

Problem 5: 2.115

Prove that the translation τa ∈ SG deﬁned by τa ∶ a ↦ ag is a regular permutation (Assume G is ﬁnite).

Proof. We need to show that τa is a product of disjoint cycles of the same length with no ﬁxed points or is the identity when SG is identiﬁed with SSGS. Observe that the elements can be reached ○n n from g by applying τa repeatedly are exactly those of the form (τa) (g) = a g for n ∈ N where in SaS 2 SaS−1 practice, a = 1, so this consists exactly of the elements g, ag, a g, . . . , a g. Hence in the disjoint 2 3 SaS−1 cycle decomposition of τa, g is in the cycle (a, a g, a g, . . . , a g). Note that this cycle coincides exactly with ⟨a⟩g, a right coset of ⟨g⟩ in G. Since a was arbitrary, any element in G appears in the disjoint cycle decomposition of τa in a cycle which is some permutation of the elements of ⟨a⟩g. Hence the cycle length for g is exactly S⟨a⟩gS. The proof to Lagrange’s theorem shows that all cosets of ⟨a⟩ have the same order, namely S⟨a⟩S = SaS. Thus we see that τa is a product of disjoint cycles of equal length.

If τa has a ﬁxed point, then ag = g for some g ∈ G, but then a = 1G. In that case, τa is the identity map on G, so τa has a ﬁxed point only if it is the identity. Hence τa is a regular permutation.

Problem 6: 2.126

Let σ, τ ∈ S5 with σ a 5 cycle and τ a transposition. Show that ⟨σ, τ⟩ = S5.

Proof. Without loss of generality, we may relabel so that τ = (12). Since σ is a 5-cycle, then there k k exists 0 < k < 5 such that σ (1) = 2. Since 5 is prime and 0 < k < 5 so that (5, k) = 1, σ is also a 5 k ` cycle. Thus (σ ) (2) ≠ 1, 2 if ` < 3. Since we only needed to fix the labelling of 1, 2, we can freely change the labelling of 3, 4, 5 and still have a group isomorphic to S5 with τ = (12). Thus we can k k relabel so that without loss of generality σ = (12345) because the first two letters of the 5 cycle σ are fixed as 1, 2, so the last 3 must be among 3, 4, 5. n −n Now observe that the element (12345) (12)(12345) = (1 + n, 2 + n) for 0 ≤ n ≤ 3 and is in ⟨σ, τ⟩ (You should actually show this calculation). Thus ⟨σ, τ⟩ contains all of the simple transpositions in S5. In a previous homework exercise, we showed that the simple transpositions in Sn generate Sn, so ⟨σ, τ⟩ must contain S5, so we conclude they are equal.