MATH 103A HW 9 SOLUTIONS (NONAUTHORITATIVE)

Chapter 6 Problems

Problem 6: We wish to find Aut(Z). So suppose φ is an automorphism of Z. Then φ is determinedby φ(1). To see this, suppose n ∈ Z, then since φ is a homomorphism, φ(n) = φ(n · 1) = nφ(1).Now note that since φ is an automorphism, φ(1) must be a generator of Z, thus φ(1) ∈ {1,−1}.There are thus 2 elements of Aut(Z)., namely the identity automorphism, and the automorphismdefined by φ(1) = −1.

Problem 7: We wish to show that S4 is not isomorphic to D12. To see this, note that D12 haselements of order 12, namely rotations by πi/6, but S4 has no elements of order 12.

Problem 30: Let G be a finite abelian group with no elements of order 2. Then the kernel of themap φ : g 7→ g2 is trivial. But an injective function from a finite set to itself is also a surjection.Thus φ is an automorphism.

Alternatively, we can see that φ is surjective as follows: Note that we must have |G| is odd, andin a group of odd order, every element is a square, by a previous homework assignment. To seethat the result is not true if G is an infinite group, consider G = Z. Then the map in question isx 7→ 2x. This is clearly not surjective since all odd numbers are not in the image.

Problem 35: Let a ∈ G have finite order, and let φa be the automorphism of G given by φa(x) =axa−1. Let k = |a| and note that for any x ∈ G, φka(x) = (axa−1)k = akxa−k = x, so φka is theidentity automorphism, and thus |φa|

∣∣k.

Alternatively, we can note that φa is actually an inner automorphism, and we have an isomorphismInn(G) ∼= G/Z(G), then use the fact that the order of aZ(G) divides the order of a.

Problem 40: Consider the group Q under addition, and let φ ∈ Aut(Q). Note first that for anym ∈ Z, we have φ(m) = φ(m · 1) = mφ(1). Also, for any 0 6= n ∈ Z we have φ(1) = φ(n/n) =nφ(1/n), so dividing by n yields φ(1/n) = (1/n)φ(1). Putting this together, for any m/n ∈ Q, wehave φ(m/n) = mφ(1/n) = (m/n)φ(1).

Chapter 10 Problems

Problem 2: That φ is a homomorphism is obvious: φ(xy) = |xy| = |x||y| = φ(x)φ(y).

Problem 6: Let φ : R[x] 7→ R[x] denote the map, where we think of R[x] as a group under addition.Then φ(f + g) =

∫f + g = φ(f) + φ(g). The kernel is {f ∈ R[x] |

∫f = 0} = {0}. Note that the

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2 MATH 103A HW 9 SOLUTIONS (NONAUTHORITATIVE)

map is not a homomorphism if instead we use the antiderivative passing through the point (0, 1),for then φ(0 + 0) = φ(0) = 1, but φ(0) + φ(0) = 2.

Problem 8: That sgn is a homomorphism follows from what we know about the parity of productsof two permutations. The kernel is the set of even permutations.

Problem 9: Let π : G⊕H → G be defined by π(g, h) = g. Then π((g, h)(g′h′)) = π(gg′, hh′) = gg′,and also π((g, h))π((g′h′)) = gg′, so it is a homomorphism. The kernel of π is the set {(e, h) | h ∈H}, which we can write more compactly as {e} ⊕H.

Problem 10: Just recall that the composition of a rotation and a reflection is (some other)reflection, and that the composition of two reflections is a rotation. The kernel is the set of rotations.

Problem 16: If φ is such a surjective homomorphism, then since the groups have the same (finite)order, φ would actually be an isomorphism. But note that Z8⊕Z2 has an element of order 8, whileZ4 ⊕ Z4 has no elements of order 8.

Problem 18: There are no homomorphisms from Z4⊕Z4 onto Z8, because if φ is a homomorphismfrom G to H, then |φ(x)|

∣∣|x| for any x ∈ G. But Z4⊕Z4 has no elements of order 8. We also can’thave any homomorphisms from Z16 onto Z2⊕Z2 since any homomorphic image of a cyclic group iscyclic.

Problem 28: Let G be a group, and φ : G→ Z6⊕Z2 a surjective homomorphism with |Ker(φ)| = 5.First note that this tells us that |G| = 5 · 12 = 60, and that Ker(φ) is a normal subgroup of order5. Since G is a normal subgroup of order 60, we just need to show that G has normal subgroups oforders 10, 15, 20 and 30. Note that we have the following surjective homomorphisms

π1 : Z6 ⊕ Z2 → Z6 given by (m,n) 7→ m

π2 : Z6 ⊕ Z2 → Z2 given by (m,n) 7→ n

π3 : Z6 ⊕ Z2 → Z3 given by (m,n) 7→ 3m

π4 : Z6 ⊕ Z2 → Z2 ⊕ Z2 given by (m,n) 7→ (2m,n)

The composite homomorphisms are surjections from G to groups of orders 6, 2, 3, and 4, respec-tively, so their kernels, which are normal subgroups of G, have orders 10, 30, 20, and 15 respectively.

Alternatively, we can do the following. Recall that the inverse image of a normal subgroup undera homomorphism is again normal. That is, if N is any normal subgroup of Z6 ⊕ Z2, and φ−1(N)denotes the set {g ∈ G | φ(g) ∈ N}, then φ−1(N) is normal in G. Now, since Ker(φ) has order 5, φis a 5-to-1 map. So if N /Z6⊕Z2, then |φ−1(N)| = 5|N |. So we just need to find normal subgroupsof Z6⊕Z2 of orders 2, 3, 4, and 6. This is easy. Since Z6⊕Z2 is abelian, every subgroup is normal.So the subgroups 〈(0, 1)〉, 〈(2, 0)〉, 〈(3, 0)〉⊕〈(0, 1)〉, and 〈(2, 0)〉⊕〈(0, 1)〉 have the prescribed orders.

MATH 103A HW 9 SOLUTIONS (NONAUTHORITATIVE) 3

Problem 38: Let M,N be normal subgroups of G with N ≤ M . We wish to prove that(G/N)/(M/N) ∼= G/M . Consider the map φ from G/N to G/M given by gN 7→ gM . Weclaim that this is a homomorphism. To see this, note that for g, h ∈ G, we have φ((gN)(hN)) =φ(ghN) = ghM and φ(gN)φ(hN) = (gM)(hM) = ghM . Now compute,

Ker(φ) = {gN ∈ G/N | gM = M} = {gN ∈ G/N | g ∈M} = M/N.

The result now follows from the first isomorphism theorem.

Problem 50: H is a subgroup. Note first that e ∈ H. Now, if x, y ∈ H, then α(xy−1) =α(x)α(y−1) = α(x)α(y)−1 = β(x)β(y)−1 = β(xy−1).

Problem 52: Let H,K be normal subgroups of G with H ∩ K = {e}. Consider the map φ :G→ G/H ⊕G/K given by φ(g) = (gH, gK). We claim that φ is a homomorphism. Note that forany x, y ∈ G, φ(xy) = (xyH, xyK) = (xH, xK)(yH, yK) = φ(x)φ(y). What’s the kernel? Well,Ker(φ) = {g ∈ G | (gH, gK) = (H,K)} = H ∩K = {e}, so φ is injective.