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 determined
by 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 automorphism
defined by φ(1) = −1.
Problem 7: We wish to show that S4 is not isomorphic to D12 . To see this, note that D12 has
elements 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 the
map φ : g 7→ g 2 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, and
in a group of odd order, every element is a square, by a previous homework assignment. To see
that the result is not true if G is an infinite group, consider G = Z. Then the map in question is
x 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 = ak xa−k = x, so φka is the
identity automorphism, and thus |φa |k.
Alternatively, we can note that φa is actually an inner automorphism, and we have an isomorphism
Inn(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 any
m ∈ 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, we
have 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 φR : R[x] 7→ R[x] denote the map, where we think of RR[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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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 products
of 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 0 h0 )) = π(gg 0 , hh0 ) = gg 0 ,
and also π((g, h))π((g 0 h0 )) = gg 0 , 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, while
Z4 ⊕ Z4 has no elements of order 8.
Problem 18: There are no
homomorphisms from Z4 ⊕ Z4 onto Z8 , because if φ is a homomorphism
from G to H, then |φ(x)||x| for any x ∈ G. But Z4 ⊕ Z4 has no elements of order 8. We also can’t
have any homomorphisms from Z16 onto Z2 ⊕ Z2 since any homomorphic image of a cyclic group is
cyclic.
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 order
5. Since G is a normal subgroup of order 60, we just need to show that G has normal subgroups of
orders 10, 15, 20 and 30. Note that we have the following surjective homomorphisms
π1
π2
π3
π4
:
:
:
:
Z6 ⊕ Z2
Z6 ⊕ Z2
Z6 ⊕ Z2
Z6 ⊕ Z2
→ Z6 given by (m, n) 7→ m
→ Z2 given by (m, n) 7→ n
→ Z3 given by (m, n) 7→ 3m
→ 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, respectively, 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 under
a 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, φ
is a 5-to-1 map. So if N / Z6 ⊕ Z2 , then |φ−1 (N )| = 5|N |. So we just need to find normal subgroups
of Z6 ⊕ Z2 of orders 2, 3, 4, and 6. This is easy. Since Z6 ⊕ Z2 is abelian, every subgroup is normal.
So the subgroups h(0, 1)i, h(2, 0)i, h(3, 0)i ⊕ h(0, 1)i, and h(2, 0)i ⊕ h(0, 1)i have the prescribed orders.
MATH 103A HW 9 SOLUTIONS (NONAUTHORITATIVE)
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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 . We
claim 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 for
any 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.