Solutions to Assignment 5
#2
1. Addition in vector spaces is abelian, so every subgroup of Rn is normal. By the definition of vector
spaces, V is closed under addition and inverses, so it is a normal subgroup of Rn .
2. Elements of Rn /V are cosets of V of the form V + ~x for ~x ∈ Rn , and the operation on Rn /V is
(V + ~x) + (V + ~y ) = V + ~x + ~y .
Commutativity and associativity follow from the commutativity and associativity of the usual
addition on Rn . The additive identity is V + ~0 = V .
We define scalar multiplication as
k(V + ~x) = V + k~x
This is well-defined, because if V +~x = V +~y , then ~x −~y ∈ V , so k~x −k~y ∈ V , and V +k~x = V +k~y .
Distributivity, multiplicative inverses, and the rest of the properties of a vector space follow from
the same properties for Rn .
3. By the First Isomorphism Theorem, it is enough to find a surjective homomorphism Rn → V ⊥
with kernel V . Let f : Rn → V ⊥ be orthogonal projection to V ⊥ . Then f is a linear map, so it’s
a homomorphism, its kernel is V , and it’s the identity of V ⊥ , so it is surjective.
4. By the First Isomorphism Theorem, fA is isomorphic to Rn /K. By the previous part of the
question, Rn /K ∼
= K ⊥.
5. Theorem 1 Let f : G → H be a homomorphism with kernel K, let b ∈ H, and let g ∈ G be an
element such that f (g) = b. Then the solution set of f (x) = b is the set Kg.
For all k ∈ K, we have f (kg) = f (k)f (g) = f (g), so Kg ⊂ f −1 (b). On the other hand, if
x ∈ f −1 (b), then
f (xg −1 ) = f (x)f (g)−1 = bb−1 = eH ,
so xg −1 ∈ K and x = (xg −1 )g ∈ Kg.
p. 64: #3 As in the hint, consider the map φ : G → G, φ(g) = g n . Since G is abelian, we have
φ(ab) = (ab)n = an bn = φ(a)φ(b),
so φ is a homomorphism.
Suppose that φ(g) = e. Then o(g) is a divisor of n. But for all g ∈ G, o(g) is a divisor of o(G), so
since n is relatively prime to o(G), we must have o(g) = 1, i.e., g = e. Therefore, ker(φ) = {e}, so φ is
one-to-one. Since G is finite, a one-to-one map is also onto, so for every g ∈ G, there’s an h ∈ G such
that φ(h) = hn = g.
#4 Suppose that f : Q → Q and f (1) = q. Then
n × f (1/n) = f (1/n) + · · · + f (1/n) = f (1/n + · · · + 1/n) = f (1) = q,
|
{z
}
n times
so f (1/n) = q/n and
f (m/n) = q/n + · · · + q/n = mq/n.
|
{z
}
m times
That is, f (x) = mx for every x ∈ Q. Therefore, every homomorphism f : Q → Q is multiplication by
q for some q ∈ Q.
#5 Suppose that f : Q → Z2 is a homomorphism and that x ∈ Q. Then f (x) = f (x/2)+f (x/2). Regardless
of whether f (x/2) = 0 or f (x/2) = 1, we have f (x) = f (x/2) + f (x/2) = 0, so f (x) = 0 for all x ∈ Q.
There is thus no surjective homomorphism from Q to the group Z2 .
Suppose that H ⊂ Q is a subgroup of index 2. Then H is normal (because Q is abelian), and Q/H
is the cyclic group of order 2. The canonical homomorphism Q → Q/H ∼
= Z2 is then a surjective
homomorphism to Z2 . This contradicts what we did above, so there is no index-2 subgroup of Q.
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#6 Let f : G → H be a homomorphism and let K be its kernel. By the First Isomorphism Theorem,
f (G) ∼
= G/K, so o(f (G)) = iG (K). By Lagrange’s Theorem, o(f (G)) is a divisor of o(G). But f (G) is
also a subgroup of H, so o(f (G)) is a divisor of o(H). Their only common divisor is 1, so f (G) = {e},
and f (g) = e for all g ∈ G.
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