Download HOMEWORK 8: GRADER`S NOTES AND SELECTED SOLUTIONS

HOMEWORK 8: GRADER’S NOTES AND SELECTED SOLUTIONS
Grader’s Notes:
● In general when you are trying to show that two groups G and H are not isomorphic, it’s never enough
to pick one particular map and show it’s not an isomorphism. For example, in showing that the rational
numbers Q under addition are not isomorphic to any proper sub-group you can’t just show that the
map φ(x) = x2 is not an isomorphism because there could, conceivably be some other map that is an
isomorphism.
● The proper way to show that G ≅/ H in general is usually either:
(1) Find some property that you can prove / have proved is preserved under isomorphism and show
one of G, H have this property while the other group does not.
(2) Assume that φ ∶ G → H is an isomorphism (and assume nothing more about φ) and somehow arrive
at a contradiction.
● The properties “one-to-one”, “onto”, and “operation preserving” are properties that functions have, it
is impossible for groups to have these properties, when groups G, H are isomorphic the property that
holds of the groups is that there exists a function f ∶ G → H such that f , the function is one-to-one,
onto, and order preserving.
Chapter 6, page 135, no 28
Prove the quaternion group is not isomorphic to the dihedral group D4 .
Example Solution:
Use Theorem 6.2.7 and note that the Quaternions have 1 element of order 2 while D4 has 5 elements of order
2. There are variants on this method of proof all related to the fact that if φ is an isomorphism then ∣x∣ = ∣φ(x)∣.
◻
Grader’s Notes:
● Note that when two groups G, H are given by Cayley tables, an isomorphism φ ∶ G → H need not take
the n × m-th entry of the Cayley table for G to the n × m-th entry of the Cayley table for H.
Chapter 6, page 136, no. 38
Let G = {0, ±2, ±4, ±6, . . .}, and H{0. ± 3, ±6, ±9, . . .}. Show that G and H are isomorphic groups under
addition. Does your isomorphism preserve multiplication? Generalize to the case where G = ⟨m⟩ and H = ⟨n⟩.
Grader’s Notes:
● Just becuase G, H are not groups under multiplication, this doesn’t neccessarily mean that a map
φ ∶ G → H cannot satisfy φ(ab) = φ(a)φ(b).
● There was a very common miscalculation of φ(ab) for the map φ(ab) used in the example solution.
Example Solution:
Let φ ∶ G → H by φ(2n) = 3n. We claim that φ is an isomorphism, it should be clear that (φ) = G and
(φ) = H.
We check that φ is one-to-one: If φ(2n) = 3n = φ(2m) = 3m then it must be the case that n = m, so φ is
one-to-one.
We check that φ is onto: Given 3n ∈ H, then φ(2n) = 3n and 2n ∈ G, so φ is onto.
And lastly, we check that φ preserves the group operation:
φ(2n + 2m) = φ(2(n + m)) = 2(n + m) = 2n + 2m = φ(2n) + φ(2m)
1
2
HOMEWORK 8: GRADER’S NOTES AND SELECTED SOLUTIONS
so φ preserves that group operation.
φ does not, however, preserve multiplation:
φ(2n2m) = φ(2(2nm)) = 3(2nm) = 6nm ≠ 9nm = 3nm(3nm) = φ(2nm)φ(2nm)
1
In general you can show that φ(kn) = km is a group isomorphism from ⟨n⟩ to ⟨m⟩, that does not preserve
multiplication when n ≠ m, this proof is essentially the same as the proof above and I’m not going to write out
the details. If this seems non-obvious you should try and write out a detailed proof.
◻
An alternate, but similar, and perhaps more elegant proof would be to show that Z ≅ ⟨n⟩ for any non-zero
n ∈ Z, by the map φ(k) = nk, then by transitivity of isomorphism (problem (6)) ⟨n⟩ ≅ ⟨m⟩.
Chapter 6, page 136 no. 43
Prove that Q+ , the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup.
Grader’s Notes:
● In general, any function f ∶ X → Y is always maps onto it’s image.
● If φ ∶ G → (G) is an isomorphism, that is φ is one-to-one, onto (by the remark above φ is neccesarily
onto), and operation preserving, then (G) must be a group.
Proof. Let H = (G), let e ∈ G be the identity for G, note that since φ is onto, any h ∈ H has the
form h = φ(g) for some g ∈ G, so we may always write elements of H in the form φ(g) where g ∈ G.
φ(e) is an identity for H, becuase φ(e)φ(g) = φ(eg) = φ(g) = φ(ge) = φ(e)φ(g). H is closed under the
operation since if φ(g), φ(h) ∈ H the φ(g)φ(h) = φ(gh) ∈ H. H has inverses because given φ(g) ∈ H,
φ(g)φ(g −1 ) = φ(gg −1 ) = φ(e) = φ(g −1 g)φ(g)φ(g −1 ) so φ(g −1 ) is an inverse for φ(g).
● If you think about it, the proof of the above gives you way to give a group structure to to any set Y
when you have any bijection f ∶ G → Y , where G is a group, by defining x ⋅ y = f (f −1 (x)f −1 (y)).
Example Solution:
Let φ(x) = x2 (actually we could take φ(x) = xn for natural number n ≥ 2), note that φ(x) is onto it’s image,
furthermore φ(x) is one to one since φ(x) = x2 = φ(y) = y 2 implise that x = y since both x and y are positive
rational numbers. Note that φ is operation preserving, φ(xy) = (xy)2 = x2 y 2 = φ(x)φ(y).
Note that φ is proper becuase numbers such as 2 with no rational square root are not in (φ). So φ is an
isomorphism from Q to a proper subgroup of Q.
◻
Chapter 6, page 136, no 44
Show that
Q, the group of rational numbers under addition, is not isomorphic to a proper subgroup of itself.
Grader’s Notes:
● I originally was going to grade this problem but I decided not to becuase there is a subtle logic error in
what I believe is the intended solution, and I didn’t want to nitpick about it.
● below I present what I believe is the “intended” solution, assuming (42) (Any automorphism φ of Q has
the form φ(x) = xφ(1)).
Example Solution:
Assume for contradiction that φ ∶ Q → H is an isomorphism of Q to a proper subgroup H ≤ Q, then by (42),
φ(x) = xφ(1) for all x ∈ Q, but clearly φ(1) ≠ 0, since φ is one-to-one and φ(0) = 0 must hold for φ to be
an isomorphism. So given any y ∈ Q ∖ H (note that since 0 is the identity for Q, 0 ∈ H, so y ≠ 0), note that
y
y
y = φ(1)
y = φ(1) φ(1)
= φ( φ(1)
), so we’ve shown that y ∈ H, but this is a contradiction, so it must be that case
φ(1)
that Q is not isomorphic to any of it’s proper subgroups.
◻
1many students incorrectly caluclated φ(2n2m) = 3nm.
HOMEWORK 8: GRADER’S NOTES AND SELECTED SOLUTIONS
3
Did you spot the error? The problem is that φ is not an automorphism of Q; it’s an isomorphism of Q onto
a proper subgroup, so it doesn’t follow from the statement of (42) that φ must have the form φ(x) = φ(1)x. In
fact this problem can be solved but you have to use this slightly strong proposition:
Proposition 0.1. If φ ∶ Q → Q is any operation preserving map (for all x, y ∈ Qφ(x + y) = φ(x) + φ(y) then φ has
the form φ(x) = φ(1)x.
Note that the proposition is stronger than (42) because it applies to maps that need not be automorphisms
of Q, like for instance the constantly 0 map: φ(x) = 0.
Proof. Let φ ∶ Q → Q be operation preserving. We need to show that φ(x) = φ(1)x for any x ∈ Q, we may assume
n
n
1
x= m
. Note that 1 = n( m
) = ∑ni=1 m
so:
1
) = ∑m
Now note that 1 = m( m
i=1
φ(
n
n
1
1
1
) = φ (∑ i = 1n ) = ∑ φ ( ) = n ⋅ φ ( )
m
m
m
m
i=1
1
m
so:
m
m
1
1
1
) = ∑ φ ( ) = m (φ ( ))
m
m
m
i=1
i=1
φ(1) = φ (∑
1
From which it follows that φ( m
)=
φ(1)
,
m
putting theses two derived identities together we get:
φ(
n
n
) = φ (1)
m
m