Download Math 325 - Dr. Miller - HW #4: Definition of Group

Math 325 - Dr. Miller - HW #4: Definition of Group - SOLUTIONS, 9/21/12
1. Once we start using the “familiar operation” argument in proofs, students mistakenly
expect that new operations are commutative simply because they are defined using only
known commutative operations such as ordinary addition and multiplication. Prove
that this expectation is WRONG by creating examples of TWO operations (on R) that
are NOT commutative but are defined using only ordinary addition and multiplication.
Prove that each of your examples indeed is not commutative.
The moral of the story is that any NEW operation needs a formal proof or disproof of
commutativity.
Example 1 - On Z, define a ∗ b = 2a + b. The formula only multiplies by 2 and adds,
but it’s NOT commutative, for 1, 2 ∈ Z and 1 ∗ 2 = 4 while 2 ∗ 1 = 5.
Example 2 - On Z, define a ∗ b = (a + 1) ∗ b. The formula uses only addition and
multiplication, but it’s NOT commutative: again, 1, 2 ∈ Z and 1 ∗ 2 = 4, which doesn’t
equal 2 ∗ 1 = 3.
2. The warning above also applies to associativity: a new operation isn’t necessarily associative just because it’s defined using only known associative operations. Create
examples of TWO operations (on R) that are NOT associative but are defined using
only ordinary addition and multiplication. Prove that each of your examples indeed is
not associative. (You may reuse your examples from Problem #1 if you like.)
Here, the point is that any NEW operation also always needs a formal proof or disproof
of associativity.
I want to try my same examples again:
Example 1 - On Z, define a ∗ b = 2a + b and consider 1, −2, 0 ∈ Z. We have
(1 ∗ −2) ∗ 0 = 0 ∗ 0 = 0 but 1 ∗ (−2 ∗ 0) = 1 ∗ −4 = −2, so ∗ is NOT associative.
Example 2 - On Z, define a ∗ b = (a + 1) ∗ b and consider −1, 0, 1 ∈ Z. Then
(−1 ∗ 0) ∗ 1 = 0 ∗ 1 = 1, which is not equal to −1 ∗ (0 ∗ 1) = −1 ∗ 0 = 0, proving
that ∗ is NOT associative.
3. Prove that each example below forms an abelian group.
(a) R+ , a ∗ b = 3ab
We already know that the product of positive reals is a positive real, so R+ is closed
under ∗. Because ordinary multiplication is commutative, a ∗ b = 3ab = 3ba = b ∗ a
for all a, b ∈ R+ , so ∗ is commutative. Associativity should be checked fully
also: given any a, b, c ∈ R+ , (a ∗ b) ∗ c = 3ab ∗ c = 9abc, which does equal
a ∗ (b ∗ c) = a ∗ 3bc = 9abc, so ∗ is associative. The identity is e = 1/3 because for all a ∈ R+ , (1/3) ∗ a = a = a ∗ (1/3) and 1/3 ∈ R+ , as we require.
Finally, given any a ∈ R+ , we claim that its inverse is 1/(9a), which is a positive
1
= (3a)/(9a) = 1/3, the identity, which also
real number since a is too, and a ∗ 9a
1
equals 9a
∗ a due to commutativity.
(b) Z, a ∗ b = a + b + 4
The integers are closed under addition, so for all a, b ∈ Z, a + b + 4 ∈ Z also.
The operation ∗ is commutative because due to that of ordinary addition: a ∗ b =
a + b + 4 = b + a + 4 = b ∗ a for all a, b ∈ Z. Given any a, b, c ∈ Z, we have
(a ∗ b) ∗ c = (a + b + 4) ∗ c = a + b + c + 8 = a ∗ (b + c + 4) = a ∗ (b ∗ c),
proving that ∗ is associative as well. Exploration yields an identity of e = −4, for
(−4) ∗ a = a = a ∗ (−4) for every a ∈ Z, and −4 ∈ Z itself, as needed. Lastly, for
each a ∈ Z, consider −8 − a, which is an integer because Z is closed under subtraction, and satisfies a ∗ (−8− a) = −4 = (−8− a) ∗ a, making −8− a the inverse of a.
(c) πZ, ordinary addition
(By definition, for any a ∈ R, the set aZ = {an | n ∈ Z}.)
Let πn, πm ∈ πZ; then πn+πm = π(n+m), which belongs to πZ because n+m ∈ Z
(it is a sum of integers). Ordinary addition is well-known to be both commutative
and associative on the set of real numbers, of which πZ is a subset. The additive
identity 0 can be expressed as the product π · 0 and is therefore an element of πZ.
Finally, for any πn ∈ πZ, the additive inverse is known to be −πn = π(−n), which
belongs to πZ because −n ∈ Z whenever n ∈ Z.
(d) F = { πa | a ∈ R+ }, ordinary multiplication
For every πa , πb ∈ F , where a, b ∈ R+ , the product πa πb can be rewritten as ab/π
,
π
which belongs to F since ab/π is the product and quotient of positive real numbers
and therefore a positive real itself. Ordinary multiplication is well-known to be both
commutative and associative on R, of which F is a subset. (Actually, F = R+ ,
which you could have proved and used alternatively for this problem.) We know that
the multiplicative identity is 1, which can be expressed as π/π to demonstrate its
membership in F . Finally, given any element πa ∈ F , its well-known multiplicative
2
inverse π/a can be rewritten as (π π/a) , which is now of the correct form to belong
to F because the expression π 2/a is a product and quotient of positive reals and
therefore also such a number itself.
4. (a) Complete the following Cayley table in such a way that ∗ is commutative and has
an identity while each element has an inverse. Explain your reasoning.
∗
w
w
y
x
z
x
y
z
x
w
y
z
w
The identity is unique and is the only member of this group that can leave EVERY
member unchanged. Since w, x, and z all “change” some other member, y must be
the identity. Next, in w’s row/column, every group member must appear exactly
once: x is currently missing, so it must equal z ∗ w = w ∗ z by commutativity.
Likewise, in the z row/column, y is not yet represented, so it equals z ∗ x = x ∗ z.
Finally, w ∗ x must be z since that is the only element not appearing in the top row,
or because by commutativity, it must equal x ∗ w.
∗
w
x
y
z
w
y
z
w
x
x
z
w
x
y
y
w
x
y
z
z
x
y
z
w
(b) Make all possible Cayley tables for an operation 4 on G = {a, b, c} such that G
could be a group with b4b = a. Are any of your options abelian? (You do not
need to check associativity, but if you did, how many cases would that require?)
We know that a equals b4b; let’s explore the possibilities for a4b. First, a cannot
equal a4b because then it would appear twice in b’s column. If we try b = a4b,
then a is the identity, and our table begins as follows:
4
a
b
c
a
a
b
c
b
b
a
c
c
But now b4c must equal c to complete the row and yet cannot be c to complete the
column. Therefore, this option is not viable.
Therefore, we must have c = a4b. It follows that neither a nor b is the identity,
making c serve as that element.
4
a
b
a
a
b
c
c
a
a
b
b
c
Now both a4b and b4a must equal c to complete their rows/columns correctly,
leaving a2 = b as the only correct final entry:
4
a
b
c
a
b
c
a
b
c
a
b
c
a
b
c
Conclusion: The previous table is the only option, and it is indeed abelian.
(In general, there would be 3 × 3 × 3 = 27 cases to check for associativity, but
because this potential group is abelian, many cases will be equivalent due to ability
to rearrange the elements.)
5. We proved in class that if G is a group, its Cayley table always has each element appearing exactly once in each row and each column. Prove that the converse is NOT true, by
demonstrating a Cayley table having this “exactly once” quality but not corresponding
to a group. Explain how you know your table is NOT that of a group.
Let G = {x, y, z} with the Cayley table below:
| x y z
x | x z y
y | z y x
z | y x z
This Cayley table cannot represent a group because it has no identity. There is no
SINGLE element that leaves ALL elements unchanged simultaneously.