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Contemporary Abstract Algebra (6th ed
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4 Find integers s and t such that 1 = 7 · s + 11 · t
...

We can choose s and t to be −3 and 2, respectively
...
Thus, we can show that
s = −3 and t = 2 are two integers that we can find
...

0
...
If a = da and b = db , show that gcd(a , b ) = 1
...
2, gcd(a, b) = as + bt where s and t are integers, but gcd(a, b) = d
as well, thus d = as + bt
...
Since we know that gcd(a, b) = d
and d = 0, it follows that 1 = a s + b t = gcd(a , b )
...
11 Let n be a fixed positive integer greater than 1
...

If a mod n = a and b mod n = b , then by definition, a = a +rn and b = b +sn
for some r, s ∈ Z
...
Also,
ab = (a + rn)(b + sn) = a b + sna + rnb + rsn2 =
a b + n(sa + rb + rsn)
Since n divides n(sa + rb + rsn) evenly, we have (ab) mod n = (a b ) mod n
...
13 Let n and a be positive integers and let d =gcd(a, n)
...

1

Firstly suppose that ax mod n = 1 has a solution
...
Also, ax − kn = 1, which also happens to be the form of gcd(a, n)
by theorem 0
...
Thus gcd(a, n) = d and gcd(a, n) = 1, then d = 1 as required
...
2, there exists x, t such that
1 = ax + nt
...

0
...

We can show using the Euclidean algorithm:
7n + 4
5n + 3
2n + 1
n+1
n

=
=
=
=
=

(5n + 3) · 1 + (2n + 1)
(2n + 1) · 2 + (n + 1)
(n + 1) · 1 + n
n·1+1
1·n

Since 1 is the last nonzero remainder, gcd(7n + 4, 5n + 3) = 1, which shows that
they are relatively prime
...
16 Use the Euclidean algorithm to find gcd(34, 126) and write it as a linear combination
of 34 and 126
...

Now to express it as a linear combination:
2 = 10 + 4(−2) = 10 + [24 + 10(−2)](−2) = 24(−2) + 10(5)
= 24(−2) + [34 + 24(−1)](5) = 24(−7) + 34(5)
= [126 + 34(−3)](−7) + 34(5) = 126(−7) + 34(26)
Hence, we can express 2 as 126(−7) + 34(26)
...
26 What is the largest bet that cannot be made with chips worth 7 dollars and 9 dollars?
Verify that your answer is correct with both forms of induction
...
It appears that the answer is 47
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This will solve the problem, since a represents
the number of 7 dollar bets and b the number of 9 dollar bets needed to make a
bet of a · 7 + b · 9
...
Let S be the set of all
integers of the form a·7+b·9, where a and b are nonnegative
...
Now assume that some integer n ∈ S, say, n = a · 7 + b · 9
...
First note that since n ≥ 48, we cannot have both a and b
less than 3
...

If b ≥ 3, then n + 1 = (a · 7 + b · 9) + (4 · 7 − 3 · 9) = (a + 4) · 7 + (b − 3) · 9
...

To prove the same statement by the Second Principle, we note that each of the
integers 48, 49, 50, 51, 52, 53, and 54 is in S
...
We must show that n ∈ S
...

But then n = (a + 1) · 7 + b · 9
...

0
...

To show that n3 mod 6 = n mod 6, for all integers n, we take into account the 6
different equivalence classes on Z
...

[0] = {
...
} and are in the form 6k, k ∈ Z
...

(6k)3 mod 6 = 216k 3 mod 6 = 6(36k 3 ) mod 6 = 0 = 6k mod 6
[1] = {
...
} and are in the form 6k + 1, k ∈ Z
...

(6k + 1)3 mod 6 = (216k 3 + 108k 2 + 18k + 1) mod 6
= 6(36k 3 + 18k 2 + 3k) mod 6 + 1 mod 6
= 1 = (6k + 1) mod 6
[2] = {
...
} and are in the form 6k + 2, k ∈ Z
...

(6k + 2)3 mod 6 = (216k 3 + 216k 2 + 72k + 8) mod 6
= 6(36k 3 + 18k 2 + 3k) mod 6 + 8 mod 6 = 0 + 2
= 2 = (6k + 2) mod 6
3

[3] = {
...
} and are in the form 6k + 3, k ∈ Z
...

(6k + 3)3 mod 6 = (216k 3 + 324k 2 + 162k + 27) mod 6
= 6(36k 3 + 54k 2 + 27k) mod 6 + 27 mod 6 = 0 + 3
= 3 = (6k + 3) mod 6
[4] = {
...
} and are in the form 6k + 4, k ∈ Z
...

(6k + 4)3 mod 6 = (216k 3 + 432k 2 + 288k + 64) mod 6
= 6(36k 3 + 72k 2 + 48k) mod 6 + 64 mod 6 = 0 + 4
= 4 = (6k + 4) mod 6
[5] = {
...
} and are in the form 6k + 5, k ∈ Z
...

(6k + 5)3 mod 6 = (216k 3 + 540k 2 + 450k + 125) mod 6
= 6(36k 3 + 90k 2 + 75k) mod 6 + 125 mod 6 = 0 + 5
= 5 = (6k + 5) mod 6
0
...
Prove that this error will not be detected by
the check digit
...

Let a9 a8 · · · a0 be the nine digit money order number
...

i=0
For every nonnegative integer, 10n mod 9 = 1 and the above reduces to just
ai · 1 = ai and each ai is going to belong to it’s own distinct equivalence class,
except for 0 and 9 because 0 mod 9 = 9 mod 9
...
Now following from above, we can get the check number,
c = (a9 + a8 + · · · + ai + · · · + a0 ) mod 9
...
Since
aj − ai = 0, c = c and thus a single position digit swap will be detected
...
48 Let S be the set of real numbers
...
Show
that ∼ is an equivalence relation on S
...

Reflexive: a − a = 0 and 0 ∈ Z
Symmetric: a − b = k where k ∈ Z, implies b − a = −k, where −k ∈ Z
4

Transitive: a−b = k and b−c = n where k, n ∈ Z implies that a−c = a+b−b−c =
(a − b) + (b − c) = k + n, where (k + n) ∈ Z
Thus ∼ is an equivalence relation
...
We might also add that the set of all equivalence
classes for this relation is {[a] : a ∈ [0, 1)}
...
5 For n ≥ 3, describe the elements of Dn (Hint: You will need to consider two cases - n
even and n odd
...
1 for example)
...
, n − 1} and we obtain the rotations by fixing a center of rotation
and rotating
...
Rotations act in the same manner as above
...

1
...

The simplest case is when the same reflection twice
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A reflection fixes
a line and then rotates about it
...
5, is a rotation
...
7 In Dn , explain geometrically why a rotation followed by a rotation must be a rotation
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5, rotating fixes a point to be the
center of rotation
...

1
...

Geometrically, we’ll describe Dn as a plane in 3-space
...
5 and
a rotation would be a 2 dimensional action (spinning) about a center of rotation
(in this case, would be at the origin)
...
If we do a reflection, then looking down from the
positive z-axis, we would be looking at the bottom of the plane or side 2
...
The same
can be said for a rotation and then a reflection, because after rotating it, from
the positive z-axis, it would still be side 1 but then a reflection would cause it to
be side 2
...

2
...

Proof by counterexample: Consider a = 1, b = 2 and c = 3, then
(1 − 2) − 3 = −1 − 3 = −1 = 2 = 1 − (−1) = 1 − (2 − 3)
Thus, subtraction on the set of integers is not associative
...
4 Show that the group GL(2, R) of Example 9 is non-Abelian by exhibiting a pair of
matrices A and B in GL(2, R) such that AB = BA
...
5 Find the inverse of the element

1 0

...


First, let us find the determinant; [(2)(5)−(6)(3)] mod 11 = −8 mod 11 = 3
...
Then using what was
2 6
said in class, example 2
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18, we know that the inverse of
3 5
5 · 4 −6 · 4
20 −24
9 9
is going to be
mod 11 =
mod 11 =
−3 · 4 2 · 4
−12 8
10 8
2
...

We will use D4 and the Cayley table from pg
...
Let a = R90 and b = H,
then that means that a−1 = R270
...
Thus a−1 ba = V = H = b
...
15 (Law of Exponents for Abelian Groups) Let a and b be elements of an Abelian group
and let n be any integer
...
Is this also true for non-Abelian
groups?

6

To show that (ab)n = an bn , first let us show the case when n = 0;
(ab)0 = e = ee = a0 b0
...

Our base case is n = 1, then (ab)1 = ab = a1 b1
...
Then we show that
(ab)n+1 = (ab)n (ab) = an bn ab = an abn b = an+1 bn+1
Now, we can show for when n is negative by using the definition that g n = (g −1 )|n| ,
but clearly we don’t need the absolute value exponent since we’re only going to be
using positive integers for n
...
16 and the fact that this group is Abelian
...
Then we show that
[(ab)−1 ]n+1 = [(ab)−1 ]n (ab)−1 = [a−1 b−1 ]n (ab)−1
= [a−1 b−1 ]n [b−1 a−1 ] = [a−1 b−1 ]n+1
No, (ab)n = an bn is not true for non-Abelian groups
...
Furthermore, we can show that this doesn’t hold
true for non-Abelian groups by counterexample
...
4, then (AB)2 =
=
= A2 B 2
...
16 (Socks-Shoes Property) In a group, prove that (ab)−1 = b−1 a−1
...
Find distinct nonidentity elements
a and b from a non-Abelian group with the property that (ab)−1 = a−1 b−1
...

Let e be the identity of a group G and a, b ∈ G be distinct
...
First, we show by multiplication on the right:
(ab)−1 (ab) = b−1 a−1 (ab)
e = b−1 (a−1 a)b
by associativity
−1
= b eb
by inverse property
−1
= b b
by identity property
= e
by inverse property
Now, on the left:

7

(ab)−1 (ab) = (ab)b−1 a−1
e = a(bb−1 )a−1
by associativity
= aea−1
by inverse property
−1
= a a
by identity property
= e
by inverse property
To give an example that shows (ab)−2 = b−2 a−2 , we will use the Cayley table
from D4 on pg
...
If we let a = R270 and b = H, then a−1 = R90
and b−1 = H
...
It’s just as well, that a−2 = (R90 )2 = R180 and
b−2 =H2 = R0
...

We can use the same example, D4 † , from which we can pick distinct nonidentity
elements a and b from the non-Abelian group such that property (ab)−1 = a−1 b−1
is satisfied
...
From the Cayley table, we get that ab = R180
and the inverse of R180 is itself, so (ab)−1 = R180
...
Hence, (ab)−1 = R180 = a−1 b−1
...
Therefore, we can let a be putting
on your socks and b be putting on your shoes, which means that the inverse, a−1
is taking off your socks while b−1 is taking off your shoes
...
From the first thing we
proved in this exercise, (ab)−1 = b−1 a−1 , which tells us that the entire process
of taking off your socks and shoes can be broken down into two actions, that is
taking off your shoes, b−1 and then taking off your socks, a−1
...
18 In a group, prove that (a−1 )−1 = a for all a
...
3, for each element a in a group G, there is a unique element a−1 in G
such that aa−1 = a−1 a = e
...

Thus by the uniqueness of a−1 in G, (a−1 )−1 must be a for all a
...
19 For any elements a and b from a group and any integer n, prove that (a−1 ba)n = a−1 bn a
...
Now, we can use induction to show all
†

We use D4 because of what was said on pg
...
Our base case is when n = 1, then (a−1 ba)1 = a−1 ba = a−1 b1 a
...
Then we show
(a−1 ba)n+1 = (a−1 ba)n (a−1 ba) = (a−1 bn a)(a−1 ba) =
a−1 bn (aa−1 )ba = a−1 bn eba = a−1 bn ba = a−1 bn+1 a
Now, we can show for when n is negative by using the definition that g n = (g −1 )|n| ,
but clearly we don’t need the absolute value exponent since we’re only going to
be using positive integers for n
...
Now, we assume
[(a−1 ba)−1 ]n = a−1 bn a holds true for n
...
1 For each group in the following list, find the order of the group and the order of each
element in the group
...


Z12
U (10) U (12)
U (20)
D4
|0| = 1
|1| = 1 |1| = 1 |1| = 1 |R0 | = 1
|1| = 12 |3| = 4 |5| = 2 |3| = 4 |R90 | = 4
|2| = 6
|7| = 4 |7| = 2 |7| = 4 |R180 | = 2
|3| = 4
|9| = 2 |11| = 2 |9| = 2 |R270 | = 4
|4| = 3
|11| = 2 |H| = 2
|5| = 12
|13| = 4 |V | = 2
|6| = 2
|17| = 4 |D| = 2
|7| = 12
|19| = 2 |D | = 2
|8| = 3
|9| = 4
|10| = 6
|11| = 12
Now the order of the groups: |Z12 | = 12, |U (10)| = 4, |U (12)| = 4, |U (20)| = 8,
and |D4 | = 8
...

3
...


9

Suppose that the order of an element a in G is n and n is finite
...
Multiply both sides (left or
right) by the inverse of an which is (a−1 )n
...
Now,
if the order of a happens to be infinite, then a−1 must also be of infinite order
otherwise it would contradict what was just proven above
...
14 If H and K are subgroups of G, show that H ∩ K is a subgroup of G
...
Then by definition, both are
non-empty and both contain the identity, e
...
Now, to prove that H ∩ K is a
subgroup, by theorem 3
...

So, let a, b ∈ H ∩ K, which means a, b ∈ H and a, b ∈ K
...
Hence
we have ab−1 ∈ H ∩ K
...
19 Prove theorem 3
...

If C(a) = G, then we are done
...
Showing that
e ∈ C(a) is trivial, since e · a = a · e = a
...
Then,
(bc)a = b(ca) = b(ac) = (ba)c = (ab)c = a(bc), which shows that bc ∈ C(a)
...
So,
b·a = a·b
a = b−1 · a · b
a · b−1 = b−1 · a

multiply on left by inverse of b
multiply on right by inverse of b

Thus b−1 ∈ C(a) whenever b is
...
27 Suppose a group contains elements a and b such that |a| = 4, |b| = 2, and a3 b = ba
...

e = ee = a4 b2 = aa3 bb = a(a3 b)b = a(ba)b = (ab)(ab) = (ab)2
First thing to note is that |a| = |b|, which means by Exercise 3
...
We get the second inequality from the order of a
and b, then we use associativity, followed by the given equality, followed by more
associativity, to conclude that |ab| = 2
...
Does your answer surprise you?

3
...
Find

−1 0
0 1
, A3 =
, and A4 =
0 −1
−1 0
1 0
0
1
−1 −1

...
Now, B =
, then B 2 =
, and
0 1
−1 −1
1
0
B 3 is the identity matrix
...
To try to find the order of AB, see
1 1
1 2
1 3
that AB =
, then (AB)2 =
, and (AB)3 =

...
The answer does not surprise me
...
30 For any positive integer n and any angle θ, show that in the group SL(2, R),
cos θ − sin θ
sin θ cos θ
Use this formula to find the order of
√
√
cos(√ 2)◦ − sin( 2)◦
√

...


and

Firstly, the trigonometric identities to be used
...
Our base case is when n = 1;
1
cos(1 · θ) − sin(1 · θ)
cos θ − sin θ
=
is clearly true
...
Then,
sin θ cos θ
sin nθ cos nθ
n+1
n
cos θ − sin θ
cos θ − sin θ
cos θ − sin θ
=
=
sin θ cos θ
sin θ cos θ
sin θ cos θ
cos nθ − sin nθ
sin nθ cos nθ

cos θ − sin θ
sin θ cos θ

=

(cos nθ)(cos θ) − (sin nθ)(sin θ) (cos nθ)(− sin θ) − (sin nθ)(cos θ)
(sin nθ)(cos θ) + (cos nθ)(sin θ) (− sin θ)(sin nθ) + (cos nθ)(cos θ)
=

cos(n + 1)θ − sin(n + 1)θ
sin(n + 1)θ cos(n + 1)θ

11

The last equality comes from row 1:column 1 and row 2:column 2 both using (1),
while row 1:column 2 and row 2:column 1 use (2)
...
The
sin 60◦ cos 60◦
order of
√
√
cos(√ 2)◦ − sin( 2)◦
√ ◦
is infinite because in order to get the identity matrix,
sin( 2)◦ cos( 2)
θ must equal 0, 360◦ , 720◦ ,
...

3
...
List them
...
44 Let H = {A ∈ GL(2, R) | det A is a power of 2}
...

First, we check if the identity is in H
...
Next, we let
A, B ∈ H and show that AB ∈ H as well
...
9 on page 45,
det(AB) = det(A) · det(B) = 2k · 2n = 2k+n
...
Lastly, we
need to show that A−1 ∈ H whenever A ∈ H to satisfy Theorem 3
...
Clearly, for
any A ∈ GL(2, R) there exists an inverse, A−1 ∈ GL(2, R) otherwise this would
contradict GL(2, R) being a group
...
Since this
is true in GL(2, R), then det(A−1 ) must be 2−k , which is a power of 2 and hence
A−1 ∈ H as well
...
45 Let H be a subgroup of R under addition
...
Prove that K is a
subgroup of R∗ under multiplication
...
Here we know that
20 = 1 ∈ K
...
Now if a, b ∈ K, then a = 2k and b = 2n for
some k, n ∈ H
...
Now if c ∈ K, then ∃z ∈ H such that c = 2z
...
Thus 2−z ∈ K
...
By the two-step
subgroup test, we are done
...
52 Let G be a finite group with more than one element
...

Let g ∈ G, g = e
...

Hence let |g| = n
...
3
...
Now we have
an = amp = (am )p = e
4
...

Exercises 4
...
5 have in common the fact that the two cyclic subgroups given,
while generated by a different element of the set, are in the fact the same cyclic
subgroup, as satisfying theorem 4
...
A generalization we can make
about the two generators of the cyclic subgroup given in each exercise is that
they are in fact inverses to each other in their own respective group, e
...
, 10 is
the inverse of 20 in Z30
...
, an−1 }, for |a| = n
...

4
...

U (8) satisfies our requirements
...

4
...
List all generators for the subgroup of order 8
...
Since we know that |ak | = | ak | from
Theorem 4
...
2
...

Again utilizing theorem 4
...
Now we let n = |a3 | = 8 in order find all the generators
to a3
...
2, Corollary 2, (a3 )k = a3 if and only if gcd(n, k) =
1
...

4
...
Find a generator for a21 ∩ a10
...
2, since a21 =
agcd(24,21) = a3 and a21 = agcd(24,10) = a2
...
, a3 , a6 , a9 ,
...
, a2 , a4 , a6 ,
...
Now, we can deduce
that a6 is actually a generator of a3 ∩ a2 since all elements of a3 are some
multiple of 3 while all elements of a2 are multiples of 2 thereby making the all
the elements in common multiples of 6 and a6 = a21 ∩ a10
...
2, then
as ∩ ar = at , where s = gcd(m, d), r = gcd(n, d), and t = lcm(s, r)
...

Proof: Let m, n ∈ Z
...
So, let d = gcd(m, n)
...
Given c, a common multiple of m and n, then c = m · k = dm · k and
c = n · l = dn · l
...
We want to show that n | k
...
If not, then we know that the gcd(m , n ) = 1 (from
our first hw
...
But n | (c/d) = m k and n does not divide m , which implies that n | k
...
With that fact, we see that d · m · n ≤ all common multiples of m
and n
...


4
...
Prove that H is a
subgroup of G
...

Let’s prove the general result first
...
First we note that e ∈ G and |e| = 1 and 1 divides any m, so that e ∈ H
and H is nonempty
...
4, which means that |a−1 | divides m as well
...
2 so let a, b ∈ H and n = |a| and k = |b|
...
It follows then that since G is Abelian that (ab)t = at bt = e, where t =
lcm(n, k)
...
Now going back and answering the
first questions
...
There is nothing special about 12 since 12 is only one integer
from the entire set of natural numbers
...

4
...
Furthermore because G is
cyclic we also know that there exists an a ∈ G such that a = G
...
Then ∃k ∈ Z such that ck = e
...
Because G is cyclic with a generator, a, ∃n ∈ N with
an = c
...
With nk < 2nk, we have
contradicted theorem 4
...
Thus there does not exist a finite element in G other
than the identity
...

T2

Given a group G and a, b ∈ G, |a| = |b| = p, p is a prime number, then if any
c ∈ a , c = e and c ∈ b , then a = b , otherwise a ∩ b = {e}
...
Further let c ∈ a , c = e
and c ∈ b
...
With n
prime, gcd(n, k) = 1, thus c = a by theorem 4
...
By the same reasoning (by
symmetry) b = c
...
If there doesn’t exist a c ∈ a with c ∈ b ,
c = e, then with a and b groups, it follows that a ∩ b = {e}
...
20 Suppose that G is an Abelian group of order 35 and every element of G satisfies the
equation x35 = e
...
Does your argument work if 35 is replaced
by 33?
First, since G is an Abelian group, we know that (ab)n = an bn from Exercise 2
...

Since every element of G satisfies x35 = e, we also know by theorem 4
...
If we know
there is some element in G that is order 35, then we are done and know that G is
cyclic
...

We know that |ab| must be a divisor of 35 and (ab)5 = a5 b5 = eb5 = b5 = e and
that (ab)7 = a7 b7 = a7 e = a7 = e
...
Now, to show that all the elements of G cannot have either order 5
or order 7, let’s assume for the sake of contradiction that ∀a ∈ G, a = e, the order
of a is 5
...
Then we have 8 groups containing 4 nonidentity
elements with the identity and 2 nonidentity elements leftover from G which have
no group to belong to of order 5, hence a contradiction by T2
...

No, the argument above would not work if 35 was replaced with 33
...

4
...

15

Theorem 3
...
Moreover, theorem 4
...
If we let e be the
identity of G, it suffices to show that |a| = | a | = |G| to show that a = G
...
Also since a = e, then |a| = 1 = | a |,
so |a| = 2 = | a |
...
Thus ab = e, hence a(ab) = ae
which implies that a2 b = a, yet a2 = e so b = a
...
Thus | a | = 3 ⇒ a = G
...
28 Let a be a group element that has infinite order
...

Suppose that i = ±j
...
When i = j
and when i = −j
...
Now, when i = −j, we know that
i = −i, which we know by definition that a−i = (a−1 )|i| , which then shows us that
(a−1 )|i| ∈ ai
...

4
...

1

FF
x
FF
xx
FF
xx
FF
x
x
F
xx
q
p RRR
RRR
RRR
RRR
RRR
RRR

p2

EE
EE
EE
EE

p2 q

yy
yy
yy
yy

pq

4
...

Suppose that a finite group, G, is the union of proper subgroups,
that is G = G1 ∪ G2 ∪ · · · ∪ Gk
...
This means that there is some a ∈ G such that a = G
...
Clearly, whatever Gi
that contains a, must also contain G, which contradicts Gi being a proper subset
...
This means that there
is no a ∈ G such that a = G, thus for every a ∈ G, a < G, by theorem 3
...

Then it must follow that G = a∈G a and we’re done
...
42 Prove that an infinite group must have an infinite number of subgroups
...
For the sake of contradiction, let’s assume
that G has a finite number of subgroups
...
Clearly, there
has to be one infinite a otherwise our finite union of subgroups would fall short
of being equal to G
...
1
that all distinct powers of a are distinct group elements
...
Thus a contradiction that G has a finite amount
of subgroups
...
60 Suppose that |x| = n
...

Just by inspecting some of the examples on page 79 and 80, we can build a
conjecture that our necessary and sufficient condition is going to be xr ⊆ xs if
and only if s|r, where s and r have to be divisors of n
...
Suppose that s|r, then for some nonzero integer k and since r|n, xn = xkr by
theorem 0
...
So, if we let
0 < k < n/r, then xkr = (xk )r = (xst )r = (xs )tr ∈ xs , for some nonzero integer
t and by theorem 0
...
Thus, we can conclude that xr ⊆ xs
...
2
on page 76, we know that ar = (as )t for some integer t
...
1 in the finite case, this is essentially, addition modulo n, so
r = st, which by theorem 0
...

4
...
If Z(G) is not trivial, prove that every
nonidentity element of G has the same order
...
Since, a ∈ Z(G), then we know that for all b ∈ G, ab = ba and (ab)n = an bn
...
Then it must be that the lcm(m, n) = mn and (ab)mn = e
...
1, corollary 2, k must divide mn
...
But
then,
(ab)m = am bm = ebm = bm = e
(ab)n = an bn = an e = an = e
which means that k must be mn, since neither m| n nor n| m
...

Therefore, |a| = |b|
...
1 Find the order of each of the following permutations
...
(12)
Since there is only one cycle of length 2, then the order must be 2 as any other
cycle must be a one cycle (or fixed) and the lcm of all the cycles is going to be 2
...
(147)
For the same reasoning as part (a); the order is 3
...
(14762)
For the same reasoning as part (a) and (b); the order is 5
...
2 What is the order of a k-cycle (a1 a2 · · · ak )?
We let (a1 a2 · · · ak ) be our k-cycle and we’ll denote the mapping of all ai in the
k-cycle to ai+1 , with the exception of ak which maps to a1 , by σ
...
Then we know
inductively that if σ(ai ) = ai+1 mod k , then σ 2 (ai ) = σ(σ(ai )) = σ(ai+1 mod k ) =
ai+1+1 mod k = ai+2 mod k and so on and so forth until we get σ k (ai ) = ai+k mod k =
ai
...

5
...
3, the lcm of these disjoint cycles is 3
...

b) (124)(3567)
Following part (a), lcm(3, 4) = 12
c) (124)(35)
” ”, lcm(2, 3) = 6
d) (124)(357869)
” ”, lcm(3, 6) = 6
e) (1235)(24567)
Need to write in disjoint cycle form, so we have (241)(5673) and the order is the
lcm(3, 4) which is 12
...

5
...

a
...
3, the lcm(2, 3) = 6,
which means that 6 is the order
...

1 2 3 4 5 6 7
7 6 1 2 3 4 5
in cycle notation, we have (1753)(264), which by Theorem 5
...

5
...
But then so are the combinations of two disjoint cycles of 8 and 2,
7 and 3, 6 and 4, 5 and 5
...
Listing the rest is negligible since the only
one that comes close to 21 is two cycles of length 5 and 3 with 2 fixed points,
which has an lcm of 15
...
13 Prove Theorem 5
...

Let’s use the two-step subgroup test
...
Also for any even permutation, σ, in Sn , then the inverse is just the
reversed transpositions which made up the even permutation in the first place
...

Now, we let σ and τ both be even permutations in Sn , then we know that σ can be
expressed as n transpositions and τ can be expressed as m transpositions, where
n and m are even
...

5
...


Compute each of the following
...
α−1
Mentioned in class and turns out that α = α−1
...
βα
Trace out a few, so as to show some underlying work; β(α(1)) = β(2) = 1 and
β(α(4)) = β(5) = 3
...
αβ
In the same fashion as part (b); α(β(1)) = α(6) = 6 and α(β(4)) = α(4) = 5
...
18 Let
α=

1 2 3 4 5 6 7 8
2 3 4 5 1 7 8 6

and β =

1 2 3 4 5 6 7 8
1 3 8 7 6 5 2 4


...
So,
αβ =

1 2 3 4 5 6 7
2 4 6 8 7 1 3

which is αβ = (12485736)
...
so using Example 5
...

5
...

Following what was said in class: If every member of H is an even permutation,
then we are done
...
Then
there is some τ which is an odd permutation and we let f be the function which
maps an even permutation, σ, to an odd by composing with τ
...
Now by the same argument, f −1 ,
which maps an odd permutation to an even one by composing it with τ , is also
one-to-one
...

5
...
1 to compute the following
...

Using table 5
...
Thus we get C(α3 ) = {α1 , α2 , α3 , α4 }
...
Following the same procedure as part (a), except
it being a little more easier and thus quicker to compare the entries as they are
on the very tip of the table, we get C(α12 ) = {α1 , α7 , α12 }
...
27 Let β ∈ S7 and suppose β 4 = (2143567)
...

Since we can clearly see that |β 4 | = 7, then we know that |β 28 | = ε and that |β|
must divide 28
...
So, since
|β 4 | = ε, then we can eliminate 1, 2, and 4 as being the order of β right away
...
3, which
means that the lcm of the disjoint cycles that make up β must be 14
...
By the same argument (different numbers, same result), 28 cannot be
the order of β either, thus it must be 7
...
Now, we can work backwards,
since β 4 = (1435672) = (β 2 )2 = (1647325)2 = [(β)2 ]2 = [(1362457)2 ]2
...

5
...
Write β 99 in disjoint cycle form
...
Then β 99 = β −1 = (32541)
...
well, we know the β is a 5-cycle,
thus the order of it is 5, since any multiple of 5 is going to give us ε, we know
that this reduces to ε · β 4 , but β 4 = β −1
...
32 Let β = (1, 3, 5, 7, 9, 8, 6)(2, 4, 10)
...
3, we know that the order of β is 21
...
But, we want the smallest, so we can use the order of β,
then β 100 = (β 21 )4 β 16 = (ε)4 β 16 = β 16
...

5
...

The cyclic subgroup can be achieved by (1234)
...

Now for our noncyclic subgroup;
{(1), (14), (23), (14)(23)}
...
37 Suppose that β is a 10-cycle
...

5
...

Let α = (12) and β = (13)
...
So, |α| = 2 = |β| and |αβ| = 3
...
41 Prove that Sn is non-Abelian for all n ≥ 3
...
1 shows that S3 is non-Abelian because there exists two permutations
which do not commute, namely α = (123) and β = (23)
...
Thus
Sn , n ≥ 3 is non-Abelian
...
42 Let α and β belong to Sn
...

From lemma 2 of the conjugacy class handout, we know that if α is a k-cycle,
then βαβ −1 is also a k-cycle and equivalently even or odd, depending on whether
α was even or odd to begin with because both k-cyles will have the same number
of transpositions
...
Hence by theorem 3 of the conjugacy class handout, βαβ −1
must also have the same cycle type and is equivalently odd or even depending on
what α was to begin with
...

5
...

Since |β 3 | = 7, then we know that β 21 = ε and that |β| must divide 21
...
So, since β 3 = ε, then we can eliminate 1
and 3 as being the order of β right away
...
3, which means that the lcm of the disjoint cycles
that make up β must be 21
...
Thus |β| must be 7
...
45 Show that every element in An for n ≥ 3 can be expressed as a 3-cycle or a product of
three cycles
...
Now, (ac)(ab) = (abc) and (ab)(cd) = (cbd)(acb), for any a, b, c, d in
A, where A is the finite set in which the permutations act upon
...
46 Show that for n ≥ 3, Z(Sn ) = { }
...
Now let α ∈
Sn , α = ε
...
1
...
, n}
...
Then α = (a1 a2
...
ak }
...
ak ) and (a1 b)α =
(a1 a2
...

Case: ii) α is an n-cycle
...
an ) with n ≥ 3, we have (a1 a2 ) ∈ Sn
...
an ) and (a1 a2 )α = (a2 a3
...

Case: iii) α is the composition of at least 2 disjoint cycles of lengths k and l ≥ 2
...
βr , where each βi is a disjoint cycle
...
ak )
and β2 = (b1 b2
...
ak )(b1 b2
...
βr
...
So, consider
α(a1 b1 ) = β1 β2 (a1 b1 )
...
bl b1 a2
...
βr
and
(a1 b1 )α = (a1 b1 )β1 β2
...
bl a1 a2
...
βr
so α ∈ Z(Sn ) with (a1 b1 )α = α(a1 b1 )
...

6
...

Since the only generators of Z under addition are 1 and -1, any possible element
of Aut(Z) must map 1 → −1, 1 → 1, −1 → 1, or −1 → −1, by theorem 6
...
We can achieve all these mapping by x → x (the identity map) and
x → −x, thus finding the only two elements of Aut(Z)
...
4 Show that U (8) is not isomorphic to U (10)
...

6
...
That is, if G, H, and K are
groups and G ∼ H and H ∼ K, then G ∼ K
...
Since φ : G → H is bijective and ψ : H → K
=
=
is bijective, then the composition ψφ : G → K is bijective
...
Hence G ∼ K
...
12 Find two groups G and H such that G ∼ H, but Aut(G) ∼ Aut(H)
...
Clearly G ∼ H since |G| = |H|, yet Aut(G) ∼
=
=
∼ U (12)
...

6
...
Determine φ(x)
...
2
...
Moreover, we see simply then, that the mapping φ(x) = φ(1)x = 9x
...

6
...
3
...
2, property 1 and since e ∈ K, then e ∈ φ(K), so φ(K) is nonempty
...
So let a, ¯ ∈ φ(K) such that
¯ b
φ(a) = a and φ(b) = ¯ Then
¯
b
...
2, property 2
...

6
...

Consider 3, an element from U (20) which has order 4
...
2, property 5
...
24 Let

√
G = {a + b 2 | a, b rational }

and
H={

a 2b
b a

| a, b rational }
...
Prove that G and H are closed
under multiplication
...

b a
To show that φ is bijective,√ φ(x) = φ(y)
...
Now, from the mapping we
a 2b
c 2d
find that
=
from which it is clear that a = c and b = d, thus
b a
d c
α 2β
x = y
...
Then we can
β α
√
find g ∈ G with g = α + 2β where φ(g) = h
...
Then there are rational
∈
√
numbers k, l, m, n with i = k + 2l and j = m + 2n
...

l+n k+m
l k
n m
To find if G and H to be closed under multiplication, √
take rational numbers
√
√
a, b, c, d and look at (a + 2b)(c + 2d) = ac + 2bd + 2(bc + ad) ∈ G, and
a 2b
c 2d
ac + 2bd 2(bc + ad)
=
∈ H
...
It is easy to see then that φ is also an isomorphism under
multiplication as the bijective property holds from before, and the homomorphism
follows trivially from the proof that G and H are closed under multiplication
...
25 Prove that Z under addition is not isomorphic to Q under addition
...

Now, we know by theorem 6
...
Hence φ(1) will
generate Q under addition
...
Hence p/q = Q under addition, yet we know that any
element in p/q is of the form k(p/q) for some k ∈ Z
...
Contradiction
...
26 Prove that the quaternion group is not isomorphic to the dihedral group D4
...
2, property 5 cannot be satisfied
...

6
...

First, we must note an algebraic property of logs, namely that log10 (xy) =
log10 (x) + log10 (y) and the fact that exp10 (log10 (x)) = x, where exp is defined
as the exponential function from which logarithms (which are the inverse of the
exponential functions) are derived from
...
Showing that φ is one-to-one, suppose that φ(a) = φ(b), then
log10 (a) = log10 (b)
...
To show φ is onto, we
take a ∈ R and since 10a ∈ R+ , then it happens that φ(10a ) = log10 (10a ) = a
...
Thus φ is an isomorphism
...
33 Suppose that g and h induce the same inner automorphism of a group G
...

Suppose that g and h induce the same inner automorphism of G, then φg = φh
...
Now multiply on the right by g and
multiply on the left by h−1 , so that we obtain h−1 gx = xh−1 g
...

6
...
} and H = {0, ±3, ±6, ±9,
...
Does your isomorphism preserve multiplication?
First, it is important to note that 2 = G and 3 = H
...
Clearly, with φ strictly monotone, it is 1-1
...
To show that φ is a homomorphism, take a, b ∈ G
...
Hence, φ(a + b) = φ(2k + 2l) = φ(2(k + l)) =
3
(2(k + l)) = 3(k + l) = 3k + 3l = φ(2k) + φ(2l) = φ(a) + φ(b)
...
Under multiplication, G and H aren’t even groups, being that there are no
inverses
...
37 Prove that Q under addition is not isomorphic to R+ under multiplication
...
Then from theorem 6
...
Furthermore, if m, p ∈ Z then φ(m/p) = φ(1/p)m = (z 1/p )m = z m/p
...
Thus no isomorphism exists
...
40 Show that every automorphism φ of the rational numbers Q under addition to itself
has the form φ(x) = xφ(1)
...
Then ∃a ∈ Q
where φ(1) = a, which is φ(n/n) = a for some n ∈ Z
...

Now, let m/n ∈ Q, then
φ(m/n) = φ((1/n)m ) = φ(1/n)m = [(1/n)φ(1)]m = (m/n)φ(1)
26

6
...

Suppose that H is a proper subgroup of Q
...
Also
let φ : Q → H be any isomorphism, then there does not exist any x ∈ Q such
that φ(x) = z
...
So φ(x) = xφ(1) by
problem 6
...
Finally xφ(1) = (z/a)φ(1) = (z/a)a = z is a contradiction
...

7
...
How many left cosets of H in S4 are there? (Determine
this without listing them
...

7
...
Find all of the left cosets of a5 in a
...
, a14 } and a5 = {e, a5 , a10 }
...
14 Suppose that K is a proper subgroup of H and H is a proper subgroup of G
...
Now by Lagrange’s theorem, |K| = 42 | |H|, which means that
|H| = 42k, for some k ∈ N, and again by Lagrange’s theorem 42k | 420 = |G|
...

Hence the two possible orders of H are 42 · 2 = 84 or 42 · 5 = 210
...
16 Recall that, for any integer n greater than 1, φ(n) denotes the number of integers less
than n and relatively prime to n
...

Firstly, a is relatively prime to n, so a ∈ U (n)
...
Since e = 1 in U (n), then
aφ(n) mod n = 1 as required
...
17 Compute 515 mod 7 and 713 mod 11
...

713 mod 11 =
=
=
=
=

(711 mod 11) · (72 mod 11)
(7 mod 11) · (49 mod 11)
(7 · 5 mod 11)
35 mod 11
2

7
...
Prove that |G| is prime
...
)
Suppose that G is a group with more than one element and G has no proper,
nontrivial subgroups
...
Then g is a
subgroup of G
...
Hence, it must follow that G can only be finite,
otherwise g 2 is a proper subgroup and a contradiction
...
3,
we know if k | n = |G|, then there is a subgroup of order k, yet this means
that k cannot be anything other than 1 or n (otherwise we would have a proper
subgroup), so it must be that the order of G is prime
...
26 Let |G| = 8
...

Let g be any nonidentity element in G
...
Since
|G| = 8, |g| must be 2, 4, or 8
...

Otherwise, |g| = 4 or |g| = 8, then |g 2 | = 2 or |g 4 | = 2, respectively and again
we’re done
...
28 Show that Q, the group of rational numbers under addition, has no proper subgroup
of finite index
...
Then there exist an element, z ∈ Q\H, which
z
also means that z ∈ H
...

2
2
2

28

z
z
Thus there exists a k ∈ N with ( z ± n ) ∈ H such that z + H = z ± k + H, which
2
2
2
we know must be true since H has a finite index
...
Thus there is no proper subgroups of Q under addition with
2
2
finite index
...
33 Let G be a group of order pn where p is prime
...

For the sake of contradiction, let |Z(G)| = pn−1
...
So, take one of those elements, let’s say g ∈ G \ Z(G), and
look at the centralizer of g
...
Furthermore, g commutes with all the elements from the center, so
Z(G) ⊆ C(g)
...
6 that C(g) ≤ G and from Lagrange’s
theorem that |C(g)| must divide G
...
This means that
G = C(g), which implies that g commutes with every element of G and must be
in the center
...

7
...

The nth roots of unity are those complex numbers, x, such that xn = 1
...

Furthermore, these roots form a cyclic subgroup in C∗ under multiplication and
we can find them by e(2πik)/n for any n ∈ N, k = 0, 1,
...
Thus, {e(2πik)/n |
k = 0, 1,
...

8
...

Z2 ⊕ Z2 ⊕ Z2 = {(0, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1), (1, 0, 0), (1, 0, 1),
(1, 1, 0), (1, 1, 1)}
...
Note that this is possible since any
nonidentity element has an order of 2 and thus must be its own inverse
...
10 How many elements of order 9 does Z3 ⊕ Z9 have?

29

By theorem 8
...
This is only possible in two ways, that is, |g| = 9
and |h| = 1 or |g| = 1 and |h| = 9
...
Now, we know that any element in Z9 that is
going to have order 9 will be relatively prime to 9, thus the number of elements in
Z9 of order 9 is just φ(9)
...

8
...
Explain why Dn cannot be isomorphic to the external direct
product of two such groups
...

=
From theorem 8
...
But then h ∈ G2 , h = e2 and |(t, h)| =
lcm(|t|, |h|) = 2|t| = 2n
...

If n is even, then notice that there exists an odd number of generators of the
rotation subgroup
...
But |(g, h)| = |(g, e2 )|, so they always come in pairs
...

=
8
...

Proof 1: Suppose that G ⊕ H is cyclic
...
1, corollary 1 and since we know that |G ⊕ H| =
|G||H|
...
1, |(g, h)| = lcm(|g|, |h|) = |G||H|, so it must follow
that |g| = |G| and |h| = |H| by theorem 8
...
Thus G and H must be cyclic
...
Now for the sake of contradiction, assume
that G is not cyclic or H is not cyclic
...

Then | (g, h) | = |G⊕H| = |G||H| = |(g, h)| by theorem 4
...
Now, by theorem 8
...
But this impossible since either |g| = |G| or |h| = |H|
...

8
...



0 0 1

(See Exercise 34 in Chapter 2 for the definition of multiplication
...
Is H isomorphic to Z9 or to Z3 ⊕ Z3 ?
30

If we take a look at exercise 2
...
With this in
mind, we show that H is Abelian, by taking any a , a, b, b ∈ Z3 and concluding
that


 

1 a b
1 a b
1 (a + a) (b + b)
 0 1 0  0 1 0  =  0
=
1
0
0 0 1
0 0 1
0
0
1

 


1 (a + a ) (b + b )
1 a b
1 a b
 0
 =  0 1 0  0 1 0 
1
0
0
0
1
0 0 1
0 0 1
since Z3 is Abelian
...
Hence, |H| = 9
...
So, for this same reason, H ∼ Z9
...
We
Z3
can further prove this by letting φ : H → Z3 ⊕ Z3 such that

1 a b
φ  0 1 0  = (a, b)
...

8
...

Z2 ⊕ Z2 = {(0, 0), (1, 0), (0, 1), (1, 1)}
...
Since this is not a very large number, we can just
3
show the mappings (elements) of Aut(Z2 ⊕ Z2 ):
(0, 0)

(0, 0)

/ (0, 0)

(0, 0)

/ (0, 0)

(0, 0)

/ (0, 0)

(1, 0)

/ (1, 0)

(1, 0)

(1, 0)

(1, 0)
4

(1, 0)

/ (1, 0)

/ (0, 1)

(0, 1)

(0, 1)

(0, 1)

(0, 1)

(1, 1)

/ (1, 1)

(1, 1)

/ (1, 1)

44

D
44



44



44

(0, 1)


44 / (0, 1)


444
44








(1, 0)

(0, 1)

extra

/ (0, 0)

(1, 1)

(1, 1)

(1, 1)

;
GG
GG www
GG
wwGG
ww G
ww
#

(1, 1)

;
GG
GG www
Gw
wG
ww GGG
ww
#

Because rotations commute we know that given any 2 rotations, R1 and R2 ,
−1
then R1 R2 R1 = R2
...
Notice that if n is even, Rπ ∈ Dn ,
−1
so [Rπ ] = {Rπ , Rπ } = {Rπ }
...
We need only consider f
and rf
...

So, if n is even, we find that [f ] = {rx f | x ∈ {0, 2, 4,
...
, n−1}}
...
, n−1}}
...


32



---