Ch 3. Groups – Solutions

1. In the group $\boldsymbol{\mathbb{Z}_8}$ compute:

(a)

6+7

.

\mathbb{Z}_8 =\{0,1,…,7\}

.

6+7 \equiv 5 \pmod{8}

.

(b)

2^{-1}

.

2^{-1}

is the inverse of

2

in

\mathbb{Z}_8

. That is,

2+2^{-1} \equiv 0 \pmod{8}

.

\implies 2^{-1}=8-2=6

.

2. In the group $\boldsymbol{U(16)}$ , perform the following computations.

Recall that

U(n)

is the “group of units” of

\mathbb{Z}_n

. It is comprised of the elements of

\mathbb{Z}_n

that have a multiplicative inverse. To figure out what

U(n)

contains we must find the elements of

\mathbb{Z}_n

that are co-prime to

n

.

In this case, we have:

U(16)=\{1,3,5,7,9,11,13,15\}

.

(a)

5 \cdot 7

.

5 \cdot 7 = 35 \equiv 3 \pmod{16}

.

(b)

3^{-1}

.

3^{-1}

is the inverse of

3

in

U(16)

. That is,

3 \cdot 3^{-1} \equiv 1 \pmod{16}

.

\implies 3^{-1} = 11

.

3. State the definition of a group.

A group is a set of elements, G, equipped with a binary operation, ⋅, satisfying the following properties:

(i) Closure. The product of any two elements is also an element of the set.

\forall a,b \in G: (a \cdot b) \in G

.
(ii) Associativity.

\forall a,b,c \in G: a \cdot (b \cdot c) = (a \cdot b) \cdot c

.
(iii) Identity. There exists an identity element. That is,

\exists \varepsilon \in G

such that

\forall a \in G: a \cdot \varepsilon = \varepsilon \cdot a = a

.
(iv) Inverses. Every element has an inverse.

\forall a \in G: \exists a^{-1} \in G

such that

a \cdot a^{-1} = a^{-1} \cdot a = \varepsilon

.

4. Explain a single method that will decide if a subset of a group is itself a subgroup.

H \subseteq G

, where

G

is a group. To determine if H is a subgroup we must check the following:

(i) Identity element. If H is a subgroup, then it will have the same identity element as G. Check if

\varepsilon \in H

.
(ii) Inverses. Every element of H must have an inverse in H.

\forall h \in H: \exists h^{-1} \in H

such that

h \cdot h^{-1} = h^{-1} \cdot h = \varepsilon

.
(iii) Closure. The product of any two elements must also belong to H.

\forall h_1,h_2 \in H: (h_1 \cdot h_2) \in H

.

Notice that we don’t need to check associativity. The property of associativity is inherited by any subset of G.

5. Explain the origin of the term “abelian” for a commutative group.

“Abelian” groups, meaning commutative, are named after Norwegian mathematician Niels Henrik Abel (1802 – 1829). Abel was the first to prove that there do not exist solutions in radicals for polynomials of degree 5 and higher. Think about the quadratic formula:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

, where

P(x)=ax^2+bx+c

. This formula gives us the roots of any polynomial of degree 2 as a function of its coefficients. Although unwieldy, there are also formulas to calculate the roots of cubic, and even quartic polynomials. However, for degree 5 and higher no such formulas exist. In order to prove this Abel invented group theory—independently of French mathematician Évariste Galois. Sadly, Abel died at the young age of 26 from tuberculosis. The Abel Prize, initially proposed by Sophus Lie due to the lack of a Nobel Prize in Mathematics, is also named in his honour.

6. Give an example of a group you have seen in your previous mathematical experience, but that is not an example in this chapter.

The 12-hour clock is a great example. The hours of a 12-hour clock can be represented by the cyclic group

(\mathbb{Z}_{12}, +)

, where

+

is addition modulo 12.

1. Find all $\boldsymbol{x \in \mathbb{Z}}$ satisfying each of the following equations.

(a)

3x \equiv 2 \pmod{7}

.

S=\{…, -4, 3, 10, 17, …\}=\{3+7z:z \in \mathbb{Z}\}

.

(b)

5x+1 \equiv 13 \pmod{23}

.

S=\{…, -16, 7, 30, 43, …\}=\{7+23z:z \in \mathbb{Z}\}

.

(c)

5x+1 \equiv 13 \pmod{26}

.

S=\{…, -8, 18, 44, 70, …\}=\{18+26z:z \in \mathbb{Z}\}

.

(d)

9x \equiv 3 \pmod{5}

.

S=\{…, -3, 2, 7, 12, …\}=\{2+5z:z \in \mathbb{Z}\}

.

(e)

5x \equiv 1 \pmod{6}

.

S=\{…, -1, 5, 11, 17, …\}=\{5+6z:z \in \mathbb{Z}\}

.

(f)

3x \equiv 1 \pmod{6}

.

S=\emptyset

.

2. Which of the following multiplication tables defined on the set $\boldsymbol{G= \{ a,b,c,d\}}$ form a group? Support your answer in each case.

(a) This is not a group because it lacks an identity element. Alternatively, we can see that the same element,

a

, appears twice in the first row. For groups, every element will always appear exactly once in each row and column. If an element appears more than once, we can derive an immediate contradiction. In this case:

aa=ad \implies a=d

(by the left cancellation property).

(b) Yes, this is a group. It is isomorphic to

K_4

, the Klein four-group.
Identity. ✓

a

is the identity element.
Inverses. ✓
Every element has a unique inverse (the identity element appears exactly once in every row).
Closed. ✓
No elements not in the group appear in the table.
Associative. ✓
Rather than checking every case, we can recognize this as being isomorphic to

K_4

, the Klein four-group.

(c) Yes, this is a group. It is isomorphic to

C_4

, the cyclic group of order 4.

Identity. ✓

a

is the identity element.
Inverses. ✓
Every element has a unique inverse (the identity element appears exactly once in every row).
Closed. ✓
No elements not in the group appear in the table.
Associative. ✓
Rather than checking every case, we can recognize this as being isomorphic to

C_4

, the cyclic group of order 4.

(d) We notice that

d

appears twice in the fourth row.

da=db \implies a=b

(by the left cancellation property). Contradiction. Therefore, this is not a group.

3. Write out Cayley tables for groups formed by the symmetries of a rectangle and for $\boldsymbol{ (\mathbb{Z}_4, +) }$ . How many elements are in each group? Are the groups the same? Why or why not?

This is the Cayley table for the symmetries of a non-square rectangle, where

h

is a reflection across the horizontal axis,

v

is a reflection across the vertical axis, and

r

is a 180° rotation:

This is the Cayley table for

\mathbb{Z}_4

, under addition:

Both groups have 4 elements. The groups are not the same, as there does not exist an isomorphism between them.

4. Describe the symmetries of a rhombus and prove that the set of symmetries forms a group. Give Cayley tables for both the symmetries of a rectangle and the symmetries of a rhombus. Are the symmetries of a rectangle and those of a rhombus the same?

Yes, the symmetries of a rhombus and a non-square rectangle are the same. This is the Cayley table for the symmetries of a rhombus, where

h

is a reflection across the horizontal axis,

v

is a reflection across the vertical axis, and

r

is a 180° rotation:

5. Describe the symmetries of a square and prove that the set of symmetries is a group. Give a Cayley table for the symmetries. How many ways can the vertices of a square be permuted? Is each permutation necessarily a symmetry of the square? The symmetry group of the square is denoted by $\boldsymbol{D_4}$ .

Let

b

represent a reflection and let

a

represent a 90° rotation. By composing rotations and reflections we can find all 8 of the square’s symmetries. The symmetries of a square form the dihedral group

D_4

. Here is the Cayley table of the group

D_4

:

Each permutation is not necessarily a symmetry of the square. Consider the following: our identity element is

\varepsilon=(1\;2\;3\;4)

and we have a permutation

x=(1\;3\;2\;4)

. Clearly

x

cannot be a symmetry of the square because vertex

1

is now opposite vertex

2

. This cannot be accomplished with any combination of reflections and rotations.

6. Give a multiplication table for the group $\boldsymbol{U(12)}$ .

Recall that

U(n)

is the “group of units” of

\mathbb{Z}_n

. It is comprised of the elements of

\mathbb{Z}_n

that have a multiplicative inverse. To figure out what

U(n)

will look like we must find the elements of

\mathbb{Z}_n

that are relatively prime to

n

.

In this case, we have:

U(12)=\{1,5,7,11\}

.

U(12)

has the following Cayley table. We note that

U(12)

is isomorphic to the Klein four-group.

7. Let $\boldsymbol{S=\mathbb{R} \backslash \{-1\}}$ and define a binary operation on $\boldsymbol{S}$ by $\boldsymbol{a*b=a+b+ab}$ . Prove that $\boldsymbol{(S,*)}$ is an abelian group.

First, we will show that

(S,*)

is in fact a group.

Identity?
Let

\varepsilon=0

. Consider arbitrary

x \in S

.

0*x=0+x+0 \cdot x=x

.

x*0=x+0+x \cdot 0=x

.
Yes, the identity element is 0. ✓

Closed?
We know the reals are closed under addition and multiplication. To show S is closed under

*

, we must show that

a*b\neq-1

for any

a,b \in S

.
Let

-1=a*b

.

\implies -1=a+b+ab

.

\implies -1=a+b(1+a)

.

\implies b=\frac{-1-a}{1+a}

.

\implies b=-1(\frac{1+a}{1+a})

.

\implies b=-1

.
Similarly,

-1=a*b \implies a=-1

. Thus,

a*b=-1 \iff a=b=-1

. Since

-1 \notin S

, we know

S

is closed.
Yes, S is closed under

*

. ✓

Inverses?
The inverse of an element

a \in S

is an element

a^{-1} \in S

such that

a*a^{-1}=a^{-1}*a=\varepsilon

. Our identity element is

\varepsilon=0

. Consider arbitrary

a \in S

.
Let

a*a^{-1}=0

.

\implies a+a^{-1}+aa^{-1}=0

.

\implies a^{-1}(1+a)=-a

.

\implies a^{-1}=\frac{-a}{1+a}

.
This is well defined

\forall a\in S

since

-1 \notin S

. Therefore,

\forall a\in S

,

\exists a^{-1} \in S

.
Yes, every element of S has an inverse. ✓

Associative?
Consider arbitrary

a,b,c \in S

.
First, let’s consider

a*(b*c)

.

a*(b*c)=a*(b+c+bc)

.

=a+b+c+bc+a(b+c+bc)

.

=(abc)+(ab+ac+bc)+(a+b+c)

.
Now, let’s consider

(a*b)*c

.

(a*b)*c=(a+b+ab)*c

.

=a+b+ab+c+(a+b+ab)c

.

=(abc)+(ab+ac+bc)+(a+b+c)

.
Thus,

\forall a,b,c\in S[latex]: [latex]a*(b*c)=(a*b)*c

.
Yes,

(S,*)

is associative. ✓

We can conclude that

(S,*)

is indeed a group.

To show that the group is abelian we must show that

a*b=b*a

for all

a,b \in S

.
Consider arbitrary

a,b \in S

.
Since addition and multiplication are commutative on the real numbers, we have:

a*b=a+b+ab=b+a+ba=b*a

.

Therefore,

(S,*)

is an abelian group. □

8. Give an example of two elements $\boldsymbol{A}$ and $\boldsymbol{B}$ in $\boldsymbol{GL_2(\mathbb{R})}$ with $\boldsymbol{AB \neq BA}$ .

Let

A = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}

and

B = \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}

.
Then:

AB = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}

.
However:

BA = \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}

.

AB \neq BA

. □

9. Prove that the product of two matrices in $\boldsymbol{SL_2 ( \mathbb{R} )}$ has determinant one.

Recall that

SL_2 ( \mathbb{R} )

is the group of linear transformations on

\mathbb{R}^{2}

that preserve area. This implies that

det(A)=1, \forall A \in SL_2 ( \mathbb{R} )

.
Consider arbitrary

A,B \in SL_2 ( \mathbb{R} )

.

det(A)=det(B)=1

.

det(A)det(B)=det(AB) \implies det(AB)=1 \implies (AB) \in SL_2 ( \mathbb{R} )

. □

10. $\boldsymbol{ H_3 = \{ \begin{pmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{pmatrix} : x,y,z \in \mathbb{R} \} }$
is known as the Heisenberg group. It is important in quantum physics. Matrix multiplication in the Heisenberg group is defined by:
$\boldsymbol{ A A' = \begin{pmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & x' & y'\\ 0 & 1 & z'\\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & x+x' & y+y'+xz'\\ 0 & 1 & z+z'\\ 0 & 0 & 1 \end{pmatrix} }$
Prove that $\boldsymbol{ H_3}$ under matrix multiplication is a group.

Closed?
Consider arbitrary

A, A' \in H_3

.

x,y,z,x',y',z' \in \mathbb{R} \implies (x+x'),(y+y'+xz'),(z+z') \in \mathbb{R}

.

\implies \begin{pmatrix} 1 & x+x' & y+y'+xz'\\ 0 & 1 & z+z'\\ 0 & 0 & 1 \end{pmatrix} \in H_3

.
Therefore,

H_3

is closed. ✓

Identity?
Let

\varepsilon = I_3 = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}

.

\forall A \in \mathbb{R}^{3 \times 3}: A I_3 = I_3 A = A

.
Yes, the identity element is

I_3

. ✓

Inverses?
Recall that a square matrix A is invertible

\iff

its columns are linearly independent.
Consider an arbitrary element

A = \begin{pmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{pmatrix} \in H_3

.
Since

A

has linearly independent columns, it follows that any element

A \in H_3

is invertible.

A^{-1} = \begin{pmatrix} 1 & -x & -(y-xz)\\ 0 & 1 & -z\\ 0 & 0 & 1 \end{pmatrix}

.

\implies A^{-1} \in H_3

.
Yes, every element in

H_3

has an inverse in

H_3

. ✓

Associative?
Matrix multiplication is associative in

\mathbb{R}^{3 \times 3}

.
Yes, matrix multiplication is associative in

H_3

. ✓

Therefore,

H_3

is a group. □

11. Prove that $\boldsymbol{ det(AB) = det(A)det(B)}$ in $\boldsymbol{GL_2(\mathbb{R})}$ . Use this result to show that the binary operation in the group $\boldsymbol{GL_2(\mathbb{R})}$ is closed; that is, if $\boldsymbol{A}$ and $\boldsymbol{B}$ are in $\boldsymbol{GL_2(\mathbb{R})}$ , then $\boldsymbol{AB \in GL_2(\mathbb{R})}$ .

GL_2 (\mathbb{R})

is the group of invertible

2 \times 2

matrices with real entries.

GL_2 (\mathbb{R}) = \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix} : ad-bc \neq 0 \}

.

Consider arbitrary

A,B \in GL_2 (\mathbb{R})

.
Let

A= \begin{pmatrix} a & b\\ c & d \end{pmatrix}

,

B= \begin{pmatrix} w & x\\ y & z \end{pmatrix}

.

det(A)=ad-bc

.

det(B)=wz-xy

.

det(A)det(B)=(ad-bc)(wz-xy)=adwz-adxy-bcwz+bcxy

.

Now, let's find

det(AB)

.

AB=\begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} w & x\\ y & z \end{pmatrix} = \begin{pmatrix} aw+by & ax+bz\\ cw+dy & cx+dz \end{pmatrix}

.

det(AB)=(aw+by)(cx+dz)-(ax+bz)(cw+dy)

.

det(AB)=acwx+adwz+bcxy+bdyz-(acwx+adxy+bcwz+bdyz)

.

det(AB)=adwz-adxy-bcwz+bcxy

.

det(AB)=det(A)det(B)

.

Now that we have established that

det(AB)=det(A)det(B)

:

det(A) \neq 0

and

det(B) \neq 0

\implies det(A)det(B) \neq 0

.

\implies det(AB) \neq 0

.

\implies (AB) \in GL_2(\mathbb{R})

.
Therefore,

GL_2(\mathbb{R})

is closed under matrix multiplication. □

12. Let $\boldsymbol{\mathbb{Z}_2^n= \{ (a_1, a_2, ..., a_n): a_i \in \mathbb{Z}_2 \} }$ .
Define binary operation $\boldsymbol{+}$ by:
$\boldsymbol{(a_1, a_2, ..., a_n)+(b_1, b_2, ..., b_n)=(a_1+b_1, a_2+b_2, ..., a_n+b_n)}$
Show that $\boldsymbol{\mathbb{Z}_2^n}$ is a group under this operation. This group is important in algebraic coding theory. Recall that $\boldsymbol{\mathbb{Z}_2= \{ 0,1 \}}$ .

Closed?
Consider arbitrary

a,b \in \mathbb{Z}_2^n

.
Addition modulo 2 is closed, meaning that

\forall a_i,b_i \in \mathbb{Z}_2: a_i+b_i \in \mathbb{Z}_2

.
Therefore, every component

a_i+b_i

in the vector

a+b=(a_1+b_1, a_2+b_2, ..., a_n+b_n)

belongs to

\mathbb{Z}_2

.

\implies a+b \in \mathbb{Z}_2^n

.
Yes,

\mathbb{Z}_2^n

is closed. ✓

Identity?
Let

\varepsilon = 0_n

and consider arbitrary

a \in \mathbb{Z}_2^n

.

a+ \varepsilon = (a_1, a_2..., a_n)+0_n = (a_1+0, a_2+0,..., a_n+0) = (a_1,a_2,...,a_n)=a

.
Yes,

0_n

is the identity element. ✓

Inverses?
We want

\forall a \in \mathbb{Z}_2^n : \exists a^{-1} \in \mathbb{Z}_2^n

such that

a^{-1}a=\varepsilon

.
Consider arbitrary

a=(a_1,a_2,...,a_n) \in \mathbb{Z}_2^n

.
Let

a^{-1}=a

.
Then:

a^{-1}+a=(a_1+a_1, a_2+a_2, ..., a_n+a_n)

.

a^{-1}+a=(2a_1,2a_2, ..., 2a_n)

.

a^{-1}+a=(0 \pmod{2},0 \pmod{2}, ..., 0 \pmod{2})

.

a^{-1}+a=\varepsilon

.
Every element is its own inverse.
Yes, every element has an inverse in

\mathbb{Z}_2^n

. ✓

Associative?
We know that addition modulo 2 is associative on

\mathbb{Z}_2

.
Consider arbitrary

a,b,c \in \mathbb{Z}_2^n

.

(a+b)+c=(a_1+b_1, ..., a_n+b_n)+(c_1, ..., c_n)=(a_1+b_1+c_1, ..., a_n+b_n+c_n)

.

a+(b+c)=(a_1, ..., a_n)+(b_1+c_1, ..., b_n+c_n)=(a_1+b_1+c_1, ..., a_n+b_n+c_n)

.

\implies (a+b)+c=a+(b+c)

.
Yes,

+

is associative. ✓

Therefore,

\mathbb{Z}_2^n

is a group under this operation. □

13. Show that $\boldsymbol{\mathbb{R}^*= \mathbb{R} \backslash \{0\}}$ is a group under the operation of multiplication.

Identity?
Let

\varepsilon = 1

and consider arbitrary

x \in \mathbb{R}^*

.

\forall x \in \mathbb{R}^*: x \varepsilon = \varepsilon x = x

.
Yes, 1 is the identity element. ✓

Closed?
Multiplication is closed on the real numbers.
We only have to check if

0=xy

for any

x, y

.
Suppose

0=xy

, where

x, y \in \mathbb{R}^*

.

0=xy \implies x=0

and/or

y=0

.

0 \notin \mathbb{R}^* \implies \nexists x,y \in \mathbb{R}^*

such that

xy=0

.
Yes, closed. ✓

Inverses?
Consider arbitrary

x \in \mathbb{R}^*

. Let

x^{-1} = \frac{1}{x}

.
Then:

xx^{-1}=x \frac{1}{x}=1=\varepsilon

.

x{^1}=\frac{1}{x}

is well-defined

\forall x \in \mathbb{R}^*

since

0 \notin \mathbb{R}^*

.
Yes, every element has an inverse. ✓

Associative?
Multiplication is associative over the real numbers.
Yes, multiplication is associative. ✓

Therefore,

\mathbb{R}^*

is a group. □

14. Given the groups $\boldsymbol{\mathbb{R}^*}$ and $\boldsymbol{\mathbb{Z}}$ , let $\boldsymbol{G = \mathbb{R}^* \times \mathbb{Z}}$ . Define a binary operation $\boldsymbol{\circ}$ on $\boldsymbol{G}$ by: $\boldsymbol{(a, m) \circ (b, n) = (ab, m + n)}$ . Show that $\boldsymbol{G}$ is a group under this operation.

Identity?
Let

\varepsilon = (1,0)

. Consider arbitrary

(a,m) \in G

.

(a,m) \circ (1,0) = (a1, m+0)= (a,m)

.
Yes,

(1,0)

is the identity element. ✓

Closed?
Consider arbitrary

(a,m), (b,n) \in G

. Recall that

a,b \in \mathbb{R}^*

and

m,n \in \mathbb{Z}

.
G is closed if

ab \in \mathbb{R}^*

and

(m+n) \in \mathbb{Z}

.

\forall a,b, \in \mathbb{R}^*: ab \in \mathbb{R}^*

, as demonstrated in the previous question.

(n+m) \in \mathbb{Z}

since addition is closed on the integers.
Yes, G is closed under

\circ

. ✓

Inverses?
Consider arbitrary

(a,m) \in G

.
Let

(a,m)^{-1} = ( \frac{1}{a}, -n)

.
Then:

(a,m) \circ (a,m)^{-1} = (a,m) \circ ( \frac{1}{a}, -m) = (a \frac{1}{a}, m-m) = (1,0) = \varepsilon

.

(a,m)^{-1} = ( \frac{1}{a}, -m)

is well-defined for all

(a,m) \in G

.
Yes, every element has an inverse. ✓

Associative?
Consider arbitrary

(a,m),(b,n),(x,y) \in G

.

(a,m) \circ [(b,n)\circ(x,y)] = (a,m) \circ (bx, n+y) = (abx, m+n+y)

.

[(a,m) \circ (b,n)]\circ(x,y) = (ab,m+n) \circ (x, y) = (abx, m+n+y)

.

\implies (a,m) \circ [(b,n)\circ(x,y)] = [(a,m) \circ (b,n)]\circ(x,y)

.
Yes,

\circ

is associative on

G

. ✓

Therefore,

G

is a group. □

15. Prove or disprove that every group containing six elements is abelian.

There are two distinct groups of order 6: the cyclic group

\mathbb{Z}_6

and the symmetric group

S_3

.

\mathbb{Z}_6

is abelian, as are all cyclic groups. However,

S_3

is not abelian. Recall that

S_3

is the group of all permutations of 3 objects.

We will disprove that every group containing six elements is abelian via a counterexample:

Consider

S_3

and

(1 \;  3 \;  2), (2 \;  1 \;  3) \in S_3

.

(1 \; 3 \; 2) \circ (2 \; 1 \; 3) = (3 \; 1 \; 2)

.

(2 \; 1 \; 3) \circ (1 \; 3 \; 2) = (2 \; 3 \; 1)

.

\implies (1 \; 3 \; 2) \circ (2 \; 1 \; 3) \neq (2 \; 1 \; 3) \circ (1 \; 3 \; 2)

.

It is not true that every group containing six elements is abelian. □

17. Give an example of three different groups with eight elements. Why are the groups different?

\mathbb{Z}_8

is the group of addition modulo 8. It is cyclic, and therefore abelian.

D_8

, the dihedral group of order 8, is the group of symmetries of a square. It is non-abelian, and non-cyclic.

\mathbb{Z}_4 \times \mathbb{Z}_2

is abelian and non-cyclic.

18. Show that there are $\boldsymbol{n!}$ permutations of a set containing $\boldsymbol{n}$ items.

A permutation is a specific ordering of elements. Suppose we have a set of elements

S=\{ x_1, x_2, ..., x_n \}

and we are given the task of ordering them any manner we wish. We begin to order them: for the first element we have

n

options (since there are

n

elements), for the second element we have

(n-1)

options, for the third we have

(n-2)

options, and so on, until eventually for the

n

th element we have only 1 option. Hence, the total number of ways to order the elements is:

n(n-1)(n-2) \cdot \cdot \cdot (2)(1) = n!

Therefore, a set containing

n

elements has

n!

permutations. □

21. For each $\boldsymbol{a \in \mathbb{Z}_n}$ find an element $\boldsymbol{b \in \mathbb{Z}_n}$ such that: $\boldsymbol{a + b \equiv b + a \equiv 0 \pmod{n}}$ .

Note that

b

has been defined as the inverse of

a

; that is,

b=a^{-1}

.
Consider arbitrary

a \in \mathbb{Z}_n

and let

a^{-1}=b=n-a

.
Then

a+b=a+n-a=n \equiv 0 \pmod{n}

. □

24. Show that multiplication distributes over addition modulo $\boldsymbol{n}$ :
$\boldsymbol{a(b + c) \equiv ab + ac \pmod{n}}$ .

We must show that

a(b + c) \pmod{n} = ab + ac \pmod{n}

,

\forall a,b,c \in \mathbb{Z_n}

.

First, we notice:

a,b,c \in \mathbb{Z}_n \implies a,b,c \in \mathbb{Z}

.
Let's view the product

a(b+c)

as an element of the integers.
The distributive property holds over the integers.
Thus,

a(b+c)=ab + ac=x

, where

x \in \mathbb{Z}

.

Next we make use of Theorem 2.9 (Division Algorithm):

x=\alpha n + r

, where

\alpha

and

r

are unique integers,

n>0

, and

0 \leq r < \alpha

.

Now, let's consider

x

as evaluated in

\mathbb{Z}_n

:

x=\alpha n + r \implies x \equiv r \pmod{n}

.

\implies a(b+c) \equiv r \pmod{n}

and

ab + ac \equiv r \pmod{n}

.
Therefore, multiplication distributes over addition modulo

n

. □

26. Let $\boldsymbol{U(n)}$ be the group of units in $\boldsymbol{\mathbb{Z}}$ . If $\boldsymbol{n > 2}$ , prove that there is an element $\boldsymbol{k \in U(n)}$ such that $\boldsymbol{k^2 = 1}$ and $\boldsymbol{k \neq 1}$ .

Consider

k=(n-1)

, where

n>2

:

\gcd(n,n-1)=1 \implies (n-1) \in U(n)

.

(n-1)^2=n^2-2n+1

.

(n-1)^2=n(n-2)+1

.

(n-1)^2 \equiv 1 \pmod{n}

. □

27. Prove that the inverse of $\boldsymbol{g_1 g_2 \cdot \cdot \cdot g_n}$ is $\boldsymbol{g^{-1}_n \cdot \cdot \cdot g^{-1}_2 g^{-1}_1}$ .

(g_1 g_2 \cdot \cdot \cdot g_n)(g^{-1}_n \cdot \cdot \cdot g^{-1}_1)

.

(g_1 g_2 \cdot \cdot \cdot g_{n-1} (g_n g^{-1}_n) g^{-1}_{n-1} \cdot \cdot \cdot g^{-1}_2 g^{-1}_1)

.

(g_1 g_2 \cdot \cdot \cdot g_{n-1} (\varepsilon) g^{-1}_{n-1} \cdot \cdot \cdot g^{-1}_2 g^{-1}_1)

.

(g_1 g_2 \cdot \cdot \cdot g_{n-2} (g_{n-1} g^{-1}_{n-1}) g_{n-2} \cdot \cdot \cdot g^{-1}_2 g^{-1}_1)

.

(g_1 g_2 \cdot \cdot \cdot g_{n-2} (\varepsilon) g^{-1}_{n-2} \cdot \cdot \cdot g^{-1}_2 g^{-1}_1)

.

And so on, until eventually:

g_1 g_2 g^{-1}_2 g^{-1}_1 = g_1 g^{-1}_1 = \varepsilon

. □

28. Prove the remainder of Proposition 3.21: if $\boldsymbol{G}$ is a group and $\boldsymbol{a,b \in G}$ , then the equation $\boldsymbol{xa = b}$ has a unique solution in $\boldsymbol{G}$ .

First, let's show that solutions exist:
Let

xa=b

.

xa=b \implies xaa^{-1}=ba^{-1} \implies x=ba^{-1}

.

Next, let's show that the solution is unique:
Suppose

x_1

is a solution of

xa=b \implies x_1a=b

.
Suppose

x_2

is also a solution of

xa=b \implies x_2a=b

.

x_1a=b \implies x_1=ba^{-1}

.

x_2a=b \implies x_2=ba^{-1}

.

\implies x_1=x_2

. □

30. Prove the right and left cancellation laws for a group $\boldsymbol{G}$ ; that is, show that in the group $\boldsymbol{G}$ : $\boldsymbol{ba = ca \implies b = c}$ and $\boldsymbol{ab = ac \implies b = c}$ for any elements $\boldsymbol{a, b, c \in G}$ .

Consider arbitrary elements

a,b,c \in G

.
First, let's prove the right cancellation law:

ba=ca \implies baa^{-1}=caa^{-1} \implies b\varepsilon = c \varepsilon \implies b=c

.

Next, let's prove the left cancellation law:

ab=ac \implies aa^{-1}b=aa^{-1}c \implies \varepsilon b= \varepsilon c \implies b=c

. □

31. Show that if $\boldsymbol{a^2 = \varepsilon}$ for all elements $\boldsymbol{a}$ in a group $\boldsymbol{G}$ , then $\boldsymbol{G}$ must be abelian.

We must show that

\forall x,y \in G: xy=yx

.

a^2 = \varepsilon, \forall a \in G \implies a^{-1}=a

(i.e. every element is its own inverse).
Consider arbitrary

x,y \in G

.

(xy)^{-1}=xy

.

(xy)^{-1}=y^{-1}x^{-1}

(by Theorem 3.23).

x^{-1}=x

and

y^{-1}=y \implies (xy)^{-1}=y^{-1}x^{-1}=yx

.

xy=yx

. □

32. Show that if $\boldsymbol{G}$ is a finite group of even order, then there is an $\boldsymbol{a \in G}$ such that $\boldsymbol{a}$ is not the identity and $\boldsymbol{a^2 \neq \varepsilon}$ .

G=\{ \varepsilon, g_1, g_2, ..., g_n\}

, where n is an odd number.
Suppose there does not exist an

a \in G

such that

a

is not the identity and

a^2 \neq \varepsilon

.
Then every non-identity element

g_i

has an inverse

g_i^{-1} \in G

,

1 \leq i \leq n, g_i \neq g^{-1}_i

.
Every non-identity element and its inverse form a pair.
However, we have an odd number of elements. This is a contradiction. □

33. Let $\boldsymbol{G}$ be a group and suppose that $\boldsymbol{(ab)^2 = a^2b^2}$ for all $\boldsymbol{a}$ and $\boldsymbol{b}$ in $\boldsymbol{G}$ . Prove that $\boldsymbol{G}$ is an abelian group.

Consider arbitrary

a,b \in G

.

(ab)^2=a^2b^2=abab

.

abab=aabb

.

\implies (aba)b=(aab)b

.

\implies aba=aab

(by right cancellation).

\implies a(ba)=a(ab)

.

\implies ba=ab

(by left cancellation). □

Sean Lawlis

Ch 3. Groups – Solutions

3.4 Reading Questions

3.5 Exercises