Non-Examples of Groups

After verifying several examples of groups, it is equally important to study non-examples. A non-example shows exactly how a group axiom can fail. This prevents students from thinking that every familiar set and operation automatically forms a group. To prove that a structure is not a group, it is enough to find one failed axiom. In this lesson, students will study common non-examples such as positive integers under addition, integers under multiplication, nonzero real numbers under addition, and invertible matrices under addition.

The fastest way to disprove a group claim is to locate a failed axiom. If closure fails, no further checking is needed. If closure and associativity hold but the identity element is missing, the structure is not a group. If the identity exists but some element has no inverse in the set, the structure is not a group. This method is rigorous because the definition of group requires all four axioms to hold at the same time.

Let

G

be a non-empty set and let

\circ

be a binary operation on

G

. If at least one group axiom fails for

(G,\circ)

, then

(G,\circ)

is not a group.

Given that

G

is a non-empty set and

\circ

is a binary operation on

G

. To prove that if at least one group axiom fails for

(G,\circ)

, then

(G,\circ)

is not a group. The definition of a group requires the following conditions: (i) closure, (ii) associativity, (iii) existence of an identity element, (iv) existence of inverse elements for all elements. If possible let

(G,\circ)

be a group and at least one group axiom fails. Since

(G,\circ)

is a group, all four group axioms must hold. Therefore no group axiom can fail. A contradiction. Hence, if at least one group axiom fails for

(G,\circ)

, then

(G,\circ)

is not a group.

\square

Let

\mathbb{N}=\{1,2,3,\ldots\}

. Then

(\mathbb{N},+)

is not a group. Closure holds because the sum of two positive integers is a positive integer. Associativity also holds because addition of integers is associative. However, an additive identity would have to be

0

, and

\begin{equation*} 0\notin\mathbb{N}. \end{equation*}

Therefore the identity axiom fails. Hence,

(\mathbb{N},+)

is not a group.

The structure

(\mathbb{Z},\cdot)

is not a group under ordinary multiplication. The identity element is

1

, and multiplication is associative. However, the inverse axiom fails. Take

2\in\mathbb{Z}

. If

2

had an inverse in

\mathbb{Z}

, then there would exist

b\in\mathbb{Z}

such that

\begin{equation*} 2b=1. \end{equation*}

Then

\begin{align*} 2b=1 &\implies b=\frac{1}{2}. \end{align*}

But

\dfrac{1}{2}\notin\mathbb{Z}

. Therefore

2

has no multiplicative inverse in

\mathbb{Z}

. Hence,

(\mathbb{Z},\cdot)

is not a group.

The structure

(\mathbb{R}-\{0\},+)

is not a group under ordinary addition. Take

1,-1\in\mathbb{R}-\{0\}

. Then

\begin{align*} 1+(-1)&=0. \end{align*}

But

0\notin\mathbb{R}-\{0\}

. Therefore closure fails. Hence,

(\mathbb{R}-\{0\},+)

is not a group.

Let

\mathbb{GL}(2,\mathbb{R})

be the set of all invertible

2\times2

real matrices. Then

\mathbb{GL}(2,\mathbb{R})

is not a group under matrix addition. Take

\begin{equation*} A=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad B=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}. \end{equation*}

Both

A

and

B

are invertible. But

\begin{align*} A+B&=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\\ &=\begin{pmatrix}0&0\\0&0\end{pmatrix}. \end{align*}

The zero matrix is not invertible. Therefore

A+B\notin\mathbb{GL}(2,\mathbb{R})

. Thus closure fails. Hence,

\mathbb{GL}(2,\mathbb{R})

is not a group under matrix addition.

The previous example is a common source of confusion. The set

\mathbb{GL}(2,\mathbb{R})

is very important in algebra, but its usual group operation is matrix multiplication, not matrix addition. Under multiplication, the product of two invertible matrices is invertible, the identity matrix acts as identity, and each invertible matrix has a multiplicative inverse. Under addition, closure fails. This shows again that the set and operation must always be specified together.

Determine whether

(\mathbb{Z}-\{0\},+)

is a group.

Let

G=\mathbb{Z}-\{0\}

and let

+

be ordinary addition. To determine whether

(G,+)

is a group. Take

1,-1\in G

. Then

\begin{align*} 1+(-1)&=0. \end{align*}

But

0\notin G

. Therefore

G

is not closed under addition. Since closure fails,

(\mathbb{Z}-\{0\},+)

is not a group.

\square

Determine whether the set

G=\{1,-1\}

is a group under ordinary multiplication.

Let

G=\{1,-1\}

and let

\cdot

be ordinary multiplication. To determine whether

(G,\cdot)

is a group. [1] To prove closure. The products of elements of

G

are

\begin{equation*} 1\cdot1=1,\quad 1\cdot(-1)=-1,\quad (-1)\cdot1=-1,\quad (-1)\cdot(-1)=1. \end{equation*}

Therefore

G

is closed under multiplication. [2] To prove associativity. Multiplication of real numbers is associative. Therefore

\begin{equation*} (ab)c=a(bc) \quad \forall\, a,b,c\in G. \end{equation*}

[3] To prove identity element. There exists

1\in G

such that

\begin{equation*} a\cdot1=1\cdot a=a \quad \forall\, a\in G. \end{equation*}

[4] To prove inverse elements. The inverse of

1

1

, and the inverse of

-1

-1

, since

\begin{equation*} 1\cdot1=1,\quad (-1)\cdot(-1)=1. \end{equation*}

Hence,

(G,\cdot)

is a group.

\square

The last solved problem is included to prevent overgeneralization. Not every small subset fails to be a group. The set

\{1,-1\}

is a group under multiplication because it is closed, contains the identity, and each element is its own inverse. A correct conclusion must come from the axioms, not from the size or appearance of the set.

Use the calculator below to test whether a finite set of integers is closed under addition or multiplication. A failed closure test immediately proves that the selected operation does not give a group on that set. Try the set

1,-1

with multiplication, and then try the same set with addition.

[1] Determine whether

(\mathbb{N},+)

is a group, where

\mathbb{N}=\{1,2,3,\ldots\}

. [2] Determine whether

(\mathbb{Q},\cdot)

is a group. [3] Determine whether

(\mathbb{R}-\{0\},+)

is a group. [4] Determine whether

\mathbb{GL}(2,\mathbb{R})

is a group under addition. [5] Determine whether

\{1,-1\}

is a group under multiplication.

[1] No. The additive identity

0

is not in

\mathbb{N}

. [2] No. The element

0

has no multiplicative inverse. [3] No. Closure fails because

1+(-1)=0

and

0\notin\mathbb{R}-\{0\}

. [4] No. The sum of two invertible matrices may be non-invertible. [5] Yes. It is closed under multiplication, has identity

1

, and each element is its own inverse.

One failed axiom is enough.

No. For example,

2

has no multiplicative inverse in

\mathbb{Z}

No. It is not closed under addition.

Yes. The operation is part of the structure.

BMLabs Mathematics

BMLabs Mathematics

Non-Examples of Groups

Non-Examples of Groups

NON-EXAMPLE TEST

ELEMENTARY PROPERTIES OF GROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation