Klein Four Group

After studying alternating groups, we now finish this section with one of the most useful small examples in group theory: the Klein four group. This group has four elements, every non-identity element has order

2

, and the group is abelian. It is small enough to understand completely from its operation table, but important enough to appear repeatedly in finite group theory, symmetry, and direct product constructions. In this lesson, students will verify the group axioms for the Klein four group and compare it with cyclic groups of order four.

The

\textbf{Klein four group}

is the set

\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}

with operation

\circ

defined by the table

\begin{equation*} \begin{array}{c|cccc} \circ & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & e & c & b\\ b & b & c & e & a\\ c & c & b & a & e \end{array} \end{equation*}

2

The

\textbf{Klein four group}

is the set

\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}

with operation

\circ

defined by the table

\begin{equation*} \begin{array}{c|cccc} \circ & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & e & c & b\\ b & b & c & e & a\\ c & c & b & a & e \end{array} \end{equation*}

The table gives the entire operation. The first row and first column show that

e

behaves as an identity element. The entries

a\circ a=e

b\circ b=e

, and

c\circ c=e

show that each non-identity element is its own inverse. The table is symmetric across the main diagonal, so the operation is commutative. The only group axiom that is not visually immediate from the table is associativity; for this example, associativity may be verified directly from the table or recognized from a standard model of

V_4

The structure

(V_4,\circ)

defined by the table above is a finite abelian group.

Given that

\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}

with the operation defined by the stated table. To prove that

(V_4,\circ)

is a finite abelian group. [1] To prove closure. Every entry in the operation table belongs to

V_4

. Therefore

V_4

is closed under

\circ

. [2] To prove associativity. We use the standard model

\begin{equation*} V_4=\mathbb{Z}_2\times\mathbb{Z}_2 \end{equation*}

with componentwise addition modulo

2

, under the correspondence

\begin{equation*} e=(\overline{0},\overline{0}),\quad a=(\overline{1},\overline{0}),\quad b=(\overline{0},\overline{1}),\quad c=(\overline{1},\overline{1}). \end{equation*}

The table is exactly the componentwise addition table under this correspondence. Since addition modulo

2

is associative in each coordinate, componentwise addition is associative. Therefore

\circ

is associative on

V_4

. [3] To prove the existence of identity element. From the table,

\begin{equation*} e\circ x=x\circ e=x \quad \forall\, x\in V_4. \end{equation*}

Therefore

e

is the identity element. [4] To prove the existence of inverse elements. From the table,

\begin{equation*} e\circ e=e,\quad a\circ a=e,\quad b\circ b=e,\quad c\circ c=e. \end{equation*}

Thus every element is its own inverse. [5] To prove commutativity. The table is symmetric, so

\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in V_4. \end{equation*}

Therefore

V_4

is abelian. Since

V_4

has four elements, it is finite. Hence,

(V_4,\circ)

is a finite abelian group.

\square

The proof of associativity is often the delicate part. Checking all triples directly would require

4^3=64

computations. The model

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

gives a cleaner verification. Under componentwise addition modulo

2

, associativity is inherited from addition modulo

2

. This also explains why every non-identity element has order

2

V_4

, the order of

a

2

because

\begin{equation*} a^2=a\circ a=e \end{equation*}

and

a\neq e

. Similarly,

\begin{equation*} o(b)=2,\quad o(c)=2. \end{equation*}

The identity element has order

1

The Klein four group is not cyclic. If

V_4

were cyclic, it would have an element of order

4

. But the orders of the elements are

\begin{equation*} o(e)=1,\quad o(a)=2,\quad o(b)=2,\quad o(c)=2. \end{equation*}

No element has order

4

. Therefore

V_4

is not cyclic.

Using the table of

V_4

, compute

a\circ b

b\circ c

, and

c\circ a

Given the Klein four group table. To compute

a\circ b

b\circ c

, and

c\circ a

. From the row of

a

and column of

b

, we get

\begin{equation*} a\circ b=c. \end{equation*}

From the row of

b

and column of

c

, we get

\begin{equation*} b\circ c=a. \end{equation*}

From the row of

c

and column of

a

, we get

\begin{equation*} c\circ a=b. \end{equation*}

Hence,

\begin{equation*} a\circ b=c,\quad b\circ c=a,\quad c\circ a=b. \end{equation*}

\square

Prove that every non-identity element of

V_4

has order

2

Let

\begin{equation*} V_4=\{e,a,b,c\}. \end{equation*}

To prove that every non-identity element of

V_4

has order

2

. The non-identity elements are

a,b,c

. From the operation table,

\begin{equation*} a\circ a=e,\quad b\circ b=e,\quad c\circ c=e. \end{equation*}

Also,

\begin{equation*} a\neq e,\quad b\neq e,\quad c\neq e. \end{equation*}

Therefore each of

a,b,c

has least positive power

2

equal to the identity. Thus

\begin{equation*} o(a)=o(b)=o(c)=2. \end{equation*}

Hence, every non-identity element of

V_4

has order

2

\square

There are two common groups of order

4

: the cyclic group of order

4

and the Klein four group. They are not the same type of group. A cyclic group of order

4

has an element of order

4

. The Klein four group has no element of order

4

. This distinction will become important when group isomorphism is introduced.

Use the calculator below to multiply elements in the Klein four group. Choose two elements and read the product from the table. The calculator also reports the order of the product.

[1] Write the elements of

V_4

. [2] What is the identity element of

V_4

? [3] Find the inverse of

a

V_4

. [4] Find

b\circ c

V_4

. [5] Prove that

V_4

is not cyclic.

[1]

V_4=\{e,a,b,c\}

. [2] The identity element is

e

. [3] The inverse of

a

a

. [4]

b\circ c=a

. [5] The orders are

1,2,2,2

, so no element has order

4

. Hence

V_4

is not cyclic.

It has four elements.

Yes. Its operation table is symmetric.

Each non-identity element has order

2

No. It has no element of order

4

BMLabs Mathematics

BMLabs Mathematics

Klein Four Group

Klein Four Group

KLEIN FOUR GROUP

KLEIN FOUR PRODUCT

CYCLIC GROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation