Even Cyclic Groups

After studying prime-order cyclic groups, we now examine a special feature of cyclic groups of even order. If a cyclic group has order

2m

, then the element halfway around the cycle, namely

a^m

, has order

2

. More strongly, it is the only element of order

2

. This result is important because elements of order

2

appear in symmetry, permutation groups, and later subgroup arguments. In this lesson, students will prove the uniqueness of the element of order

2

in an even cyclic group.

Let

(G,\circ)

be a cyclic group with identity element

e

. If

\begin{equation*} |G|=2m \end{equation*}

for some

m\in\mathbb{N}

, then

G

has exactly one element of order

2

After studying prime-order cyclic groups, we now examine a special feature of cyclic groups of even order. If a cyclic group has order

2m

, then the element halfway around the cycle, namely

a^m

, has order

2

. More strongly, it is the only element of order

2

. This result is important because elements of order

2

appear in symmetry, permutation groups, and later subgroup arguments. In this lesson, students will prove the uniqueness of the element of order

2

in an even cyclic group.

Let

(G,\circ)

be a cyclic group with identity element

e

. If

\begin{equation*} |G|=2m \end{equation*}

for some

m\in\mathbb{N}

, then

G

has exactly one element of order

2

Given that

(G,\circ)

is a cyclic group with identity element

e

and

|G|=2m

for some

m\in\mathbb{N}

. To prove that

G

has exactly one element of order

2

. Since

G

is cyclic, there exists

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

Since

|G|=2m

, we get

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

Now

\begin{align*} (a^m)^2 &=a^{2m}\\ &=e. \end{align*}

Also

\begin{equation*} a^m\neq e, \end{equation*}

because

m<2m

and

o(a)=2m

. Therefore

\begin{equation*} o(a^m)=2. \end{equation*}

Thus

G

has at least one element of order

2

. Now let

x\in G

such that

\begin{equation*} o(x)=2. \end{equation*}

Since

G=\langle a\rangle

, there exists

r\in\mathbb{Z}

such that

\begin{equation*} x=a^r. \end{equation*}

Since

o(x)=2

, we have

\begin{equation*} x^2=e. \end{equation*}

Therefore

\begin{align*} x^2=e &\implies (a^r)^2=e\\ &\implies a^{2r}=e. \end{align*}

Since

o(a)=2m

, the divisibility criterion gives

\begin{equation*} 2m\mid2r. \end{equation*}

Therefore

\begin{equation*} m\mid r. \end{equation*}

Thus there exists

q\in\mathbb{Z}

such that

\begin{equation*} r=mq. \end{equation*}

q

is even, then

q=2t

for some

t\in\mathbb{Z}

, and

\begin{align*} x &=a^r\\ &=a^{mq}\\ &=a^{2mt}\\ &=e, \end{align*}

which contradicts

o(x)=2

. Therefore

q

is odd. Then

q=2t+1

for some

t\in\mathbb{Z}

, and

\begin{align*} x &=a^{mq}\\ &=a^{m(2t+1)}\\ &=a^{2mt+m}\\ &=(a^{2m})^t\circ a^m\\ &=a^m. \end{align*}

Therefore every element of order

2

is equal to

a^m

. Hence,

G

has exactly one element of order

2

\square

The proof has two parts. First, it constructs the element

a^m

and shows that its square is the identity. Second, it proves uniqueness by writing any element of order

2

as a power

a^r

and forcing

r

to be congruent to

m

modulo

2m

. The cyclic assumption is essential because it lets every element be written as a power of one generator.

Let

G=\langle a\rangle

and

|G|=8

. Then

\begin{equation*} m=4. \end{equation*}

The unique element of order

2

\begin{equation*} a^4. \end{equation*}

Indeed,

\begin{equation*} (a^4)^2=a^8=e \end{equation*}

and

a^4\neq e

In the additive group

(\mathbb{Z}_{10},+_{10})

, the unique element of order

2

\overline{5}

. Indeed,

\begin{equation*} 2\overline{5}=\overline{10}=\overline{0} \end{equation*}

and

\overline{5}\neq\overline{0}

. No other element has additive order

2

Find the unique element of order

2

in a cyclic group

G=\langle a\rangle

of order

18

Let

G=\langle a\rangle

be a cyclic group of order

18

. To find the unique element of order

2

. Since

\begin{equation*} 18=2\cdot9, \end{equation*}

we have

m=9

. By the theorem, the unique element of order

2

\begin{equation*} a^9. \end{equation*}

Indeed,

\begin{align*} (a^9)^2 &=a^{18}\\ &=e. \end{align*}

Since

o(a)=18

, we have

a^9\neq e

. Hence, the unique element of order

2

a^9

\square

Let

(G,\circ)

be a cyclic group with only one generator. Prove that either

|G|=1

|G|=2

Let

(G,\circ)

be a cyclic group with only one generator. To prove that either

|G|=1

|G|=2

. Since

G

is cyclic, there exists

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

By the inverse generator theorem,

\begin{equation*} G=\langle a^{-1}\rangle. \end{equation*}

Since

G

has only one generator, we get

\begin{equation*} a=a^{-1}. \end{equation*}

Therefore

\begin{align*} a=a^{-1} &\implies a^2=e. \end{align*}

Thus

\begin{equation*} o(a)=1 \end{equation*}

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

o(a)=1

, then

a=e

, so

\begin{equation*} G=\langle a\rangle=\{e\}. \end{equation*}

Therefore

|G|=1

. If

o(a)=2

, then

\begin{equation*} G=\langle a\rangle=\{e,a\}. \end{equation*}

Therefore

|G|=2

. Hence, either

|G|=1

|G|=2

\square

An even cyclic group has exactly one element of order

2

, but a non-cyclic group may behave differently. The Klein four group has three elements of order

2

. This gives another way to see that

V_4

is not cyclic.

Use the calculator below to find the unique element of order

2

in a cyclic group of even order

n

. Enter an even integer

n

. The calculator returns the exponent

n/2

, so the element is

a^{n/2}

[1] State the theorem about elements of order

2

in an even cyclic group. [2] Find the unique element of order

2

in a cyclic group

\langle a\rangle

of order

14

. [3] Find the unique element of order

2

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

. [4] Explain why

V_4

is not cyclic using elements of order

2

. [5] If a cyclic group has only one generator, prove that its order is

1

2

[1] If

G=\langle a\rangle

and

|G|=2m

, then

a^m

is the unique element of order

2

. [2] The element is

a^7

. [3] The element is

\overline{6}

. [4] The group

V_4

has three elements of order

2

, while an even cyclic group has exactly one. [5] If

a

is the only generator, then

a^{-1}

is also a generator, so

a=a^{-1}

. Hence

a^2=e

, and

|G|=1

|G|=2

The unique element of order

2

a^m

Any element is a power

a^r

, and the condition

(a^r)^2=e

forces

r

to be congruent to

m

modulo

2m

for a non-identity element.

No. This theorem is for cyclic groups.

It has three elements of order

2

, but an even cyclic group has exactly one.

BMLabs Mathematics

BMLabs Mathematics

Even Cyclic Groups

Even Cyclic Groups

EVEN CYCLIC ORDER TWO

SUBGROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation