Order of the Inverse

After defining finite and infinite order, we now compare the order of an element with the order of its inverse. Since inverse elements undo the original element, it is natural to expect that they return to the identity after the same number of repetitions. This is correct. If

a

has finite order

n

, then

a^{-1}

also has order

n

. If

a

has infinite order, then

a^{-1}

also has infinite order. In this lesson, students will prove the equality

o(a)=o(a^{-1})

and use it in examples.

Let

(G,\circ)

be a group with identity element

e

and let

a\in G

. Then

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

a

has finite order

n

, then

a^{-1}

also has order

n

. If

a

has infinite order, then

a^{-1}

also has infinite order. In this lesson, students will prove the equality

o(a)=o(a^{-1})

and use it in examples.

Let

(G,\circ)

be a group with identity element

e

and let

a\in G

. Then

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

Given that

(G,\circ)

is a group with identity element

e

and

a\in G

. To prove that

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

First suppose that

o(a)=n

, where

n\in\mathbb{N}

. Then

\begin{equation*} a^n=e \end{equation*}

and

n

is the least positive integer with this property. Taking inverses on both sides,

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

Therefore

a^{-1}

has finite order, and

\begin{equation*} o(a^{-1})\leq n. \end{equation*}

Let

o(a^{-1})=m

. Then

\begin{equation*} (a^{-1})^m=e. \end{equation*}

Taking inverses on both sides,

\begin{align*} (a^{-1})^m=e &\implies \big((a^{-1})^m\big)^{-1}=e^{-1}\\ &\implies a^m=e. \end{align*}

Since

o(a)=n

, the least positive exponent giving

e

n

. Hence,

\begin{equation*} n\leq m. \end{equation*}

Also

m=o(a^{-1})\leq n

. Therefore

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

Thus

o(a^{-1})=o(a)

when

o(a)

is finite. Now suppose that

o(a)

is infinite. If possible let

o(a^{-1})=k

for some

k\in\mathbb{N}

. Then

\begin{equation*} (a^{-1})^k=e. \end{equation*}

Taking inverses gives

\begin{equation*} a^k=e. \end{equation*}

This contradicts that

a

has infinite order. Therefore

a^{-1}

has infinite order. Hence,

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

\square

The proof has two parts because order may be finite or infinite. In the finite case, the argument shows two inequalities and then concludes equality. In the infinite case, the proof uses contradiction. A common mistake is to prove only that

(a^{-1})^n=e

and then immediately claim the order is

n

. That is incomplete, because one must also prove that no smaller positive exponent works.

In the group

S=\{1,i,-1,-i\}

under multiplication, the inverse of

i

-i

. Since

\begin{equation*} i^4=1 \end{equation*}

and no smaller positive power of

i

1

, we have

o(i)=4

. Also,

\begin{equation*} (-i)^1=-i,\quad (-i)^2=-1,\quad (-i)^3=i,\quad (-i)^4=1. \end{equation*}

Thus

\begin{equation*} o(-i)=4. \end{equation*}

Hence,

\begin{equation*} o(i)=o(i^{-1}). \end{equation*}

(\mathbb{Z},+)

, the inverse of

5

-5

. Since

5

has infinite order under addition, the theorem says that

-5

also has infinite order. Indeed, there is no positive integer

n

such that

\begin{equation*} n(-5)=0. \end{equation*}

Hence,

o(5)=o(-5)=\infty

(\mathbb{Z},+)

Let

(G,\circ)

be a group and let

a\in G

. If

\begin{equation*} o(a)=6, \end{equation*}

find

o(a^{-1})

Let

(G,\circ)

be a group and let

a\in G

. Given that

\begin{equation*} o(a)=6. \end{equation*}

To find

o(a^{-1})

. By the theorem,

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

Therefore

\begin{equation*} o(a^{-1})=6. \end{equation*}

Hence,

o(a^{-1})=6

\square

Let

(G,\circ)

be a group and let

a\in G

. If

\begin{equation*} o(a^{-1})=9, \end{equation*}

find

o(a)

Let

(G,\circ)

be a group and let

a\in G

. Given that

\begin{equation*} o(a^{-1})=9. \end{equation*}

To find

o(a)

. Since

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

we get

\begin{equation*} o(a)=9. \end{equation*}

Hence,

o(a)=9

\square

The equality

o(a)=o(a^{-1})

is often used silently in later problems. For example, if an element has order

2

, then

a^{-1}=a

because

a^2=e

. If an element has order

3

, then

a^{-1}=a^2

, and both

a

and

a^2

have order

3

. The theorem gives a reliable way to compare an element and its inverse without recomputing everything.

Use the calculator below to compare the order of an element and the order of its inverse in the additive group modulo

n

. In additive notation, the inverse of

a

-a

. The calculator computes both orders and displays them side by side.

[1] State the theorem relating

o(a)

and

o(a^{-1})

. [2] If

o(a)=10

, find

o(a^{-1})

. [3] If

o(a^{-1})=15

, find

o(a)

. [4] If

a

has infinite order, prove that

a^{-1}

has infinite order. [5] In

\{1,i,-1,-i\}

, find

o((-i)^{-1})

[1] For every element

a

of a group,

o(a)=o(a^{-1})

. [2]

o(a^{-1})=10

. [3]

o(a)=15

. [4] If

(a^{-1})^k=e

for some

k\in\mathbb{N}

, then taking inverses gives

a^k=e

, contradicting infinite order. [5] Since

(-i)^{-1}=i

and

o(i)=4

, we get

o((-i)^{-1})=4

Yes. In any group,

o(a)=o(a^{-1})

Then

a^{-1}

also has infinite order.

No. To prove the order is

n

, one must also prove that no smaller positive exponent gives the identity.

o(a)=2

, then

a^{-1}=a

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