Power Equations

After learning power laws and conjugation power formulas, we now apply them to group equations. These problems show how a small relation involving two elements can force a high power to become the identity. The key method is to rewrite conjugation expressions as powers, apply the same conjugation in two different ways, and then compare the results. In this lesson, students will solve representative power-equation problems and learn how to keep the order of factors correct in a non-commutative group.

The main tool is the following observation. If

a^2=e

, then

a^{-1}=a

. Therefore an expression such as

a\circ b^r\circ a

may be treated as

a\circ b^r\circ a^{-1}

, which is a conjugation. Conjugation respects powers:

\begin{equation*} a\circ b^{rn}\circ a^{-1}=(a\circ b^r\circ a^{-1})^n. \end{equation*}

This allows one relation involving

b^r

to produce relations involving higher powers of

b

The main tool is the following observation. If

a^2=e

, then

a^{-1}=a

. Therefore an expression such as

a\circ b^r\circ a

may be treated as

a\circ b^r\circ a^{-1}

, which is a conjugation. Conjugation respects powers:

\begin{equation*} a\circ b^{rn}\circ a^{-1}=(a\circ b^r\circ a^{-1})^n. \end{equation*}

This allows one relation involving

b^r

to produce relations involving higher powers of

b

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. If

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

and

\begin{equation*} a\circ b^2\circ a=b^3, \end{equation*}

prove that

\begin{equation*} b^5=e. \end{equation*}

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. Given that

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

and

\begin{equation*} a\circ b^2\circ a=b^3. \end{equation*}

To prove that

b^5=e

. Since

a^2=e

, we get

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

Therefore

\begin{equation*} a\circ b^2\circ a^{-1}=b^3. \end{equation*}

Apply conjugation by

a

to both sides. Since conjugation by

a

is its own inverse when

a^2=e

, we get

\begin{equation*} a\circ b^3\circ a^{-1}=b^2. \end{equation*}

Now compute

a\circ b^6\circ a^{-1}

in two ways. First,

\begin{align*} a\circ b^6\circ a^{-1} &=a\circ(b^2)^3\circ a^{-1}\\ &=(a\circ b^2\circ a^{-1})^3\\ &=(b^3)^3\\ &=b^9. \end{align*}

Second,

\begin{align*} a\circ b^6\circ a^{-1} &=a\circ(b^3)^2\circ a^{-1}\\ &=(a\circ b^3\circ a^{-1})^2\\ &=(b^2)^2\\ &=b^4. \end{align*}

Therefore

\begin{equation*} b^9=b^4. \end{equation*}

Multiplying both sides by

b^{-4}

, we get

\begin{align*} b^9=b^4 &\implies b^9\circ b^{-4}=b^4\circ b^{-4}\\ &\implies b^5=e. \end{align*}

Hence,

\begin{equation*} b^5=e. \end{equation*}

\square

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. If

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

and

\begin{equation*} a\circ b^4\circ a=b^7, \end{equation*}

prove that

\begin{equation*} b^{33}=e. \end{equation*}

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. Given that

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

and

\begin{equation*} a\circ b^4\circ a=b^7. \end{equation*}

To prove that

b^{33}=e

. Since

a^2=e

, we get

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

Therefore

\begin{equation*} a\circ b^4\circ a^{-1}=b^7. \end{equation*}

Apply conjugation by

a

to both sides. Since

a^{-1}=a

, conjugation by

a

is its own inverse. Hence,

\begin{equation*} a\circ b^7\circ a^{-1}=b^4. \end{equation*}

Now compute

a\circ b^{28}\circ a^{-1}

in two ways. First,

\begin{align*} a\circ b^{28}\circ a^{-1} &=a\circ(b^4)^7\circ a^{-1}\\ &=(a\circ b^4\circ a^{-1})^7\\ &=(b^7)^7\\ &=b^{49}. \end{align*}

Second,

\begin{align*} a\circ b^{28}\circ a^{-1} &=a\circ(b^7)^4\circ a^{-1}\\ &=(a\circ b^7\circ a^{-1})^4\\ &=(b^4)^4\\ &=b^{16}. \end{align*}

Therefore

\begin{equation*} b^{49}=b^{16}. \end{equation*}

Multiplying both sides by

b^{-16}

, we get

\begin{align*} b^{49}=b^{16} &\implies b^{49}\circ b^{-16}=b^{16}\circ b^{-16}\\ &\implies b^{33}=e. \end{align*}

Hence,

\begin{equation*} b^{33}=e. \end{equation*}

\square

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. If

\begin{equation*} a^4=e \end{equation*}

and

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

then

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

Given that

(G,\circ)

is a group with identity element

e

and

a,b\in G

. Given that

\begin{equation*} a^4=e \end{equation*}

and

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

To prove that

a=e

. From

a^2\circ b=b\circ a

, multiply on the right by

b^{-1}

to get

\begin{equation*} a^2=b\circ a\circ b^{-1}. \end{equation*}

Now

\begin{align*} b^2\circ a\circ b^{-2} &=b\circ(b\circ a\circ b^{-1})\circ b^{-1}\\ &=b\circ a^2\circ b^{-1}\\ &=(b\circ a\circ b^{-1})^2\\ &=(a^2)^2\\ &=a^4\\ &=e. \end{align*}

Multiplying

b^2\circ a\circ b^{-2}=e

on the left by

b^{-2}

and on the right by

b^2

, we get

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

Hence,

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

\square

The last theorem illustrates why conjugation is useful. The relation

a^2\circ b=b\circ a

is rewritten as

b\circ a\circ b^{-1}=a^2

. Conjugating again by

b

produces

b^2\circ a\circ b^{-2}

, and the right-hand side becomes

(a^2)^2=a^4=e

. Since conjugation of an element is

e

only when the original element is

e

, the conclusion

a=e

follows.

Power-equation problems require careful side control. Multiplying by an inverse on the wrong side may produce a false equation in a non-commutative group. In the theorem above, the equation

a^2\circ b=b\circ a

gives

a^2=b\circ a\circ b^{-1}

by multiplying on the right by

b^{-1}

. It does not give

a^2=b^{-1}\circ b\circ a

Use the calculator below to check the arithmetic behind the exponent comparisons in the first two solved problems. Choose two positive integers

r

and

s

. The calculator compares the two ways of computing the conjugate of

b^{rs}

and returns the exponent difference

s^2-r^2

. This models why equations such as

b^9=b^4

and

b^{49}=b^{16}

force a power of

b

to be the identity.

[1] If

a^2=e

and

a\circ b^3\circ a=b^5

, what is

a^{-1}

? [2] If

a^2=e

and

a\circ b^3\circ a=b^5

, prove that

a\circ b^5\circ a=b^3

. [3] If

a^2=e

and

a\circ b^3\circ a=b^5

, prove that

b^{16}=e

. [4] If

a^2=e

and

a\circ b\circ a=b^{-1}

, prove that

a\circ b^n\circ a=b^{-n}

. [5] In the theorem above, explain why

b^2\circ a\circ b^{-2}=e

implies

a=e

[1]

a^{-1}=a

. [2] Since conjugation by

a

is its own inverse, applying it to

a\circ b^3\circ a=b^5

gives

a\circ b^5\circ a=b^3

. [3] Compute

a\circ b^{15}\circ a

(a\circ b^3\circ a)^5=b^{25}

and also as

(a\circ b^5\circ a)^3=b^9

. Thus

b^{25}=b^9

, so

b^{16}=e

. [4] Since

a=a^{-1}

, use the conjugation power formula:

a\circ b^n\circ a=(a\circ b\circ a)^n=(b^{-1})^n=b^{-n}

. [5] Multiply on the left by

b^{-2}

and on the right by

b^2

to get

a=e

Because

a\circ a=e

, so

a

is its own inverse.

They convert conjugates of powers into powers of conjugates.

No. In a non-commutative group, the side of multiplication matters.

Compute the same conjugated power in two ways and compare the resulting powers.

BMLabs Mathematics

BMLabs Mathematics

Power Equations

Power Equations

POWER EQUATION EXPONENT TEST

ORDER OF AN ELEMENT

Coordinates and Locus

Invariants under Orthogonal Transformation