Equations in Groups

After proving cancellation laws and the idempotent property, we now study equations in groups. One of the strongest consequences of the group axioms is that equations of the forms

a\circ x=b

and

y\circ a=b

always have unique solutions. This result shows why groups are algebraic systems where division-like operations are always possible, even when the operation is not ordinary multiplication. In this lesson, students will learn how to solve left and right group equations, why the two formulas differ in non-commutative groups, and how uniqueness follows from cancellation laws.

Let

(G,\circ)

be a group and let

a,b\in G

. An equation in which the unknown element belongs to

G

and is combined with known group elements using

\circ

is called a

\textbf{group equation}

. Two basic forms are

\begin{equation*} a\circ x=b \end{equation*}

and

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

After proving cancellation laws and the idempotent property, we now study equations in groups. One of the strongest consequences of the group axioms is that equations of the forms

a\circ x=b

and

y\circ a=b

Let

(G,\circ)

be a group and let

a,b\in G

. An equation in which the unknown element belongs to

G

and is combined with known group elements using

\circ

is called a

\textbf{group equation}

. Two basic forms are

\begin{equation*} a\circ x=b \end{equation*}

and

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

In ordinary arithmetic, the equation

ax=b

is solved by multiplying by

a^{-1}

when

a\neq0

. In a group, the same idea works, but the side matters. To solve

a\circ x=b

, multiply by

a^{-1}

on the left. To solve

y\circ a=b

, multiply by

a^{-1}

on the right. If the group is not commutative, these two operations cannot be interchanged. This side discipline is a major theme in abstract algebra.

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. Then the equation

\begin{equation*} a\circ x=b \end{equation*}

has a unique solution in

G

, namely

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

Given that

(G,\circ)

is a group with identity element

e

and

a,b\in G

. To prove that the equation

a\circ x=b

has a unique solution in

G

. Let

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

Since

a^{-1}\in G

b\in G

, and

G

is closed under

\circ

, we get

x\in G

. Now

\begin{align*} a\circ x &=a\circ(a^{-1}\circ b)\\ &=(a\circ a^{-1})\circ b \quad [\because\, \circ \text{ is associative}]\\ &=e\circ b \quad [\because\, a\circ a^{-1}=e]\\ &=b \quad [\because\, e \text{ is the identity element}]. \end{align*}

Therefore

x=a^{-1}\circ b

is a solution of

a\circ x=b

. Let

x_1,x_2\in G

be two solutions of

a\circ x=b

. Then

\begin{equation*} a\circ x_1=b \end{equation*}

and

\begin{equation*} a\circ x_2=b. \end{equation*}

Therefore

\begin{align*} a\circ x_1=b,\quad a\circ x_2=b &\implies a\circ x_1=a\circ x_2\\ &\implies x_1=x_2 \quad [\because\, \text{left cancellation law}]. \end{align*}

Hence,

a\circ x=b

has a unique solution in

G

\square

Let

(G,\circ)

be a group with identity element

e

and let

a,b\in G

. Then the equation

\begin{equation*} y\circ a=b \end{equation*}

has a unique solution in

G

, namely

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

Given that

(G,\circ)

is a group with identity element

e

and

a,b\in G

. To prove that the equation

y\circ a=b

has a unique solution in

G

. Let

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

Since

b\in G

a^{-1}\in G

, and

G

is closed under

\circ

, we get

y\in G

. Now

\begin{align*} y\circ a &=(b\circ a^{-1})\circ a\\ &=b\circ(a^{-1}\circ a) \quad [\because\, \circ \text{ is associative}]\\ &=b\circ e \quad [\because\, a^{-1}\circ a=e]\\ &=b \quad [\because\, e \text{ is the identity element}]. \end{align*}

Therefore

y=b\circ a^{-1}

is a solution of

y\circ a=b

. Let

y_1,y_2\in G

be two solutions of

y\circ a=b

. Then

\begin{equation*} y_1\circ a=b \end{equation*}

and

\begin{equation*} y_2\circ a=b. \end{equation*}

Therefore

\begin{align*} y_1\circ a=b,\quad y_2\circ a=b &\implies y_1\circ a=y_2\circ a\\ &\implies y_1=y_2 \quad [\because\, \text{right cancellation law}]. \end{align*}

Hence,

y\circ a=b

has a unique solution in

G

\square

The two formulas should be memorized with their sides:

\begin{equation*} a\circ x=b \quad \Rightarrow \quad x=a^{-1}\circ b \end{equation*}

and

\begin{equation*} y\circ a=b \quad \Rightarrow \quad y=b\circ a^{-1}. \end{equation*}

In a commutative group, these formulas may look similar, but in a non-commutative group they are genuinely different. Students often write

x=b\circ a^{-1}

for the first equation. That is generally wrong unless commutativity is known.

In the additive group

(\mathbb{Z},+)

, the equation

\begin{equation*} 5+x=12 \end{equation*}

has solution

\begin{align*} x&=(-5)+12\\ &=7. \end{align*}

This matches the group formula

x=a^{-1}\circ b

, because the additive inverse of

5

-5

In the multiplicative group

(\mathbb{Q}-\{0\},\cdot)

, the equation

\begin{equation*} 3x=10 \end{equation*}

has solution

\begin{align*} x&=3^{-1}\cdot10\\ &=\frac{1}{3}\cdot10\\ &=\frac{10}{3}. \end{align*}

Hence, the solution is

x=\dfrac{10}{3}

Let

(G,\circ)

be a group. Solve the equation

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

for

x\in G

, where

a,b,c\in G

Let

(G,\circ)

be a group with identity element

e

, and let

a,b,c\in G

. To solve

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

for

x\in G

. Starting with the equation,

\begin{align*} a\circ x\circ b=c &\implies a^{-1}\circ(a\circ x\circ b)=a^{-1}\circ c\\ &\implies (a^{-1}\circ a)\circ x\circ b=a^{-1}\circ c \quad [\because\, \circ \text{ is associative}]\\ &\implies e\circ x\circ b=a^{-1}\circ c\\ &\implies x\circ b=a^{-1}\circ c. \end{align*}

Now multiply both sides on the right by

b^{-1}

. Then

\begin{align*} x\circ b=a^{-1}\circ c &\implies (x\circ b)\circ b^{-1}=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x\circ(b\circ b^{-1})=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x\circ e=(a^{-1}\circ c)\circ b^{-1}\\ &\implies x=(a^{-1}\circ c)\circ b^{-1}. \end{align*}

Hence, the solution is

\begin{equation*} x=(a^{-1}\circ c)\circ b^{-1}. \end{equation*}

\square

Use the calculator below to solve equations in the additive group of integers modulo

n

. Enter

n

a

, and

b

. The calculator solves

a+x=b

modulo

n

using the group formula

x=-a+b

. This is a finite concrete model of the general group equation

a\circ x=b

[1] In a group

(G,\circ)

, solve

a\circ x=b

for

x

. [2] In a group

(G,\circ)

, solve

y\circ a=b

for

y

. [3] In

(\mathbb{Z},+)

, solve

8+x=-3

. [4] In

(\mathbb{Q}-\{0\},\cdot)

, solve

5x=2

. [5] In a group, solve

p\circ x\circ q=r

[1]

x=a^{-1}\circ b

. [2]

y=b\circ a^{-1}

. [3]

x=-11

. [4]

x=\dfrac{2}{5}

. [5]

x=(p^{-1}\circ r)\circ q^{-1}

Because

a

has an inverse, and

x=a^{-1}\circ b

is in the group.

If two solutions exist, the left or right cancellation law forces them to be equal.

Not in general. That replacement requires commutativity.

The first equation is solved by multiplying on the left by

a^{-1}

, while the second is solved by multiplying on the right by

a^{-1}

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