Recall these facts about the groups

Let A be a set, ∗ a binary operation on A, and a ∈ A. Suppose that there is an identity element e for the operation. Then

- an element b is a left inverse for a if b∗a=e,

- an element c is a right inverse for a if a∗c=e,

- an element is an inverse(or two-sided inverse) for a if it is both a left and right inverse for a.

So in an abstract algebra equation about groups, taking the inverse of both sides of an equation is a valid statement. Since the inverse of an element exist iff its right inverse and left inverse exist. Consider an example.

Let G be the group with identity element e and binary operation *.

Let a,b in G consider the equation a*x=b. To solve the equation, take inverse of a both sides.

Case 1

\(\displaystyle{a}^{{-{{1}}}}\cdot{a}\cdot{x}={a}^{{-{{1}}}}\cdot{b}\)

\(\displaystyle\Rightarrow{e}\cdot{x}={a}^{{-{{1}}}}\cdot{b}\)

\(\displaystyle\Rightarrow{x}={a}^{{-{{1}}}}\cdot{b}\)

Case 2:

\(\displaystyle{a}\cdot{x}\cdot{a}^{{-{{1}}}}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{a}\cdot{a}^{{-{{1}}}}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{e}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{x}={b}\cdot{a}^{{-{{1}}}}\)

Let A be a set, ∗ a binary operation on A, and a ∈ A. Suppose that there is an identity element e for the operation. Then

- an element b is a left inverse for a if b∗a=e,

- an element c is a right inverse for a if a∗c=e,

- an element is an inverse(or two-sided inverse) for a if it is both a left and right inverse for a.

So in an abstract algebra equation about groups, taking the inverse of both sides of an equation is a valid statement. Since the inverse of an element exist iff its right inverse and left inverse exist. Consider an example.

Let G be the group with identity element e and binary operation *.

Let a,b in G consider the equation a*x=b. To solve the equation, take inverse of a both sides.

Case 1

\(\displaystyle{a}^{{-{{1}}}}\cdot{a}\cdot{x}={a}^{{-{{1}}}}\cdot{b}\)

\(\displaystyle\Rightarrow{e}\cdot{x}={a}^{{-{{1}}}}\cdot{b}\)

\(\displaystyle\Rightarrow{x}={a}^{{-{{1}}}}\cdot{b}\)

Case 2:

\(\displaystyle{a}\cdot{x}\cdot{a}^{{-{{1}}}}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{a}\cdot{a}^{{-{{1}}}}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{e}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)

\(\displaystyle\Rightarrow{x}={b}\cdot{a}^{{-{{1}}}}\)