# In an abstract algebra equation about groups, is "taking the inverse of both sides of an equation" an acceptable operation? I know you can right/left multiply equations by elements of the group, but was wondering if one can just take the inverse of both sides?

In an abstract algebra equation about groups, is "taking the inverse of both sides of an equation" an acceptable operation? I know you can right/left multiply equations by elements of the group, but was wondering if one can just take the inverse of both sides?

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Recall these facts about the groups
Let A be a set, $$\times a$$ binary operation on A, and $$a \in A$$. Suppose that there is an identity element e for the operation. Then
- an element b is a left inverse for a if $$b\times a=e$$,
- an element c is a right inverse for a if $$a\times c=e$$,
- 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 $$\times$$.
Let a,b in G consider the equation $$a\times x=b$$. 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:
$$\displaystyle{a}\cdot{x}\cdot{a}^{{-{{1}}}}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{a}\cdot{a}^{{-{{1}}}}\cdot{x}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{e}\cdot{x}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{x}={b}\cdot{a}^{{-{{1}}}}$$