 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? Clifland 2021-01-31 Answered
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:
$$\displaystyle{a}\cdot{x}\cdot{a}^{{-{{1}}}}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{a}\cdot{a}^{{-{{1}}}}\cdot{x}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{e}\cdot{x}={b}\cdot{a}^{{-{{1}}}}$$
$$\displaystyle\Rightarrow{x}={b}\cdot{a}^{{-{{1}}}}$$