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?

Question
Abstract algebra
asked 2021-01-31
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?

Answers (1)

2021-02-01
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:
\(\displaystyle{a}\cdot{x}\cdot{a}^{{-{{1}}}}={b}\cdot{a}^{{-{{1}}}}\)
\(\displaystyle\Rightarrow{a}\cdot{a}^{{-{{1}}}}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)
\(\displaystyle\Rightarrow{e}\cdot{x}={b}\cdot{a}^{{-{{1}}}}\)
\(\displaystyle\Rightarrow{x}={b}\cdot{a}^{{-{{1}}}}\)
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