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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Expert Answer

svartmaleJ
Answered 2021-02-01 Author has 16142 answers

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

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