Group Product/Examples/b a^-1 x a b^-1 = b a

Examples of Operations on Product Elements

Solve for $x$ in:

$b a^{-1} x a b^{-1} = b a$

Solution

\(\ds b a^{-1} x a b^{-1}\)

\(=\)

\(\ds b a\)

\(\ds \leadsto \ \ \)

\(\ds b^{-1} b a^{-1} x a b^{-1}\)

\(=\)

\(\ds b^{-1} b a\)

Product of both sides with $b^{-1}$

\(\ds \leadsto \ \ \)

\(\ds b^{-1} b a^{-1} x a b^{-1} b\)

\(=\)

\(\ds b^{-1} b a b\)

Product of both sides with $b$

\(\ds \leadsto \ \ \)

\(\ds a^{-1} x a\)

\(=\)

\(\ds a b\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \leadsto \ \ \)

\(\ds a a^{-1} x a\)

\(=\)

\(\ds a^2 b\)

Product of both sides with $a$

\(\ds \leadsto \ \ \)

\(\ds a a^{-1} x a a^{-1}\)

\(=\)

\(\ds a^2 b a^{-1}\)

Product of both sides with $a^{-1}$

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds a^2 b a^{-1}\)

Group Axiom $\text G 3$: Existence of Inverse Element

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Exercise $2 \ \text{(d)}$