Question

For any elements a and to from a group and any integer n, prove that (a^-1ba)^n=a^-1b^na.

Exponential growth and decay
ANSWERED
asked 2020-11-07
For any elements a and to from a group and any integer n, prove that \(\displaystyle{\left({a}^{{-{{1}}}}{b}{a}\right)}^{{n}}={a}^{{-{{1}}}}{b}^{{n}}{a}\).

Expert Answers (1)

2020-11-08

This is a direct computation: \(\displaystyle{\left({a}^{{-{{1}}}}{b}{a}\right)}^{{n}}={\left({a}^{{-{{1}}}}{b}{a}\right)}{\left({a}^{{-{{1}}}}{b}{a}\right)}\ldots{\left({a}^{{-{{1}}}}{b}{a}\right)}\rbrace{n}\) times
\(=a^{-1}b(aa^{-1})b(aa^-1)...ba =a^-1bebe...ba =a^-1(bb*b)n \times)a =a^{-1}b^na\)
This is just a durect computation

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