Properties of groups with integer exponents

Question

For any elements a and b from a group $\lt A,\ * \gt$ and any integer n, prove that

$\begin{align} & (a^{-1}\cdot b\cdot a)^n = a^{-1}\cdot b^n\cdot a \end{align}$

This is what I tried:

LHS:

$\begin{align} & \quad \ \ \ a^{-n}\cdot b^n\cdot a^n=(a^{-n}\cdot a^n) \cdot b^n=(a^{-1}\cdot a)^n\cdot b^n \\ &= \ e^n\cdot b^n\qquad \text{where $e$ is the identity element of the group } \end{align}$

And this is where I get stuck. I know that the identity elements are usually 1 and 0 and the power n should disappear but I don't know how to show that logically.

Please show me a better way to solve this.

Thanks

Answer 1

Try induction on $n$.

Beware, groups may be non-commutative. In general $(a^{-1}ba)^n$ is not equal to $a^{-n}b^na^n$.

Answer 2

Writing out the product in full $$ (a^{-1} b a )^{n} = \underbrace{(a^{-1} b a ) (a^{-1} b a ) \ldots (a^{-1} b a )}_{\large n\text{ factors of } a^{-1}ba} $$

Groups are associative so we can rewrite $$ a^{-1} b (a a^{-1}) b \ldots b ( a a^{-1}) b a $$

$a a^{-1} = e$ so we get $$ a^{-1} b e b \ldots b e b a $$

$be = b$ so we can replace each of these with $b$ in the above $$ a^{-1} b b \ldots b b a $$ where there are $n$ $b$'s

which gives finally $$ a^{-1} b^{n} a $$

Answer 3

$(a^{-1} b a)^n = (a^{-1}ba) \dots (a^{-1} b a) = a^{-1} (b a a^{-1})(b a a^{-1}) \dots(b a a^{-1}) a $

This is just

$a^{-1}(b e)( be ) \dots (be) a = a^{-1} (b \dots b) a = a^{-1} b^n a $