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Prove $(a^{-1}ba)^n = a^{-1}b^na$ for all $n \in \mathbb Z$ and $a, b$ in a group.

Assume $n \ge 1$. The identity is true for $n = 0, 1.$ Proof for $n + 1: (a^{-1}ba)^n = (a^{-1}ba)^{n + 1} = (a^{-1}ba)^n(a^{-1}ba) = (a^{-1}b^na)(a^{-1}ba) = a^{-1}b^na$

Assume $n \le -1.$ Since $-n \ge 1,$ we have $(a^{-1}ba)^n = [(a^{-1}ba)^{-1}]^{-n} = (a^{-1}b^{-1}a)^{-n} = a^{-1}({b^{-1}})^{-n}a = a^{-1}b^na.$

How are we inducting on $n$ in the second part of the proof for negative integers? Are we doing $n - 1$ or $-n - 1$?

Also is $[(a^{-1}ba)^{-1}]^{-n} = (a^{-1}b^{-1}a)^{-n} $ by inductive hypothesis?

user324367
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    You assume that it holds for $n\in\mathbb Z$ and show that it holds for $n-1$. Or: You assume that it holds for $-n$, $n\ge 0$, and show that it holds for $-n-1$. BTW: The first and last equality in your first line (starting with "$n+1:$") are false. – Friedrich Philipp Mar 23 '16 at 01:34
  • @FriedrichPhilipp, I see about the first equality, but isn't the secnd equality because $aa^{-1} = e$ in a group? – user324367 Mar 23 '16 at 01:39
  • Second equality should end with $b^{n+1}$, not $b^n$. –  Mar 23 '16 at 01:40
  • @tilper, I see. Thanks, guys. I got no more questions. – user324367 Mar 23 '16 at 01:41
  • The proof doesn't do induction on the negative values although it easily could. Instead it relies on having proved ot for positive values, a negative value is a positive power of the inverse and so it would apply as the inverse is also of form cdc^-1 – fleablood Mar 23 '16 at 05:02
  • Isn't $(a^{-1}ba)^n = b^n$? $a^{-1}a=1$... – TheRandomGuy Mar 23 '16 at 05:09
  • @Dhruv, I think the group might not be commutative. – user324367 Mar 23 '16 at 12:52

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