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p(1): $(bab^{-1})^1 = ba^1b^{-1}.$

p(k + 1): $(bab^{-1})^{k + 1} = (bab^{-1})^k(bab^{-1}) = b^ka^kb^{-k}bab^{-1} = ba^{k + 1}b^{-1}.$

Would that work?

Marko
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    You have a grave typo or miscalculation. $(bab^{-1})^k = b^1 a^k b^{-1}\neq b^k a^k b^{-k}$. Your induction hypothesis should be what you are trying to prove. It also helps to explicitly state what the induction hypothesis is and where you use it during the algebraic manipulation of the $(k+1)^{st}$ case. – JMoravitz Feb 16 '15 at 01:50
  • @ JMoravitz, is $(bab^{-1})^k = a^k?$ – Marko Feb 16 '15 at 01:53
  • not quite $(bab^{-1})^k = ba^kb^{-1}$, which is what you are trying to prove. – JMoravitz Feb 16 '15 at 01:58
  • Can you, please explain the algebra behind $(bab^{-1})^k = ba^kb^{-1}?$ I don't get it. – Marko Feb 16 '15 at 02:01
  • the purpose of the question is to prove exactly that statement. We begin with a base case, $k=1$ and show that it is true for that. We assume it to be true for some $k\geq 1$ (which is true since we already checked the case when $k=1$), and we show that it follows that is also true for $k+1$. So, it is true for $k=1\Rightarrow$ it is true for $k=2\Rightarrow$ it is true for $k=3\Rightarrow\dots$ – JMoravitz Feb 16 '15 at 02:04
  • The reason why we can't say $(bab^{-1})^k=b^ka^kb^{-k}$ is because there is no guarantee that the group is abelian (commutative). If it turns out that it is not abelian, then $b^2a^2b^{-2}\neq ba^2b^{-1}$ – JMoravitz Feb 16 '15 at 02:08

2 Answers2

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You have the majority of the argument, with a presumed typo, so the rest becomes trying to write it with proper grammar and structure.

Claim: $(bab^{-1})^n = ba^n b^{-1}$ for all $n\in\mathbb{N}$

Base case: let $n=1$. Then $(bab^{-1})^1 = bab^{-1}$

Assume for our induction hypothesis that $(bab^{-1})^n = ba^n b^{-1}$ for some $n\geq 1$.

We show then that it must also be true for $n+1$.

$(bab^{-1})^{n+1} = (bab^{-1})^n(bab^{-1}) =^{I.H.} ba^nb^{-1}bab^{-1}$

$=ba^n(b^{-1}b)ab^{-1} = ba^n 1 a b^{-1} = ba^nab^{-1}=ba^{n+1}b^{-1}$

Thus, proving the claim.

JMoravitz
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Hint $\ $ The inductive step arises by scaling both sides by $\,bab^{-1}$ then removing $\,bb^{-1}$ on RHS.

Bill Dubuque
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