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Let $a, b \in G$. Suppose $aba^{-1} = b^{i}$ . Show that $a^{r}ba^{-r} = b^{i^r}$

I tried raising both sides to the r, but it gave me $a^{r} b^{r} a^{-r}$ = $b^{ir}$ . Any suggestions?

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    i is any integer and$r$is any integer greater than or equal to 0.2011-10-13

1 Answers 1

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$(aba^{-1})^r$ surprisingly is not (usually) equal to $a^r b^r a^{-r}$. Take $r=2$ for instance:

$(aba^{-1})^2 = (aba^{-1})(aba^{-1}) = (ab)(a^{-1}a)(ba^{-1}) = (ab)(ba^{-1}) = a(b)^2 a^{-1}$

On the other hand, if you put two $a$s on the outside, you get:

$(a)^2 b (a^{-1})^2 = a(aba^{-1})a^{-1} = a(b^i)a^{-1} = (aba^{-1})^i = (b^i)^i = b^{i^2}$

If you work out a few powers like this, I think you'll see the pattern.

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    Well you do still have to prove it for general $r$, but the key is when you have $r$ copies of $aba^{-1}$ all multiplied together, the same trick will work.2011-10-14