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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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What is $r$ and what is $i'$? –  Mariano Suárez-Alvarez Oct 13 '11 at 22:36
9  
Lily: you are reasoning as if the group was abelian but the whole point of the exercise is that it is not in general. Hence you should avoid replacing some $xy$ by $yx$, these are not the same in general. –  Did Oct 13 '11 at 22:39
    
i is any integer and r is any integer greater than or equal to 0. –  Lily Oct 13 '11 at 22:41

1 Answer 1

up vote 10 down vote accepted

$(aba^{-1})^r$ 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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Okay, I understand what you're saying. But, how to I get there starting with aba^{-1} –  Lily Oct 13 '11 at 22:59
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@Lily, You seem to have misinterpreted the answer. Notice he actually proved the case when $r = 2$. Try convincing yourself it is true for some small values of $r$, just a few should be enough (say $r = 2,3$ and 4. It's important when dealing with groups to get familiar with the fact that $aa^{-1} = a^{-1}a$ and are both the identity. –  Alex Oct 14 '11 at 0:36
    
I realize that he proved it for r=2. I just wasn't sure if using actual r terms was alright. I thought I needed to prove it in the general case. Thank you! –  Lily Oct 14 '11 at 0:39
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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. –  Alex Oct 14 '11 at 0:43

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