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?
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?
$(aba^{-1})^r$ 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.