<s>[INST] order $a$ = 5, $a^3b = ba^3$. show that that $ab = ba$. Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.
Here is what I think: 
We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. Now, 
$a^3b = ba^3$ then $a^{-2}b = ba^{-2}$ then $ba^2 = a^2b$ then $ba^{-4} = a^{-4}b$ then $a^{-1}b = ba^{-1}$ then multiply on the left and right by $a$ to get $ba=ab$. Is this correct or is there a simpler approach? 
 [/INST] $b=a^2a^3b=a^2ba^3=a^{-1}a^3ba^3=a^{-1}ba^3a^3=a^{-1}ba$. So you have you result.
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