Let $G$ be a group and if $a,b \in G$ such that $a^4 =e$ and $a^2 b=ba$ then prove that $a=e$.
My attempt: I know from the generators and relation of the dihedral group of order $2n$ is $\langle r^n=s^2=1,rs=sr^{-1} \rangle$ so in the given problem if I take $r=a^2$ and $s=b$ then the problem is clear that $a^4=a$ but in the given problem it is not given that $b^2=e$ so how can I prove that $a=e$ in the given problem.
If you know enough about conjugacy, then this can be done without writing down lots of equations. The second relation, $a^2b=ba$ says that $a$ is conjugate to $a^2$. So $a^2$ is conjugate to $a^4$. But $a^4=e$ and $e$ is conjugate only to itself.
If you don't know that much about conjugacy, then there's the following proof, which I got by following the preceding paragraph. Using three times that $ba=aab$ and once that $aaaa=e$, we get $$ bba=baab=aabab=aaaabb=bb. $$ Cancelling $bb$, we get $a=e$.
$a^2b=ba ... (1)$
$b=a^4b=a^2 (a^2 b) =(a^2 b) a= b a a= b a^2$.
Then $a^2=e$.
Put it in (1), $b=ba$. Hence $a=e$.
Start with $a^2b = ba$. Multiplying on the left by $a^2$, we get $a^4b = a^2ba$. However, because $a^4 = e$, this equation reduces to $b = a^2ba$. But because $a^2ba = ba$, the RHS of $b = a^2ba$ becomes $ba^2$. Hence, $$b = ba^2$$
Multiplying both sides on the left by $b^{-1}$, we have $a^2 = e$. But then $ba = a^2b = eb = b$. Multiplying on the left by $b^{-1}$ again, we obtain $a = e$.
