<s>[INST] If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I can prove it.
$$ab^2aab^3a=ab^5a$$ so I'm thinking something here. we get $b^5=ab^5a$.
 [/INST] If $a^2 = e$ and $a^{-1}  b^2  a = b^3$, we can substitute $a^2 = e$ into the second equation to get $$a^{-1}  b^2  a = b^3 = (a^2)^{-1}  b^2  a = e^{-1}  b^2  a = b^2  a$$
Then, multiplying both sides of this equation by $b^{-2}$, we get $$b^{-2}  b^2  a = b^{-2}  (b^2  a) = a$$
Now, we can substitute this result, $a = b^{-2}  (b^2  a)$ into the equation $a^{-1}  b^2  a = b^3$ to get $$b^3 = (b^{-2}  (b^2  a))^{-1}  b^2  (b^{-2}  (b^2  a)) = a^{-1}  b^2 a = b^3$$
We can also substitute $a = b^{-2}  (b^2  a)$ into the equation $a^2 = e$ to get $(b^{-2}  (b^2  a))^2 = e$, which simplifies to $b^{-4}  b^4  a^2 = e$.
Finally, we can substitute $a^2 = e$ into this equation to get $b^{-4}  b^4  e = e$, which simplifies to $b^4 = e^{-1} = e$.
Therefore, if $a^2 = e$ and $a^{-1} b^2  a = b^3$, we can conclude that $b^5 = e$.
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