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 $b^5=ab^5a$.
 A: We are given that $b^3 = ab^2a$. If we cube this equation, we obtain that $b^9 = ab^6a = ab^3 \cdot b^3a$. Since from the given equation we deduce that $ab^3 = b^2a$ and $b^3a = ab^2$, we can make these substitutions to obtain $b^9 = b^2a \cdot ab^2 = b^4$. Hence, $b^5 = 1$, as desired.
Edit: The idea above shows that if $b^n$ and $b^m$ are conjugated in a group by an element of order $2$, then $b^{n^2-m^2} = 1$, which is a pretty neat and fairly general result.
A: 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 $b^5 = e$.
