For a,b in a group G, $|a| = 2$, $b ≠ e$, and $aba = b^2$. What is $|b|$? If it was $aba^{-1}$ = $b^2$ I could use the theorem that if $aba^{-1}$ = $b^n$ then $a^kba^{-k}$ = $b^{n^k}$ so that $b$ = $ebe$ = $a^2ba^{-2}$ = $b^4$ so that $|b|$ = 3. Is there a similar method here?
 A: As suggested in the comments, $|a| = 2$ implies $a = a^{-1}$.
But say for some reason (!) you do not want to use this fact...  Consider
\begin{equation}
\begin{aligned}
(aba)^2 
&=(aba)(aba)\\
&=aba^2ba=ab^2a \\
&=a(aba)a = a^2ba^2 \\
&=b
\end{aligned}
\end{equation}
but we also have $(aba)^2=(b^2)^2$ so that $b^4 = b$. Hence |b|=3 since $b\neq e$.
Note that in fact we are still using $a=a^{-1}$, since that is equivalent to $|a|^2$. We are simply not invoking your (general) theorem and simply proving the result for the case $k=1$, $n=2$.
A: Since $a^2=e$, we have $a=a^{-1}$. From the equation $aba=b^2$
\begin{align*}
ba &= ab^2 \text{(composing a on the left)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
ab &= b^2a \text{(composing a on the right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\
ab^2&=b^2ab \text{(composing b on the right of (2))}\,\,\,\,\,\,(3)\\
\text{Thus,} ba&=b^2ab\\
b&=b^2aba\\
b&=b^4\text{(}aba=b^2\text{)}\\
b^3&=e
\end{align*}
Thus $|b|$ divides $3$. But since $b \neq e$, we have $|b|=3$
