Free groups have been hurting my brain for months and this one in particular, I cannot comprehend what is going on with logic
$H= \langle a,b\mid a^{-1}ba=b^2,b^{-1}ab=a^2\rangle$. Show that $H$ is a trivial group. Hint: express $a^{-2}ba^2$ in two different ways to show that $a,b$ commute.
Solution
Using the first relation, $a^{-2}ba^2=a^{-1}b^2a=(a^{-1}ba)^2=b^4$
Ok, whoa, hold it right there because that doesn't make sense to me. $a^{-2}ba^2=a^{-1}b^2a$? What, I can concatenate $a^{-1}$ to $a^{-1}b^2a$ and it's still considered "equal"? How? clearly $a^{-2}ba^2 \neq a^{-1}b^2a$.
Then using the second relation, $a^{-2}ba^2=a^{-2}bb^{-1}ab=a^{-1}b$. Thus $a^{-1}b=b^4$.
I understand this $a^{-2}ba^2=a^{-2}bb^{-1}ab=a^{-1}b$ bit. But I don't understand how $a^{-1}b=b^4$ came about. What, so $a^{-2}ba^2=(a^{-1}ba)^2$?
I really don't understand this, it's like arithmetic but not arithmetic and it's just is hard as anything I could imagine. What is going on?
We start with "$a^{-2} b a^2$". Remember this because we're going to start the second step with the same thing and whatever we get here will be equal to whatever we get there.
\begin{align} a^{-2} b a^2 &= a^{-1} a^{-1} b a^1 a^1 \\ &= a^{-1} (a^{-1} b a^1) a^1 \\ &= a^{-1} (b^2) a^1 & &\text{relation in the presentation} \\ &= a^{-1} b ()b a^1 \\ &= a^{-1} b (a a^{-1}) b a^1 & &\text{because you can always introduce $1$} \\ &= (a^{-1} b a) (a^{-1} b a^1) \\ &= (a^{-1} b a)^2 \\ &= (b^2)^2 & &\text{relation in the presentation} \\ &= b^4 \text{.} \end{align} In the second step, we find $a^{-2}ba^2$ is also $a^{-1}b$. So $a^{-1}b = a^{-2}ba^2 = b^4$, using the result of the detailed explanation above.