Algebra Solutions: Deduce some properties of a group from its presentation

Let $Y = \langle a, b \ |\ a^4 = b^3 = 1, ab = b^2 a^2 \rangle$ .

From the relation $b^3 = 1$ it follows that $b^{-1} = b^2$ .
Note first that

$b^2 a^3 b$ = $(bbaa)(ab)$

= $(ab)(bbaa)$

= $(a)(b^3)(aa)$

= $a^3$

So that, left multiplying by $b$ , we have $a^3 b = b a^3$ .
Note that $a^9 = a^4 a^4 a = a$ . Hence $a b = a^9 b = (a^3)^3 b = b (a^3)^3 = ba$ , using part 2.
We have $ba = ab = b^2 a^2$ , so that $ba = 1$ . Thus $ab = 1$ .
Since $u^4 = v^3 = 1$ , we have $u^4 v^3 = 1$ . Thus $u (u^3 v^3) = u = 1$ , so that $v = 1$ . Hence $Y = 1$ .

Algebra Solutions