I'm considering two presentations for quotients of the 3-strand braid group $B_3$ that I believe should be equivalent (i.e. yield the same quotient). There is an integer parameter $d \ge 7$ involved, and I'll write $\Delta^2 = (\sigma_1\sigma_2)^3$ for the full twist braid. Then we have two quotient groups given by the presentations $$Q_1 = \langle \sigma_1, \sigma_2 \mid \sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2, \sigma_1^d = \sigma_2^d = 1, (\Delta^2)^\ell = 1\rangle$$ where $\ell = 2d/\gcd(12,d+6)$ and $$Q_2 = \langle \sigma_1, \sigma_2 \mid \sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2, \sigma_1^d = \sigma_2^d = 1, (\sigma_1^{-2}\Delta^2)^m = 1\rangle$$ where $m=2d/\gcd(4,d+2)$. Can anyone help elucidate whether these groups are in fact isomorphic?
Here's what I have so far. First, an observation relating the powers $\ell$ and $m$, proof left to the reader.
Lemma $m/\ell = 3$ when $3 \mid d$ and $m/\ell = 1$ otherwise.
Now given the typical braid relation and the relations $\sigma_1^d = \sigma_2^d=1$, we want to show that the remaining relations are equivalent.
Case 1: $3 \not\mid d$. Then $m=\ell$ and $\ell = 2d/1, 2d/2,$ or $2d/4$. Notice then that $\sigma_1^{-2\ell}=1$. Now one finds $$(\sigma_1^{-2}\Delta^2)^m = (\sigma_1^{-2}\Delta^2)^\ell = \sigma_1^{-2\ell} (\Delta^2)^\ell = (\Delta^2)^\ell.$$ Thus $(\sigma_1^{-2}\Delta^2)^m=1$ if and only if $(\Delta^2)^\ell=1$.
Case 2: $3 \mid d$. Then $m=3\ell$ and $\ell=2d/12,2d/6,$ or $2d/3$. Notice now that $\sigma_1^{-6\ell}=1$. Now one finds $$(\sigma_1^{-2}\Delta^2)^m = (\sigma_1^{-2}\Delta^2)^{3\ell} = \sigma_1^{-6\ell} (\Delta^2)^{3\ell} = (\Delta^2)^{3\ell}.$$ If $\Delta^2$ has order $\ell$, then $(\sigma_1^{-2}\Delta^2)$ has order $m$. But the converse does not seem to follow.
Can anyone clear up this last remaining hole in the argument?