Given group $D_3= \langle r,s | r^3 =s^2=(rs)^2= 1\rangle$
The following are $3$ central extensions of $D_3$ by $\mathbb{Z}_2\cong \langle b|b^2=1\rangle$. I want to know which one of $G_2$ and $G_3$ is isomorphic to $G_1$? Or $G_2$ is isomorphic to $G_3$?
$$G_1=\langle r,s, b | r^3 =s^2=(rs)^2= 1,b^2=1, [r,b]=[s,b]=1 \rangle \cong D_3\times \mathbb{Z}_2 $$
$$G_2=\langle r,s, b | r^3 =s^2= 1, (rs)^2=b ,b^2=1, [r,b]=[s,b]=1 \rangle$$
$$G_3=\langle r,s, b | r^3 =1, s^2=(rs)^2=b , b^2=1, [r,b]=[s,b]=1 \rangle$$
with commutator $[,]$ is defined by $[a,b]=a^{-1}b^{-1}ab$.
It relates the group cohomology $H^2(D_3,\mathbb{Z}_2)=\mathbb{Z}_2$. It says that they are totally two kinds of central extension. $G_1$ belongs to the trivial central extension. One of $G_2$ and $G_3$ should belong to the trivial case or $G_2\cong G_3$. I don't know how to show explicitly.
Here is a proof that $b=1$ in $G_2$, so $|G_2|=6$.
$$rsrs=b \Rightarrow r^{-2}srs=b \Rightarrow r^{-1}srsr^{-1}=rbr^{-1}=b \Rightarrow r^{-1}sr=brs = (rs)^3.$$
So then $s^2=1 \Rightarrow (rs)^6=1$, but $(rs)^4=b^2=1$, so $(rs)^2=b=1$.