How to prove two of following central extensions are isomorphic?

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.

• Are you sure these are isomorphic? The second looks like the double cover of $D_3$ corresponding to the nontrivial element of $H^2$. – David Hill Mar 27 '18 at 19:00
• $G_2$ is of order $6$ not $12$. The relations force $b=1$. If you put $s^2=b$ in $G_2$ instead of $s^2=1$ then you get a group of order $12$ but it is not isomorphic to $G_1$. – Derek Holt Mar 27 '18 at 19:02
• @DavidHill Sorry, I edit the question. – maplemaple Mar 27 '18 at 19:04
• @DerekHolt Sorry, I correct the question. – maplemaple Mar 27 '18 at 19:09
• I think the key takeaway is that you can always get any central extension of any finite group $G$ by $C_2$ following this recipe: add the generator "$b$", make it commute with all of $G$'s generators, and then set every relator of $G$ either to $b$ or to $1$. The problem is (and why this is not a trivial theory) that not every such presentation gives you an actual central extension (and sometimes two different ones give you the same group). – Steve D Mar 28 '18 at 20:13

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 $(rs)^2=b=1$.
• So that means $G_2$ is not a central extension? $G_1$ and $G_3$ are in different extension classes? – maplemaple Mar 28 '18 at 5:24