The quaternion group as a cocycle central extension of C2 x C2

Direct computation gives $I^2=J^2=-1_2$, $IJ=K=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$, $JI=-K$ and $K^2=-1_2$, so the eight matrices $\pm 1_2,\pm I,\pm J,\pm K$ are distinct and form a subgroup $M\le {\mathrm{GL}}_2(\mathbb{C})$. Listing them, $1_2,I,J,K$ have second columns $(0,1),(0,-i),(1,0),(i,0)$ up to sign, so no two of the eight coincide; the assignment $i\mapsto I$, $j\mapsto J$, $k\mapsto K$, $-1\mapsto -1_2$ preserves the relations of [F1], so $M\cong Q_8$ with centre $\{\pm 1_2\}$ and quotient $M\mathrm{/}\{\pm 1_2\}\cong Q$ via $I\mapsto x$, $J\mapsto y$.

Define $P:Q\to {\mathrm{GL}}_2(\mathbb{C})$ by $P(x^a{y}^b):=I^a{J}^b$ for $a,b\in \{0,1\}$; this is well defined because every element of $Q$ is uniquely $x^a{y}^b$. Step 1.1 further gives $IK=I(IJ)=I^{2}J=-J$, $KI=J$, $JK=I$ and $KJ=-I$. Hence, using $P(xy)=K$, the products on basis elements are $P(x{)}^2=I^2=-1_2$, $P(y{)}^2=J^2=-1_2$, $P(xy{)}^2=K^2=-1_2$, $P(x)P(y)=IJ=K=P(xy)$, $P(y)P(x)=JI=-K=-P(xy)$, $P(x)P(xy)=IK=-J=-P(y)$, $P(y)P(xy)=JK=I=P(x)$, $P(xy)P(x)=KI=J=P(y)$, $P(xy)P(y)=KJ=-I=-P(x)$, and $P(1)P(q)=P(q)=P(q)P(1)$. Comparing with $qr$ in $Q$ shows $P(q)P(r)=\alpha (q,r)P(qr)$ for all $q,r\in Q$, where $\alpha (1,q)=\alpha (q,1)=1$, $\alpha (x,x)=\alpha (y,y)=\alpha (xy,xy)=\alpha (y,x)=\alpha (xy,y)=\alpha (x,xy)=-1$ and $\alpha (x,y)=\alpha (y,xy)=\alpha (xy,x)=1$.

The function $\alpha$ is a normalized two-cocycle: comparing $(P(q)P(r))P(s)=\alpha (q,r)\alpha (qr,s)P(qrs)$ with $P(q)(P(r)P(s))=\alpha (r,s)\alpha (q,rs)P(qrs)$ using step 2.1 and cancelling the invertible matrix $P(qrs)$ gives $\alpha (q,r)\alpha (qr,s)=\alpha (r,s)\alpha (q,rs)$ for all $q,r,s\in Q$, and $\alpha (1,q)=\alpha (q,1)=1$ holds by definition, matching [F4].

The cocycle central extension $E_{\alpha }$ of [F2] contains the eight elements $Q\times \{\pm 1\}=\{(q,z):q\in Q,z=\pm 1\}$; the map $\phi (q,z):=s(q)z$ with $s(1)=1_2$, $s(x)=I$, $s(y)=J$, $s(xy)=K$ satisfies $\phi ((q,z)(r,w))=s(qr)\alpha (q,r)zw=s(q)s(r)zw=\phi (q,z)\phi (r,w)$, because $P(q)P(r)=\alpha (q,r)P(qr)$ by step 2.1; it is injective on the eight elements and its image is $M$, so $Q\times \{\pm 1\}\cong M\cong Q_8$. Also $P$ is faithful, since $P(q)=1_2$ forces $q=1$ by the distinctness of the eight matrices in step 1.1.

The factor set $\alpha$ is not a coboundary, so its class in $H^2(Q,{\mathbb{C}}^{\times })$ is nonzero: if $\alpha =\delta c$ for some $c:Q\to {\mathbb{C}}^{\times }$, then [A1] would give $\alpha (y,x)=c(y)c(x)c(yx{)}^{-1}=c(x)c(y)c(xy{)}^{-1}=\alpha (x,y)$, contradicting $\alpha (y,x)=-1\mathrm{\ne }1=\alpha (x,y)$ from step 2.1. Consequently the lifts of the two generators anticommute, $P(y)P(x)=-P(x)P(y)$, the representation $P$ of step 2.1 is a faithful projective representation of $C_2\times C_2$ with nontrivial factor set, and it corresponds by [F3] to the representation $D(q,z)=zP(q)$ of $E_{\alpha }$ whose restriction to $Q\times \{\pm 1\}\cong Q_8$ is the faithful two-dimensional representation of $Q_8$.

The example is complete: $Q_8\mathrm{/}\{\pm 1\}\cong C_2\times C_2$ by step 1.1, the faithful two-dimensional representation of $Q_8$ restricts on $Q\times \{\pm 1\}\cong Q_8$ to the projective representation $P$ of step 2.1 whose factor set has $\alpha (y,x)=-1\mathrm{\ne }1=\alpha (x,y)$ and is therefore not a coboundary by step 5.1, and the $\pm 1$-valued subgroup $Q\times \{\pm 1\}$ of the cocycle central extension $E_{\alpha }$ is isomorphic to $Q_8$ by step 4.1.