The p4 and p4m square-lattice symmetry groups

The matrices satisfy ${\rho }^4=I$, ${\sigma }^2=I$, and $\sigma \rho \sigma ={\rho }^{-1}$. The rotation matrices are $I,\rho ,-I,-\rho$, hence distinct; their products add exponents modulo $4$, and $({\rho }^j{)}^{-1}={\rho }^{-j}$. The four matrices $\sigma ,\rho \sigma ,-\sigma ,-\rho \sigma$ are distinct, have determinant $-1$, and differ from the rotations, which have determinant $1$. Thus $C_4$ has four elements and $D_4$ has eight. Writing elements of $D_4$ as ${\rho }^j{\sigma }^{ϵ}$ with $ϵ\in \{0,1\}$, the relation $\sigma {\rho }^j={\rho }^{-j}\sigma$ shows products stay in $D_4$, and $({\rho }^j{\sigma }^{ϵ}{)}^{-1}={\rho }^{-(-1{)}^{ϵ}j}{\sigma }^{ϵ}$ also lies in $D_4$. Both matrix groups preserve ${\mathbb{Z}}^2$ because $\rho$ and $\sigma$ are signed permutation matrices.

For either matrix group $A$, the pair multiplication is closed on ${\mathbb{Z}}^2\times A$. It is associative because both parenthesizations of three factors $(t,M),(s,N),(r,Q)$ equal $(t+Ms+MNr,MNQ)$. Its identity is $(0,I)$. For every pair $(t,M)$, the pair $(-M^{-1}t,M^{-1})$ belongs to the same set by 1.1, and $(t,M)(-M^{-1}t,M^{-1})=(0,I)=(-M^{-1}t,M^{-1})(t,M)$. Thus $P_{C_4}$ and $P_{D_4}$ satisfy the group axioms in [F1].

The identity pair acts identically because $(0,I)\cdot x=x$. The affine maps respect pair multiplication because $M(Nx+s)+t=MNx+Ms+t$, so $((t,M)(s,N))\cdot x=(t,M)\cdot ((s,N)\cdot x)$ and [F2] holds. The translations $(t,I)$ are distinct as $t$ ranges over ${\mathbb{Z}}^2$, so each full group is infinite.