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Every discrete additive subgroup of ${\mathbb{R}}^2$ has the form $\mathbb{Z}{v}_1+\cdots +\mathbb{Z}{v}_r$ for linearly independent vectors, with $r=0,1,$ or $2$. A full translation lattice has $r=2$, and conversely every such integer span is a discrete cocompact subgroup. This classifies planar lattices up to a choice of basis.

A frieze group is a discrete group of planar Euclidean isometries whose translation subgroup is infinite cyclic and has finite index. Consider the glide reflection $g(x,y)=(x+1,-y)$. Its square is translation by $(2,0)$, whereas every odd power reverses orientation. Therefore $⟨g⟩\cong \mathbb{Z}$, but none of its translation elements generates the whole group: they form the index-two subgroup $⟨g^2⟩$.

To realize exactly this frieze group, repeat a small scalene triangular motif at the points $(n,(-1{)}^n)$, reflecting the motif when $n$ is odd. Choose the motif with three distinct side lengths and no internal symmetry. The displayed pattern has glide reflection symmetry $g$ and translation period $2$, but no pure horizontal reflection or vertical mirror. Distinct side lengths force any symmetry to preserve corresponding vertices, which rules out extra motif permutations; the sequence of heights then leaves exactly the powers of $g$.

A frieze group is a discrete subgroup of Euclidean isometries whose translation subgroup is generated by one nonzero vector and has finite index. A plane crystallographic group is a discrete subgroup of Euclidean isometries with a rank-two translation lattice of finite index; equivalently its action has a compact fundamental region.

Here are three frieze groups, with explicit generators. The translation $T(x,y)=(x+1,y)$ alone gives $\mathbb{Z}$. Adding the vertical reflection in mathematics $S(x,y)=(-x,y)$ gives $\mathbb{Z}⋊C_2$, the infinite dihedral group, since $STS=T^{-1}$. Instead adding the horizontal reflection in mathematics $H(x,y)=(x,-y)$ gives $\mathbb{Z}\times C_2$, since $HT=TH$. Their translation parts are exactly $⟨T⟩$ and their point parts have respectively one, two and two elements.

For plane crystallographic groups, let $T_1(x,y)=(x+1,y)$ and $T_2(x,y)=(x,y+1)$. These generate ${\mathbb{Z}}^2$. Adding the half-turn $R_2(x,y)=(-x,-y)$ gives ${\mathbb{Z}}^2⋊C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives ${\mathbb{Z}}^2⋊C_4$, with $R_4{T}_1{R}_4^{-1}=T_2$ and $R_4{T}_2{R}_4^{-1}=T_1^{-1}$. These are the translation, half-turn and square-rotation wallpaper groups. Integer translation vectors and a finite point part make each group a discrete subgroup; the unit square meets every orbit, establishing a cocompact group action.

For the final claim, suppose a Euclidean isometry $R$ has order five. An orientation-reversing planar isometry has finite order only when it is a reflection in mathematics, so $R$ is a rotation through $\theta =2\pi j/5$, $j=1,2,3,4$, about some centre. Write $A$ for its linear rotation part. If $t_v\in G$ is a nonzero translation, conjugation gives $R{t}_v{R}^{-1}=t_{Av}$. Thus $G$ contains translations by $v$ and by $(A+A^{-1})v=cv$, where $c=2\cos \theta$.

For a primitive fifth root $\zeta$, dividing $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$ by ${\zeta }^2$ gives $c^2+c-1=0$. Hence $c$ is irrational. The pigeonhole principle, applied to $N+1$ fractional parts of $0,c,\dots ,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\le 1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the discrete subgroup hypothesis. There is no element of order five. This argument needs only the given nonzero translation, not a rank-two lattice.