= Solution
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 $\mathbb Z\rtimes C_2$, 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 $\langle T\rangle$ 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 $\mathbb Z^2\rtimes C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives $\mathbb Z^2\rtimes C_4$, with $R_4T_1R_4^{-1}=T_2$ and $R_4T_2R_4^{-1}=T_1^{-1}$. 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 $Rt_vR^{-1}=t_{Av}$. 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 $0,c,\ldots,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\leq1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the <discrete subgroup> hypothesis. \b[There is no element of order five.] This argument needs only the given nonzero translation, not a rank-two lattice.
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