Let $S$ be the plane, that is, $S=\{(x,y)|x,y∈R\}$ and consider $f,g∈A(S)$ defined by $f(x,y)=(−x,y)$ and $g(x,y)=(−y,x)$; $f$ is the reflection about the $y−$axis and $g$ is the rotation through $90^{\circ}$ in a counterclockwise direction. define $G=\{f^ig^j|i=0,1;j=0,1,2,3\}$ and let group operation in $G$ be the product of elements in $A(S)$. This group is called the dihedral group of order 8. Try to find a formula for $(f^ig^j)∗(f^sg^t)=f^ag^b$ that expresses $a,b$ in terms of $i,j,s$ and $t$.
first of all, I list out all the elements of $G$ as $\{\require{cancel} e,f,g,g^2,g^3,fg,fg^2,fg^3,\xcancel{gf},\xcancel{g^2f},\xcancel{g^3f}\}$ but I don't understand why those function are not consider as the elements of $G$(given that order 8)? As $fg≠gf$, they should be consider right? Next I tried to list out all of $(f^ig^j)∗(f^sg^t)$ cases to get some insight but didn't get any connection. because the result is $(±x,±y)$ then how could I connect them with $f^?g^?$
If in $$(f^{i}g^{j})*(f^{s}g^{t})=f^{a}g^{b}$$ we have $s = 0$ then $a = i \mod{2}$ and $b= j+t \mod{4}$.
Else we have that for $0\leq i \leq 4$ $$fg^{i}=g^{4-i}f$$
so $s=1$ gives us $$(f^{i}g^{j})*(fg^{t})=(f^{i}f)(g^{4-j}g^{t})=f^{i+1}g^{4-j+t}$$ so here $a= i+1 \mod{2}$ and $b = 4-j+t \mod{4}$