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In the group of all the bijections of the Euclidean plane onto itself, let $f(x,y) \colon = (-x,y)$ and $g(x,y) \colon = (-y,x)$ for all points $(x,y)$ in the plane. Let $$G:= \{f^i g^j | i=0,1; \ g=0,1,2,3 \}.$$ Then $G$ is the dihedral group of order $8$. Now given $i$, $j$, $s$, $t$, how to find $a$ and $b$ in terms of $i$, $j$, $s$, $t$ such that $$(f^i g^j)(f^s g^t) = f^a g^b?$$

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Noticing that

  • $fg=-gf$,
  • $- \mathrm{Id}$ commute with $f$ and $g$,
  • $g^2=- \operatorname{Id}$,

you should be able to show that $(f^ig^j)(f^sg^t)=f^{i+s}g^{(2s+1)j+t}$.

For an example, $$ \begin{array}{ll} f^3gf^2g^4 & =f^3(gf)fg^4=-f^3(fg)fg^4, \\ & =-f^3f(gf)g^4 = f^3f(fg)g^4, \\ & = f^5g^{5}. \end{array}$$

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Seirious, I'm not able to come up with this formula. So could you please elaborate? – Saaqib Mahmuud Sep 18 '13 at 10:18
    
I added an example. – Seirios Sep 18 '13 at 10:27

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