Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$(f^i g^j)(f^s g^t) = f^a g^b?$$

share|improve this question

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.