What is the order of the subgroup of G (rubiks cube group) generated by <FF,RR>?

Question

I got:

=1,FF,RR,FFRR,RRFF

But in my text book the answer is 12? Does any one else know the other elements?

I assumed since FFFFRRRR=1 there were no more?

Answer 1

Let $f=FF$ and $g=RR$. Note that $f$ and $g$ have order $2$, and $fg$ and $gf$ have order $6$, i.e., we have $(fg)^6=e$ and no smaller power of it. This still is not enough to show that the group generated by $f$ and $g$ has order $12$. We also have $$ g(fg)^2=(fg)^3f. $$ So the $12$ elements are $$ e,f,fg,(fg)f,(fg)^2,(fg)^2f,g,gf,(gf)g,(gf)^2,(gf)^2g,(gf)^3=(fg)^3. $$

Answer 2

using $f=FF$ and $g=RR$ (borrowing from @dietrich) isn't the answer just: $$ e,f,fg,fgf,(fg)^2,(fg)^2f,(fg)^3,(fg)^3f,(fg)^4,(fg)^4f,(fg)^5,(fg)^5f $$

and then $(fg)^6=e$?