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'Simplifying Transformations' printed from https://nrich.maths.org/
Jannis Ahlers (Long Bay Primary) found 8
transformations:
"The answer is 8. I found this by finding all the possible
positions the shape could end in by only using R, S and there
inverses."
The 8 possible transformations are:
$I$, $S$, $S^2$, $S^3=S^{-1}$, $R$, $R S=S^{-1}R$, $R S^2=S^2R$, $
R S^3=S R$.
There are eight transformations made up only of $R$, $S$ and their
inverses. Neat way to see this: draw the eight that you think
exist, then note that applying $R$ or $S$ to any of them gives
another of them, so we can't `escape' from these eight. The
simplest expressions for the eight are:
$I$, $S$, $S^2$, $S^3=S^{-1}$, $R$, $R S=S^{-1}R$, $R S^2=S^2R$, $
R S^3=S R$.
Notice that $R S R^{-1}=S^{-1}$. (Of course, $R^{-1}=R$, so $R S
R=S^{-1}$, and this can also be written as $S R=R S^{-1}$.)
So the two expressions simplified are:
$S S R S R^{-1} S R S R^{-1} = S S(R S R^{-1})S(R S R^{-1})= S S
S^{-1}S S^{-1} = S$
and
$S^{-1}R R S R S R R^{-1} S R^{-1} = S^{-1}(R R)S R S(R R^{-1})S
R^{-1} = S^{-1} S R S S R^{-1}=(S^{-1}S)R S S R^{-1}=R S(S R)=R S R
S^{-1}= (R S R)S^{-1}=S^{-1}S^{-1}=S^{-2}=S^2$.