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I define $r$ to be one rotation clockwise, and s to be reflection on the 'horizontal' line (see the figure).

enter image description here

So I can make these bijections: (in clockwise order) $$\begin{align*} 1,2,3,4,5,6 &\longleftrightarrow \mathrm{id}\\ 1,6,5,4,3,2 &\longleftrightarrow s\\ 6,1,2,3,4,5 &\longleftrightarrow r\\ 5,6,1,2,3,4 &\longleftrightarrow r^2 \end{align*}$$ Note that each 'action' is taken separately on $\mathrm{id}$.

As an example, I want to evaluate $rsr^2$. In order to arrive at the same final figure, on the number arrangements I do $r^2$ first, then $s$ and finally $r$, but on the figure I must do $r$ first, then $s$ and finally $r^2$.

WHY? That is my questions

In other words, on the number arrangements, $$1,2,3,4,5,6 \rightarrow 5,6,1,2,3,4 \rightarrow 3,2,1,6,5,4 \rightarrow 2,1,6,5,4,3,$$ and on the figure one rotation clockwise $\rightarrow$ one reflection on the line $\rightarrow$ two rotations clockwise.

Well I know the fact that in permutation elements actions behave like in composing functions (right to left), but this is a superficial fact to understand intuitively and mathematically why 'actions' in $S_6$ and $D_6$ are in reverse manner?!

One interesting fact is that in $S_6$ actions are 'independent' (all come from $\mathrm{id}$) but in $D_6$ actions are 'dependent' (one happens 'on' the previous event). I can't figure out how this fact can help to solve my questions (both mathematically and intuitively understanding of what's going on).

EDIT - The text above doesn't use any complicated concepts in group theory (isomorphism, ...) and $D_6$ and $S_6$ are used only because of common use, and text is simpler than that. Having said that and since my knowledge is not high in group theory, I truly appreciate simple and clear explanation.

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  • $\begingroup$ To clarify, is $S_6$ the symetric group of order 6, or that of order 6!=720? The latter one is standard, but it seems weird in this case. $\endgroup$ – wythagoras Jul 24 '15 at 19:23
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    $\begingroup$ What does "each 'action' is taken separately on $\mathrm{id}$" or "all come from $\mathrm{id}$" mean? $\endgroup$ – Zev Chonoles Jul 24 '15 at 19:25
  • $\begingroup$ @wythagoras - $S_6$ is of order 6!=720. $\endgroup$ – L.G. Jul 24 '15 at 19:33
  • $\begingroup$ @AlphaE Then I am afraid I don't understand the question. $\endgroup$ – wythagoras Jul 24 '15 at 19:33
  • $\begingroup$ @ZevChonoles - Thank you for your edit. I removed parentheses because they may be confusing with permutation (the numbers in order are numbers clockwise on the figure). -- it means that e.g. $5,6,1,2,3,4 \longleftrightarrow r^2$ is directly re-arrangements of numbers from $id$ and $1,6,5,4,3,2 \longleftrightarrow s$ is also from $id$, nothing to do with $r^2$ (I mean in defining the bijections). $\endgroup$ – L.G. Jul 24 '15 at 19:34
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When you write a "number arrangement" $a,b,c,d,e,f$, you are interpreting it as, the number $a$ winds up in position $1$, $b$ winds up in position $2$, and so on.

But here's a different, equally valid representation: $a,b,c,d,e,f$ is now interpreted as, vertex $1$ winds up at position $a$, vertex $2$ winds up at position $b$, and so on. With this interpretation, $r$ corresponds to $2,3,4,5,6,1$ instead of $6,1,2,3,4,5$. I think that if you set things up that way, you'll find that $rsr^2$ works out by doing $r$, then $s$, then $r^2$.

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