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NOTE - I didn't receive any answer in here and I think because my first post is not clear, so I entirely made another example:

$K={\{id,r^2,r^4,s,r^2s,r^4s}\}$ is a proper subgroup of the dihedral group $D_6$. As it is shown here by Gerry Myerson, $K$ is isomorphic to $S_3$. We label the vertices of the hexagon 1 through 6. Forget about 2, 4, and 6, and see what $K$ does to 1, 3, and 5. You will see that the 6 elements of $K$ are precisely the 6 permutations of 1, 3, 5, thus, precisely $S_3$; i.e: $$\begin{align*} \mathrm{id} &\longleftrightarrow \mathrm{id}\\ \tau_s=(35) &\longleftrightarrow s\\ (135) &\longleftrightarrow r^4\\ (13) &\longleftrightarrow r^4s\\ \tau_{r^2s}=(15) &\longleftrightarrow r^2s\\ \tau_{r^2}=(153) &\longleftrightarrow r^2 \end{align*},$$ $r$ means one rotation clockwise and $s$ means flip on horizontal line. In order to achieve the following new state of hexagon,

enter image description here

we first do $r^2$ then $s$. But in $S_3$ it happens in reversed way, i.e. first $\tau_s$ (the permutation bijective to $s$) then $\tau_{r^2}$ (the permutation bijective to $r^2$) or mathematically

$(15)(\mathrm{id})=(153)(35)(\mathrm{id}) \Leftrightarrow \tau_{r^2s} (\mathrm{id}) = \tau_{r^2} \circ \tau_s (\mathrm{id})$.

My question is that to reach to the same figure why (both in mathematical and intuitive explanations) in $K$ it is right-to-left (first $r^2$ then $s$) but in $S_3$ it is left-to-right (first $\tau_s$ (bijective of $s$) then $\tau_{r^2}$ ((bijective of $r^2$))?

EDIT - My knowledge of group theory is limited to few first chapters of C.C.Pinter's Abstract Algebra, and I highly appreciate easy-to-understand explanations. My question is on the reason of reversed order of 'actions', intuitively/geometrically/mathematically.

Thank you.

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  • $\begingroup$ This example is entirely different from the previous post. If it is appropriate to not have the previous post, please comment I will delete the previous one; or should we wait to some answer? Thank you $\endgroup$ – L.G. Jul 25 '15 at 3:19
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    $\begingroup$ The order you choose for composition is a convention; as long as you pick a consistent convention, it doesn't matter. Any group $G$ is naturally isomorphic to the same group $G^{op}$ with reversed composition, with the isomorphism given by $g \mapsto g^{-1}$. $\endgroup$ – Qiaochu Yuan Jul 25 '15 at 4:47
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    $\begingroup$ @QiaochuYuan - Sorry I didn't understand (I know very primary group theory). And my question is why isomorphic to reversed way to group-acting? $\endgroup$ – L.G. Jul 25 '15 at 4:50
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    $\begingroup$ It will be easier to follow your post if you include the "original" labeling of the hexagon. $\endgroup$ – Ted Jul 28 '15 at 2:07
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There is absolutely no reason at all. It is entirely up to the author of the textbook which way these things go, and different textbooks will use different conventions for both the symmetric group and the dihedral group.

Whenever you define a group of transformations or operations, you have the choice as to whether $gh$ should mean "do $g$ first, and then $h$", or whether it should mean "do $h$ first and then $g$". These two choices correspond to two possible binary operations on the group, and you have to pick one of these operations.

Now, you might think it's obvious that $gh$ should mean "do $g$ first, and then $h$". However, that's not how composition of functions works. Recall that if $f$ and $g$ are functions, the composition $f\circ g$ is defined by the formula $$ (f\circ g)(x) = f(g(x)). $$ As you can see, in the composition $f\circ g$, the function $g$ is applied to $x$ first, and then $f$ is applied to the result.

So when you're defining a group, you have to decide between using the "obvious" order, and using the order that agrees with function composition. There isn't a good choice here, so some books use the "obvious" order, some books use the "backwards" order, and some books (such as yours, apparently) use different orders for different groups.


Incidentally, the reason that function composition is "backwards" is that everybody writes functions to the left of the input, which is a little strange when you think about it. Everything would be so much simpler if people wrote $(x)f$ instead of $f(x)$, for then function composition could be defined as $$ (x)(f\circ g) = ((x)f)g, $$ which would mean that $f$ would go first. But no one does this, and everyone is so used to functions being on the left that equations like this are actually a little hard to read.

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    $\begingroup$ If my memory serves me, back when I was in grad school (an earlier geological era), there were a couple of algebra text books that used the notation $(x)f$ — “take $x$ and $f$ it”. $\endgroup$ – Lubin Jul 28 '15 at 2:45
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I assume you're using the original labeling here. I think you made some mistake in your calculation, or are using inconsistent conventions. At the top of your post, you indicate that $r^2 s$ corresponds to (15). Now (15) is what you get by doing $s$ first, then $r^2$. But then later in your post, you seem to think that $r^2 s$ means "do $r^2$ first, then $s$".

If you use a consistent convention, then everything works out.

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  • $\begingroup$ I never said first $s$ then $r^2$. I said first $\tau_s$ then $\tau_{r^2}$.. $\endgroup$ – L.G. Jul 28 '15 at 7:07
  • $\begingroup$ You wrote that (15) corresponds to $r^2s$. This is only correct if $r^2s$ means "first $s$, then $r^2$". $\endgroup$ – Ted Jul 28 '15 at 8:16
  • $\begingroup$ Or, possibly you are assuming that the numbers move with the hexagon when you rotate it, so that $s$ isn't always "reflection across the horizontal" but instead "reflection across the 1-4 axis, wherever it may currently be." I am assuming that the numbers stay put when the hexagon moves. In general, these 2 conventions (labels move vs labels stay) cause a right-to-left vs left-to-right reversal (see "alias vs alibi transformations".) $\endgroup$ – Ted Jul 28 '15 at 8:30
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    $\begingroup$ Alias vs alibi: isites.harvard.edu/fs/docs/icb.topic446952.files/aliasalibi.pdf $\endgroup$ – Ted Jul 28 '15 at 8:37
  • $\begingroup$ I read your answer again and now I got what you mean: Yes, by rotation or/and reflection I mean that numbers stuck to the shape. Anyway, I am reading through the link you commented and I trying to figure out what's going on, hopefully. Becasue the shape-without-numbers is the same by $r^n$ and/or $s$, that's why I stuck numbers to it while moving it. Thank you $\endgroup$ – L.G. Jul 28 '15 at 11:31
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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

As an example, I want to evaluate $rsr^2$. Without lose of generality, I choose right-to-left as order of action that is first $r$ then $s$ and $r^2$ acting on id (first configuration) on the figure. Pictorially:

enter image description here

So the final equivalence is $(12)(36)(45)$.

I am going to do the same 'direction of action' but now on permutation, that is first $r$ then $s$ and $r^2$ acting on id ($(1)$) on the number arrangements or permutation. Mathematically:

$\tau_1=r\circ\text{id}=(123456)(1)=(123456)\rightarrow \tau_2=s\circ\tau_1=(26)(35)(123456)=(16)(25)(34)\rightarrow\tau_3=r^2\circ\tau_2=(135)(246)(16)(25)(34)=(12)(36)(45)$.

Conclusion: The contradiction came from labeling one of 'outside and inside' of vertices not two of them. And, there is no revered order of actions comparing permutation and dihegral group.

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