MHB Does This Cayley Table Accurately Represent the Dihedral Group \(D_6\)?

[Guest2]

This is an attempt to create the Cayley table for dihedral group $D_6$:

$$\begin{aligned} \begin{array}{cc|c|c|c|}
* && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f\\
\\
\hline
e && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline
r && r & r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f \\
\hline
r^2 && r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf \\
\hline
r^3 && r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f \\
\hline
r^4 && r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f\\
\hline
r^5 && r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f \\
\hline
f && f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f \\
\hline
rf && rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e \\
\hline
r^2f && r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r\\
\hline
r^3f && r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf \\
\hline
r^4f && r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf & r^2f \\
\hline
r^{5}f && r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf & r^2f & r^3f \\
\hline
&& & &
\hline
\end{array} \end{aligned}$$

Is this how it should be done? $r$ denotes rotations and $f$ denotes reflections.

 

[topsquark]

This is an attempt to create the Cayley table for dihedral group $D_6$:

$$\begin{aligned} \begin{array}{cc|c|c|c|}
* && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f\\
\\
\hline
e && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline
r && r & r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f \\
\hline
r^2 && r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf \\
\hline
r^3 && r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f \\
\hline
r^4 && r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f\\
\hline
r^5 && r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f \\
\hline
f && f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f \\
\hline
rf && rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e \\
\hline
r^2f && r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r\\
\hline
r^3f && r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf \\
\hline
r^4f && r^{4}f& r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf & r^2f \\
\hline
r^{5}f && r^{5}f & f & rf & r^2f & r^3f &r^4f & r^5f &e & r & rf & r^2f & r^3f \\
\hline
&& & &
\hline
\end{array} \end{aligned}$$

Is this how it should be done? $r$ denotes rotations and $f$ denotes reflections.

You've got a boo-boo. I didn't check the rest of the chart but {e, r, r^2, r^3. r^4, r^5} is a subgroup, but you have r*r^5 = f. (etc) At least some of the f's in the chart should be e's.

-Dan

 

[Guest2]

You've got a boo-boo. I didn't check the rest of the chart but {e, r, r^2, r^3. r^4, r^5} is a subgroup, but you have r*r^5 = f. (etc) At least some of the f's in the chart should be e's.

-Dan

lol, boo-boo indeed! Any better now?

$$\begin{aligned} \begin{array}{cc|c|c|c|}
* && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f\\
\\
\hline
e && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline
r && r & r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e \\
\hline
r^2 && r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r \\
\hline
r^3 && r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 \\
\hline
r^4 && r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3\\
\hline
r^5 && r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 \\
\hline
f && f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 \\
\hline
rf && rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 & f\\
\hline
r^2f && r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 & f & rf\\
\hline
r^3f && r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 & f & rf & r^2f\\
\hline
r^4f && r^{4}f& r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 & f & rf & r^2f & r^3 f\\
\hline
r^{5}f && r^{5}f & e & r & r^2 & r^3 &r^4 & r^5 & f & rf & r^2f & r^3 f & r^4f \\
\hline
&& & &
\hline
\end{array} \end{aligned}$$

 

[topsquark]

Gah! You got the wrong f's! (Shake)

I mentioned r*r^5 = e. You still have it equal to f.

This is the {e, r, r^2, r^3, r^4, r^5} subgroup. It's cyclic so there should be no f's in there.
[math]
\begin{array}{c|c|c|c|c|c|c|}
* & e & r & r^2 & r^3 & r^4 & r^5 \\
\hline e & e & r & r^2 & r^3 & r^4 & r^5 \\
\hline r & r & r^2 & r^3 & r^4 & r^5 & e \\
\hline r^2 & r^2 & r^3 & r^4 & r^5 & e & r \\
\hline r^3 & r^3 & r^4 & r^5 & e & r & r^2 \\
\hline r^4 & r^4 & r^5 & e & r & r^2 & r^3 \\
\hline r^5 & r^5 & e & r & r^2 & r^3 & r^4 \\
\end{array}
[/math]

The f's you changed should be f's.

-Dan

 

[Deveno]

The rest of the table isn't much better off...some things that are true in $D_6$:

A rotation times a rotation is a rotation (this is topsquark's sub-table).

A rotation times a reflection is a reflection.

A reflection times a reflection is a rotation.

All reflections have order 2.

 

[Guest2]

Thanks @Deveno and @topsquark for the feedback. Here's take #3: $\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|}
\circ & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline e & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline r & r & r^2 & r^3 & r ^4 & r ^5 & e & rf & r^2f & r^3f & r^4f & r^5f & f\\
\hline r^2 & r^2 & r^3 & r ^4 & r ^5 & e & r & r^2f & r^3f & r^4f & r^5f & f & rf\\
\hline r^3& r^3 & r ^4 & r ^5 & e & r & r^2 & r^3f & r^4f & r^5f & f & rf & r^2f\\
\hline r^4& r ^4 & r ^5 & e & r & r^2 & r^3 & r^4f & r^5f & f & rf & r^2f & r^3f\\
\hline r^5 & r ^5 & e & r & r^2 & r^3 & r^4 & r^5f & f & rf & r^2f & r^3f & r^4f\\
\hline f & f& rf& r^2f& r^3f& r^{4}f& r^{5}f & e & r & r^2 & r^3 & r^4 & r^5 \\
\hline rf & rf& r^2f& r^3f& r^{4}f& r^{5}f & f & r & r^2 & r^3 & r^4 & r^5 & e \\
\hline r^2f & r^2f& r^3f& r^{4}f & r^{5}f & f & rf & r^2 & r^3 & r^4 & r^5 & e & r \\
\hline r^3f & r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3 & r^4 & r^5 & e & r & r^2 \\
\hline r^4f & r^{4}f& r^{5}f & f & rf & r^2f & r^3f & r^4 & r^5 & e & r & r^2 & r^3 \\
\hline r^5f & r^{5}f & f & rf & r^2f & r^3f & r^4f & r^5 & e & r & r^2 & r^3 & r^4 \\ \hline
\end{array}$

 

[Opalg]

There is still quite a lot wrong. For a start, $r$ and $f$ do not commute. So you should not have the product $fr$ the same as $rf$. In fact, it should be $r^5f$. Also, you have not implemented Deveno's comment that all reflections have order 2. For example, $rf$ is a reflection, so $rf\,rf$ should be $e$.

 

[Guest2]

There is still quite a lot wrong. For a start, $r$ and $f$ do not commute. So you should not have the product $fr$ the same as $rf$. In fact, it should be $r^5f$. Also, you have not implemented Deveno's comment that all reflections have order 2. For example, $rf$ is a reflection, so $rf\,rf$ should be $e$.

Thanks for the feedback, Opalg. I tried to take all the suggestions into account, and here's what I get:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|}
\circ & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline e & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline r & r & r^2 & r^3 & r ^4 & r ^5 & e & rf & r^2f & r^3f & r^4f & r^5f & f\\
\hline r^2 & r^2 & r^3 & r ^4 & r ^5 & e & r & r^2f & r^3f & r^4f & r^5f & f & rf\\
\hline r^3& r^3 & r ^4 & r ^5 & e & r & r^2 & r^3f & r^4f & r^5f & f & rf & r^2f\\
\hline r^4& r ^4 & r ^5 & e & r & r^2 & r^3 & r^4f & r^5f & f & rf & r^2f & r^3f\\
\hline r^5 & r ^5 & e & r & r^2 & r^3 & r^4 & r^5f & f & rf & r^2f & r^3f & r^4f\\
\hline f &f &r^5f & r^4f &r^3f &r^2f & rf& e& r^5& r^4 &r^3 &r^2 & r \\
\hline rf & rf & f & r^5 f & r^4f& r^3f & r^2f& r& e & r^5 & r^4 & r^3 & r^2 \\
\hline r^2f & r^2f & rf &f &r^5f & r^4f & r^3f &r^2 & r & e & r^5& r^4 & r^3 \\
\hline r^3f & r^3f & r^2f & rf& f & r^5f & r^4f& r^3& r^2 & r & e& r^5& r^4\\
\hline r^4f & r^4f & r^3f &r^2f &rf & f & r^5f & r^4 & r^3 & r^2& r&e & r^5\\
\hline r^5f & r^5f&r^4f & r^3f & r^2f &rf &f &r^5 &r^4 &r^3 &r^2 &r & e\\ \hline
\end{array}$$

 

[Deveno]

Thanks for the feedback, Opalg. I tried to take all the suggestions into account, and here's what I get:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|}
\circ & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline e & e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline r & r & r^2 & r^3 & r ^4 & r ^5 & e & rf & r^2f & r^3f & r^4f & r^5f & f\\
\hline r^2 & r^2 & r^3 & r ^4 & r ^5 & e & r & r^2f & r^3f & r^4f & r^5f & f & rf\\
\hline r^3& r^3 & r ^4 & r ^5 & e & r & r^2 & r^3f & r^4f & r^5f & f & rf & r^2f\\
\hline r^4& r ^4 & r ^5 & e & r & r^2 & r^3 & r^4f & r^5f & f & rf & r^2f & r^3f\\
\hline r^5 & r ^5 & e & r & r^2 & r^3 & r^4 & r^5f & f & rf & r^2f & r^3f & r^4f\\
\hline f &f &r^5f & r^4f &r^3f &r^2f & rf& e& r^5& r^4 &r^3 &r^2 & r \\
\hline rf & rf & f & r^5 f & r^4f& r^3f & r^2f& r& e & r^5 & r^4 & r^3 & r^2 \\
\hline r^2f & r^2f & rf &f &r^5f & r^4f & r^3f &r^2 & r & e & r^5& r^4 & r^3 \\
\hline r^3f & r^3f & r^2f & rf& f & r^5f & r^4f& r^3& r^2 & r & e& r^5& r^4\\
\hline r^4f & r^4f & r^3f &r^2f &rf & f & r^5f & r^4 & r^3 & r^2& r&e & r^5\\
\hline r^5f & r^5f&r^4f & r^3f & r^2f &rf &f &r^5 &r^4 &r^3 &r^2 &r & e\\ \hline
\end{array}$$

That looks correct (I honestly don't have time to verify all 144 products). There's a "trick" to this:

There is a rule for "getting the $r$'s in front":

$fr^k = r^{-k}f$.

Sometimes this is written as: $fr^kf = r^{-k}$, and can be proven by noting that:

$frf = frf^{-1}= r^{-1}$, and that:

$fr^kf = fr^kf^{-1} = (frf^{-1})(frf^{-1})\cdots(frf^{-1}) = (frf^{-1})^k = (frf)^k$.

(This same trick works for any dihedral group).

Notice that your Cayley table divides nicely into four 6x6 blocks, the northwest block is the subgroup table for the rotation subgroup $\langle r\rangle$, the northeast and southwest blocks can be taken as "representative" of $f\langle r\rangle$ as a whole (the "other" coset of $\langle r\rangle$ besides the rotation group itself), the southeast block represents my statement that "a reflection times a reflection is a rotation" -which you can see at a glance, since it has no $f$'s in it.

If we call the rotation group $R$, this gives us a "over-table" (whose elements are the 6x6 "blocks"):

$$\begin{array}{c|c|c|}
\circ&R&fR\\
\hline R&R&fR\\
\hline fR&fR&R\\ \hline
\end{array}$$

In other words, a visual demonstration that $D_4/R \cong C_2$ (a cyclic group of order 2).

Geometrically, $r$ is, of course, a rotation (typically counter-clockwise to agree with trigonometry), and $f$ is a "flip" about an agreed-upon axis between two opposite vertices or midpoints. If you *really* want to give yourself a work-out, you can try to think of 2x2 matrices that represent $r$ and $f$, and verify your Cayley table that way, as well (although this may seem like a lot of work, it is part of the process of creating a *character table*, something chemists actually use to classify molecular symmetry-$D_6$ is the symmetry group of the benzene molecule-which plays an important part in many organic compounds).

In any case, it is obvious that if you flip a hexagon, rotate it, and flip it again, the second flip means your rotation now counts in the opposite direction, which is why $frf = r^{-1}$ (because the "flipped plane" turns counter-clockwise to clockwise, and vice versa: it has the opposite *orientation*).

 

[Guest2]

...

Many thanks, Deveno. From what you have said it seems there's all kinds of information encoded into the table!