Is D3 Abelian Based on Its Cayley Table?

[karush]

$\textsf{Write out a complete Cayley table for $D_3$. Is $D_3$ Abelian?}$
$$
\begin{array}{ l | l l l l l l}
&R_0 &R_{120} &R_{240} &V &D_L &D_R\\
\hline
R_0 & R_0 & R_{120} &R_{240} & V &D_L &D_R\\
R_{120} & R_{120} & R_{240} &R_0 &D_L &D_R & V\\
R_{240} & R_{240} & R_0 &R_{120} &D_R & V &D_L\\
V & V &D_R &D_L & R_0 &R_{240} & R_{120}\\
D_L &D_L & V &D_R &R_{120} & R_0 &R_{240}\\
D_R &D_R &D_L & V &R_{240} & R_{120}& R_0
\end{array}
%$$
$\textsf{Is $D_3$ Abelian?}$

all I know is if it is Abelian then $AB=BA$ in all choices... so ?

 

[topsquark]

$\textsf{Write out a complete Cayley table for $D_3$. Is $D_3$ Abelian?}$
$$
\begin{array}{ l | l l l l l l}
&R_0 &R_{120} &R_{240} &V &D_L &D_R\\
\hline
R_0 & R_0 & R_{120} &R_{240} & V &D_L &D_R\\
R_{120} & R_{120} & R_{240} &R_0 &D_L &D_R & V\\
R_{240} & R_{240} & R_0 &R_{120} &D_R & V &D_L\\
V & V &D_R &D_L & R_0 &R_{240} & R_{120}\\
D_L &D_L & V &D_R &R_{120} & R_0 &R_{240}\\
D_R &D_R &D_L & V &R_{240} & R_{120}& R_0
\end{array}
%$$
$\textsf{Is $D_3$ Abelian?}$

all I know is if it is Abelian then $AB=BA$ in all choices... so ?

Correct. So what is \(\displaystyle V \cdot R_{120}\)? What is \(\displaystyle R_{120} \cdot V\)?

This isn't always true but usually if you have an inversion or mirror plane the group is non-Abelian. They are usually the first things you want to check.

-Dan

 

[karush]

Correct. So what is \(\displaystyle V \cdot R_{120}\)? What is \(\displaystyle R_{120} \cdot V\)?

This isn't always true but usually if you have an inversion or mirror plane the group is non-Abelian. They are usually the first things you want to check.

-Dan

the last 3 rows seem to indicate inversion

 

[topsquark]

the last 3 rows seem to indicate inversion

Yup. Mind you, it doesn't always mean that the group is non-Abelian. Consider the group \(\displaystyle \{ I, R_{180}, i, \sigma _z \}\). (I don't recall the name for this one.) I is the identity, \(\displaystyle R_{180}\) is the rotatin of 180 degrees, i is inversion \(\displaystyle (x, y, z) \to (-x, -y, -z)\), and \(\displaystyle \sigma _z\) is the mirror plane over the xy plane.

It's got an inversion as well as a mirror plane, yet it's Abelian. This group is isomorphic to \(\displaystyle Z_4\).

-Dan

 

[karush]

Yup. Mind you, it doesn't always mean that the group is non-Abelian. Consider the group \(\displaystyle \{ I, R_{180}, i, \sigma _z \}\). (I don't recall the name for this one.) I is the identity, \(\displaystyle R_{180}\) is the rotatin of 180 degrees, i is inversion \(\displaystyle (x, y, z) \to (-x, -y, -z)\), and \(\displaystyle \sigma _z\) is the mirror plane over the xy plane.

It's got an inversion as well as a mirror plane, yet it's Abelian. This group is isomorphic to \(\displaystyle Z_4\).

-Dan

what is \(\displaystyle \sigma _z\) and \(\displaystyle Z_4\)
I shud know but don't

 

[topsquark]

what is \(\displaystyle \sigma _z\) and \(\displaystyle Z_4\)
I shud know but don't

The group elements are operations; they operate on a set of points. So say we have a point at (1, 1, 1). Then

Identity: I (1, 1, 1) = (1, 1, 1)
Rotation about 180 degrees: \(\displaystyle R_{180} (1, 1, 1) = (-1, -1, 1)\)
Inversion: i (1, 1, 1) = (-1, -1, -1)
Mirror plane: This a reflection of the point over the xy plane: \(\displaystyle \sigma _z (1, 1, 1) = (1, 1, -1)\)

So we have 4 points in space. (I'll leave it to you to check that this is actually a group.)

\(\displaystyle Z_4\) is an Abelian group of four elements, the binary operation usually taken to be addition. The elements of this group are 0, 1, 2, 3. This is isomorphic to the modular group (mod 4), \(\displaystyle \mathbb{Z} / 4 \mathbb{Z}\). (If you don't know what that is, don't worry.)

\(\displaystyle \begin{array}{c||c|c|c|c}
Z_4^+ & 0 & 1 & 2 & 3 \\
\hline \hline
0 & 0 & 1 & 2 & 3 \\
1 & 1 & 2 & 3 & 0 \\
2 & 2 & 3 & 0 & 1 \\
3 & 3 & 0 & 1 & 2 \\
\end{array}
\)

-Dan

 

[topsquark]

I'm sorry, I gave you the wrong group. It's not \(\displaystyle Z_4\) that is isomorphic to the point group I brought up, it's V, the Klein 4-group.
\(\displaystyle \begin{array}{c||c|c|c|c|}
V & e & a & b & c \\
\hline \hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a \\
c & c & b & a & e
\end{array}
\)

where e is the identity element for the group. The way you can tell that this point group isn't \(\displaystyle Z_4\) is that each element is its own inverse.

Again, my apologies for the confusion!

-Dan