## PascGalois Triangles for *D*_{3}, the

Symmetry Group for an Equliateral Triangle.

As was the case with *Z*_{2} x *Z*_{2}, there is no
single element that can generate all of *D*_{3}. However, *a*
(or any one of the flips) together with any second element other than *e*
will generate it, so we can still construct PascGalois triangles that display
all of the group elements eventually. We choose to put a rotation, *p*, down
the right side of the triangle and a flip, *a*, down the left side. Here
is what you get with the first 66 rows:

Nothing seems to jump right out as a pattern here, let's see what a few more rows might do:
It is still difficult to discern a pattern. How about if we start shifting colors around? Let's swap the colors for *p* and *c*:

That is a little better. How about if we swap the colors for *a* and *e* now:

Now, what if we moved the blue color over in the spectrum towards green a little:

Remember what the PascGalois Triangle for *Z*_{2} looked like? Lets
redraw it using green for 0 and purple for 1:

Putting these together, we see that the only difference, aside from the variation
in colors in the *D*_{3} triangle, is the extra row of green along
the right hand edge. This occurs because we are using two generators for the *D*_{3}
triangle. If we were to start the *Z*_{2} triangle with 1's down
the left edge and 0's down the right, the result would be essentially the same
as the *D*_{3} triangle with *e, p* and *q* green and *a,
b,* and *c* purple:

Why is this? Consider *e, p,* and *q* as rotations, and relabel them
all *r*. Consider *a,b,* and *c* as flips and relabel them all
as *f*. Remember the multiplication table for *D*_{3}:

* |
e |
p |
q |
a |
b |
c |

e |
e |
p |
q |
a |
b |
c |

p |
p |
q |
e |
b |
c |
a |

q |
q |
e |
q |
c |
a |
b |

a |
a |
c |
b |
e |
q |
p |

b |
b |
a |
c |
p |
e |
q |

c |
c |
b |
a |
q |
p |
e |

Notice that a flip times a flip is always a rotation while a rotation times
a rotation is a rotation, and a flip times a rotation or a rotation times a
flip gives a flip. So identifying all the rotations and all the flips gives:

But the multiplication table for *Z*_{2} is:

So they are essentially the same or
*isomorphic*.

Click to see PascGalois triangles for
other dihedral groups.

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