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The n-gon in question is a 3-gon. It is an equilateral triangle to be exact. This is a Dihedral group of order 6 (3 reflections and 3 rotations)

I have plotted the Cayley's table.

The set of elements in $$D_3$$ is $${R_0,R_{120},R_{240},F,F',F"}$$

I know that in order to show this set is an Abelian group, 4 properties must be satisfied: 1) Closure (but under what binary condition?) 2)Identity 3)Inverse 4)Commutativity

How should I apply to property test?

Mathematicing
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  • It isn't abelian. We have $D_3\cong S_3$, which is non-abelian. Just compute some products to show a counter-example – Hayden May 21 '15 at 13:42
  • @HaydenWhat do you mean by 'product'? Are you referring to functional composition? I.e., AB is the function A of B or do you mean A*B? – Mathematicing May 21 '15 at 13:44
  • @Hayden I think you are implying the former. – Mathematicing May 21 '15 at 13:45
  • Functional Composition. In general when we're talking about (not necessarily-abelian) groups, we say "product" to mean the result of applying the group law to a pair of elements. If the group is abelian, it's sometimes common to call them "sums" instead. – Hayden May 21 '15 at 13:47
  • You say you've calculated the Cayley table. In general, an operation on a finite set is commutative if and only if the Cayley table is symmetric about the diagonal (the one going from upper-left corner to lower-right corner). – Hayden May 21 '15 at 13:48
  • @Hayden That certainly helps. It isn't abelian. The rotation of 120 followed by a reflection F is not the same as a reflection F followed by a rotation of 120. – Mathematicing May 21 '15 at 13:49
  • Precisely, and this same idea applies to all the other Dihedral groups (except for $D_2$ or $D_1$, where every element is its own inverse). Moreover, $D_3\cong S_3$ is the smallest non-abelian group, and can be used to show that in general $S_n$ is non-abelian for $n\geq 3$ (one can embed $S_3$ in $S_n$ for $n\geq 3$) – Hayden May 21 '15 at 13:51
  • @Hayden Thank you. One last but important question for this topic. How should I check to test if this set is a group? – Mathematicing May 21 '15 at 13:51
  • The identity part shouldn't be too hard (what does $R_0$ do?), and the inverses also can be found from geometric reasoning. As far as associativity is concerned, remember that in general functional composition is associative, so you don't really need to do anything than notice this fact. – Hayden May 21 '15 at 13:53
  • Well, $$R_0$$ maps any point contained within the triangle to the exact same point. ( It does nothing) – Mathematicing May 21 '15 at 13:56
  • Precisely, so what happens when you compose with anything else? (I've made an answer that explains this a bit further and ensures an answer is actually associated with the question. If you have any further questions I recommend you ask them on the answer) – Hayden May 21 '15 at 14:02

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(Just so there's an answer)

A general fact that you can use to test whether a binary operation $\ast$ on a set $S$ is commutative given the Cayley table has already been computed is by seeing whether or not it is symmetric about the main diagonal (i.e. the one going from the upper-left corner to the lower-right corner). This fact follows immediately from the definition of what it means to be commutative and from what the Cayley table is.

As far as the identity element is concerned, remember that for a function $f:S\rightarrow S$, we have $\mathrm{id}_S\circ f=f\circ \mathrm{id}_S$, so that once you've found that the identity map is in your set, then you have an identity for functional composition.

For inverses, the best idea is to try to think of $D_3$ geometrically as the set of rotational symmetries for the equilateral triangle. Your elements consist of rotations about the center and composition with a reflection about one of the (fixed) axes of symmetry. What would you think of as the inverse of a rotation about some angle $\theta$? Would would you think of as the inverse of a reflection? Note that once you've found $x^{-1}$ and $y^{-1}$ for some elements $x,y$, the inverse of $xy$ is given by $y^{-1}x^{-1}$, allowing you to compute the inverses of the compositions once you've calculated the inverses of the reflection and of the rotations.

Finally, for associativity, all you need to realize that functional composition is always associative.

Hayden
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  • To show inverse, would I have to show that for some function f and function g, $$fg = gf = R_0$$? More concretely, a rotation 120 followed by a rotation 240 is equal to a rotation of 240 followed by a rotation of 120. Both are inverse of each other-and by the properties of groups there can be only one inverse-that it, inverses are unique. – Mathematicing May 21 '15 at 14:11
  • Yep, precisely. Alternatively, you can think of it as performing a rotation with angle $-120$ degrees, which is the same thing as a rotation by $360-120=240$ degrees (really two ways of looking at the same thing). – Hayden May 21 '15 at 14:16
  • And in order to determine the inverse of a reflection, I find some function h such that this function when compose with another reflection function returns a point to the original point – Mathematicing May 21 '15 at 14:18
  • @user Yes, in general a reflection composed with itself is the identity. – Hayden May 21 '15 at 14:20
  • Tremendously helpful. Although if it isn't too much of a hassle, I'm curious as to why we would be interested to find, for instance if $$x^-1$$ is the inverse of a rotation function and $$y^-1$$ is the inverse of a reflection function,the functional composition xy? – Mathematicing May 21 '15 at 14:21
  • Because, if $\rho$ is one of the rotations (which one doesn't really matter) and $\tau$ is the reflection, then $D_3={1,\rho,\rho^2,\tau,\tau \rho,\tau\rho^2}$. You've calculated $\rho^{-1}$ (and consequently $(\rho^2)^{-1}$) and $\tau^{-1}$. But what of $\tau \rho$ and $\tau\rho^2$? By using the fact that you know $\tau^{-1}=\tau$ and $\rho^{-1}=\rho^2$, we know that $(\tau \rho)^{-1}=\rho^{-1}\tau^{-1}=\rho^2\tau$ (which is actually $\tau\rho$ again). You can do likewise for $\tau\rho^2$ – Hayden May 21 '15 at 14:25