Simplifying f=uRuR in D3: A Beginner's Guide

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In the discussion about simplifying the expression f = u R u R in the context of dihedral groups, particularly D3, participants clarify the notation and group operations involved. The relations defining the dihedral group, such as u^2 = 1 and R^n = 1, are emphasized to aid in simplification. The solution provided indicates that R = 1, which aligns with the properties of the group. Additionally, there is confusion regarding the roles of u and R, prompting questions about identifying rotations and reflections. Overall, the conversation focuses on understanding the structure and operations of D3 to simplify the given expression correctly.
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Homework Statement



In D3, simplify f = u R u R


Homework Equations






The Attempt at a Solution



u(Ru)R=(uR)uR
u(uR^2)R=(uR)u2
uR=R

I am doing this for independent study, and the website I am using says the solution is 1=R

I just got started on dihedral groups and have no idea what in the heck to do. How do you tell which is a rotation or reflection before the solution arises and/or arrive with the above answer of R=1

Thanks
 
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I have no idea what the notation "u R u R" means. Are u and R members of the group or something else?
 
I have no idea what the notation "u R u R" means. Are u and R members of the group or something else?

That is the way the problem is stated. Here are some relevant equations and an example problem.

Relevant equations:
To describe the group operation of Dn, we need only note the following rules (called "relations"):
u^2 = 1 (in fact, uk^2 = 1 for every reflection uk)
Rn = 1 (so R-1 = R^n-1)
R u = u R-1 = u R^n-1

To make calculations unambiguous, we agree that every element of Dn will have a standard form:

Rotations: Rk (or 1 for the identity = R0)
Reflections: u Rk (or just u itself if k = 0)
Since Rn = 1, we can take k between 1 and n-1

Example problem:
Example 1: In D3, the symmetry group of the equilateral triangle, calculate a = u R2 u.

Here is a sequence of steps using the relations in D3 along with associativity:
a = u R2 u = (u R2) u = u (R2 u)
= u (R R u) = u R (u R2)
(note we are using relation 3 above, for n = 3)
= u (R u) R2 = u u R4
(now use relations 1 and 2)
= u 2 R4 = R
So we conclude: a = u R2 u = R in D3
 
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