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Hello, I'm a bit confused on how to go about doing these calculations. For example, I am asked to compute the following in D6:

(μρ^2)(μρ^3) I know that the answer is ρ, but I'm completely unsure of how my professor arrived to that answer.

We are also given problems to work in D5 and then doing the same in D8, but I'm trying to at least understand this example before I can even begin to start the rest.

Thanks!

2007-09-11 11:37:30 · 2 answers · asked by The Light One 3 in Science & Mathematics Mathematics

2 answers

I can help you here. I'll assume that ρ and μ
are your generators for D6.
If you study the group properties you will find that
ρ^6 = 1, μ² = 1 and μρ = ρ^-1μ.
Using these properties,
let's compute μρ²μρ³.
General idea. The associative law let's you work
problems like from the middle outward.
Let's write our product as
μρ ρ μρ ρ².
(I've left spaces here to show the parts
that I'm working with.)
Now replace each μρ by ρ^-1μ.
We get
ρ^-1μ ρ ρ^-1μ ρ².
Next, cancel the ρρ^-1 in the center to get
ρ^-1 μ² ρ².
But
μ² = 1, so we are left with
ρ^-1ρ²,
but that's just ρ and we are done!

2007-09-11 15:55:40 · answer #1 · answered by steiner1745 7 · 0 0

Okay, I'm assuming that μ is a reflection and ρ is a rotation. Remember that in any dihedral group (not just D6), ρμ = μρ⁻¹. We can use that to simplify the expression you were given:

μρ²μρ³
μρρμρ³
μρμρ⁻¹ρ³ (here we just swapped ρμ for μρ⁻¹)
μρμρ²
μμρ⁻¹ρ²
μ²ρ
ρ (since μ² is the identity element).

2007-09-11 19:11:53 · answer #2 · answered by Pascal 7 · 0 0

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