Isomorphic to subgroup

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In summary, the discussion shows that D5 is isomorphic to a subgroup of S5, with a function f: D5->S5, where f(a,b) = f(a)f(b), f is one to one and onto. The elements of D5 can be determined by first picking a vertex and then a new vertex to move it to, with two options for orienting it. This results in a total of 10 elements in D5. Writing out the permutations explicitly shows the isomorphism between D5 and S5.
  • #1

Homework Statement

By considering the vertices of the pentagon, show that D5 is isomorphic to a subgroup of S5.

Write all permutations corresponding to the elements of D5 under this isomorphism.

The Attempt at a Solution

To show isomorphic, need to find a function f: D5->S5, where f(a,b) = f(a)f(b), f is one to one and onto.

I'm having a hard time determining the elements of of D5. D4 was trivial, but this one doesn't make sense to me.
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  • #2
If you understand D4 then it shouldn't be too hard to understand D5. Any element in D5 can be determined in the following way: first pick a vertex and then a new vertex to which you want to move it. You then have two options of orienting it by flipping it over. This means there is a total of 10 elements in D5 (this argument works in general to show that Dn has order 2n).

Since you can think of a symmetry of a pentagon as a permutation of its 5 vertices, it follows that D8 is isomorphic to a subgroup of S5 (of course not all permutations are allowed since adjacent vertices must remain adjacent. Indeed, S5 has order 5! = 120). Note that for this reason your function f cannot be one-to-one.
  • #3
Well, you could imagine flip flopping a pentagon around to get the elements of [itex]D_5[/itex], or you could look at Section 2.8 of the following book to see the dihedral groups of arbitrary even order written out. Note that in this book the author uses the naming convention [itex]D_{2n}[/itex], [itex]n\in \mathbb{N}[/itex]. So your [itex]D_5[/itex] is his [itex]D_{10}[/itex] (that is, the subscript is the actual order of the group).

Book:, by Mark Steinberger.
  • #4
So the permutations of D5= (1); (12345); (15432); (12)(53)(4); (54)(13)(2); (43)(25)(1); (53);(12);(15)(24)(3); (23)(14)(5) Is this correct?

How can I show this is isomorphic to D5?
  • #5
You can just write out the elements of D5 explicitly (usually people give them names like r, r2, ... and s, s2 - see your favorite book), consider each one and write it out as a permutation. That will explicitly give you an isomorphism.
For example, the identity 1 of D5 is sent to the identity (1) of S5;
The rotation r around an angle [itex]2\pi / 5[/itex] is sent to (12345) in S5;
The rotation r2 around an angle [itex]2 \times 2\pi / 5[/itex] is sent to (13524) = (12345)(12345) in S5, etc.

Then you will see that you get a bijective function and you can explicitly check that it preserves the group operation, although that is quite obvious from the construction.

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