Isomorphism of dihedral with a semi-direct product

In summary, the conversation discusses a problem involving groups and the goal is to show that Dm is isomorphic to the semidirect product of Zm and Z2, with a specific function phi. The conversation includes attempts at solving the problem and the suggestion to write it in terms of elements r and s. It is also mentioned that the function phi needs to be defined for both Zm and Z2.
  • #1
AllRelative
42
2

Homework Statement


Let m ≥ 3. Show that $$D_m \cong \mathbb{Z}_m \rtimes_{\varphi} \mathbb{Z}_2 $$
where $$\varphi_{(1+2\mathbb{Z})}(1+m\mathbb{Z}) = (m-1+m\mathbb{Z})$$

Homework Equations


I have seen most basic concepts of groups except group actions. Si ideally I should not use them for this problem.

The Attempt at a Solution


So I've been thinking about this problem for a couple of days and I just can't seem to arrive to the proof I am looking for.

And so we have that <s> is cyclical of order 2 and <r> is cyclical of order m. Therefore, $$\langle s \rangle \cong \mathbb{Z}_2$$
and $$\langle r \rangle \cong \mathbb{Z}_m$$
I feel I have to use the fact that <r> is normal in Dm and that <r>∩<s> = {e}. I unsure where to go from there

Thanks for the help
 
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  • #2
I suggest to write it in terms of ##r## and ##s## first, and then see how it translates to ##\mathbb{Z}_m##, resp. ##\mathbb{Z}_2##. And I will write the representatives, aka elements of ##\mathbb{Z}_m## as ##[k] = k+m\mathbb{Z}## which is more convenient and easier to read.

In general a semidirect product ##G = N \rtimes_\varphi H## goes by ##(n_1\, , \,h_1)\cdot (n_2\, , \,h_2) \stackrel{(*)}{=} (n_1 \cdot \varphi(h_1)(n_2)\, , \,h_1\cdot h_2)##, see e.g. https://en.wikipedia.org/wiki/Semidirect_product. We have ##\varphi\, : \,H \longrightarrow \operatorname{Aut}(N)##.

Here we have products ## P =(r^k\; , \;s^\varepsilon)\cdot (r^n\; , \;s^\eta) = r^k\cdot s^\varepsilon \cdot r^n \cdot s^\eta## with ##\varepsilon \in \{\,0,1\,\}##.
Now write ##P## in the form ##P=r^l \cdot s^\mu=(r^l,s^\mu)## and compare that with ##(*)## to see how ##\varphi(s^\varepsilon)(r^n)## has to be defined.
If you're done, you can write this as ##\varphi([\varepsilon])([n])##.

You have only defined ##\varphi([1])([1])##. What are ##\varphi([0])## and ##\varphi([1])([n])## if ##n>1## or short: What is ##\varphi([\varepsilon])## with ##\varepsilon \in \{\,0,1\,\} = \{\,[0],[1]\,\}\,?##
 

Related to Isomorphism of dihedral with a semi-direct product

1. What is an isomorphism?

Isomorphism is a mathematical concept that describes a relationship between two mathematical structures that preserves their essential properties. In simpler terms, it means that two structures are essentially the same even though they may look different.

2. What is dihedral?

Dihedral refers to a group of symmetries that can be applied to a regular polygon, such as a square or equilateral triangle. It is represented by the symbol Dn, where n is the number of sides of the polygon.

3. What is a semi-direct product?

A semi-direct product is a mathematical operation that combines two groups in a specific way to form a new group. It is denoted by the symbol ⋉ and is used to describe the structure of certain groups, such as the dihedral group Dn.

4. How is dihedral related to a semi-direct product?

The dihedral group Dn can be expressed as a semi-direct product of two groups - a cyclic group of rotations and a cyclic group of reflections. This means that the structure and properties of Dn can be understood through the structure and properties of these two groups.

5. Why is the isomorphism of dihedral with a semi-direct product important?

The isomorphism between dihedral and a semi-direct product allows us to better understand and analyze the properties of the dihedral group. It also helps us to connect dihedral with other mathematical concepts and structures, making it a useful tool in various fields of mathematics and science.

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