The discussion centers on the structure of the dihedral group Dm, specifically the relationship between its subgroups . It is established that Dm can be represented as Dm = ⟨r, s | r^m = s^2 = srsr = 1⟩, indicating that all elements can be expressed in terms of the generators r and s. The participants clarify that the product of the subgroups ⟨r⟩ and ⟨s⟩ does not form a direct product due to the non-normality of one of the subgroups. Consequently, no theorem is required to assert the relationship between Dm and its subgroups.
Mathematicians, particularly those studying abstract algebra, students working on group theory assignments, and educators looking to clarify the structure of dihedral groups.
There is no theorem needed. If you look at the representation ##D_m=\langle r,s\,|\,r^m=s^2=srsr=1 \rangle ## then can you say whether all words over the alphabet ##\{\,r,s\,\}## can be written as ## r^k \cdot 1## or ##r^k\cdot s\,?## Or what do you mean by ##\langle r\rangle \langle s\rangle\,?##AllRelative said:Homework Statement
This is only a step in a proof I am trying to make.
Let Dm be the dihedral group.
r is the rotation of 2π/m around the origin and s is a reflexion about a line passing trough a vertex and the origin.
Let<s> and <r> be two subgroups of Dm.
Is there a theorem that states that Dm = <r><s>
Homework Equations
The Attempt at a Solution
Thanks for the help
fresh_42 said:There is no theorem needed. If you look at the representation ##D_m=\langle r,s\,|\,r^m=s^2=srsr=1 \rangle ## then can you say whether all words over the alphabet ##\{\,r,s\,\}## can be written as ## r^k \cdot 1## or ##r^k\cdot s\,?## Or what do you mean by ##\langle r\rangle \langle s\rangle\,?##
As groups, and if you consider ##\langle r\rangle \langle s\rangle = \langle r\rangle \times \langle s\rangle## as a direct product, then this is not true. One of those subgroups is not normal.