Sub groups of the dihedral group

[AllRelative]

Homework Statement

This is only a step in a proof I am trying to make.

Let D_m 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 D_m.

Is there a theorem that states that D_m = <r><s>

Homework Equations

The Attempt at a Solution

Thanks for the help

 

[fresh_42]

Homework Statement

This is only a step in a proof I am trying to make.

Let D_m 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 D_m.

Is there a theorem that states that D_m = <r><s>

Homework Equations

The Attempt at a Solution

Thanks for the help

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.

 

[AllRelative]

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.

Thanks again for the response you rock.
You made me realize I was not on the right track. But I think I found it. Thanks!

By the way, it would probably be easier if I wrote the whole problem I'm working on but I'm concerned about plagiarism since this is an assignment haha.