Normality of the Subgroup ( langle r^2 rangle ) in the Dihedral Group ( D_n )

The dihedral group ( D_n ) is the group of symmetries of a regular ( n )-gon, which includes ( n ) rotations and ( n ) reflections. This article aims to explore why the subgroup ( langle r^2 rangle ) of ( D_n ) is a normal subgroup, emphasizing the fundamental principles of group theory and the properties of the dihedral group.

Understanding the Dihedral Group ( D_n )

The dihedral group ( D_n ) can be presented as:

D_n langle r, s mid r^n e, s^2 e, srs r^{-1} rangle

Here, ( r ) represents a rotation by ( 360^circ/n ), and ( s ) represents a reflection. The group ( D_n ) thus consists of the symmetries of a regular ( n )-gon.

The Subgroup ( langle r^2 rangle )

The subgroup ( langle r^2 rangle ) is generated by the rotation ( r^2 ). This subgroup consists of the elements:

langle r^2 rangle { e, r^2, r^4, ldots, r^{n/2} }

where ( n/2 ) is the largest integer such that ( 2k leq n ).

Normality of ( langle r^2 rangle )

A subgroup ( H ) of a group ( G ) is normal if for every ( g in G ) and ( h in H ), the element ( g h g^{-1} ) is also in ( H ). We need to show that this holds for ( g in D_n ) and ( h in langle r^2 rangle ).

Cases for Elements of ( D_n )

Consider the elements of ( D_n ): ( g ) can be either a rotation ( r^k ) or a reflection ( s r^k ).

Case 1: ( g r^k ) (rotation)

- For ( h r^{2m} in langle r^2 rangle ): [ r^k h r^{-k} r^k r^{2m} r^{-k} r^{2m} in langle r^2 rangle ]This shows that ( langle r^2 rangle ) is invariant under conjugation by any rotation.

Case 2: ( g s r^k ) (reflection)

- For ( h r^{2m} in langle r^2 rangle ): [ s r^k h s s r^k r^{2m} s s r^{2m} r^{-k} s r^{2m} s r^{-2m} ]Since ( r^{-2m} r^{2n/2 - m} ) which is also in ( langle r^2 rangle ) as it is a rotation by an even multiple of ( 360^circ/n ), we have: [ s r^k r^{2m} s in langle r^2 rangle ]

Conclusion

Since ( langle r^2 rangle ) is invariant under conjugation by both rotations and reflections in ( D_n ), we conclude that ( langle r^2 rangle ) is a normal subgroup of ( D_n ). Thus, we can state:

langle r^2 rangle trianglelefteq D_n

Generalization to ( langle r^d rangle )

The approach to solving this may depend on your axioms, but whatever axioms used are equivalent to: The dihedral group ( D_n ) has generators ( r ) (rotation) and ( s ) (flip) with relations where ( e ) is the identity element:

1. ( r^n e ) 2. ( s^2 e ) 3. ( s r r^{-1} s )

To simplify arbitrary "words" to one of the two forms ( r^k ) or ( r^k f ) by moving and cancelling extra copies of ( s ) as needed.

Normal subgroup - You just need to conjugate an arbitrary element of ( langle r^2 rangle ) by an arbitrary element of ( D_n ) and get another element of ( langle r^2 rangle ). The relations should make it straightforward to cancel out copies of ( s ), leaving you with an even power of ( r ).

Question and Further Discussion

Was there anything special about the exponent 2 or does this argument work for the subgroup ( langle r^d rangle ) in general?

The exponent 2 is special in this proof because it corresponds to rotations by ( 180^circ ). However, the argument can be generalized to any even exponent ( d ) as the subgroup ( langle r^d rangle ) will still be normal. The key is that ( d ) must be even for the subgroup to be normal under the given relations.