Subgroups generated by powers of r are normal in Dih(2n)

Let D_{2n} = \langle r,s \ |\ r^n = s^2 = 1, rs = sr^{-1} \rangle be the usual presentation of the dihedral group of order 2n and let k be a positive integer dividing n.

  1. Prove that \langle r^k \rangle is normal in D_{2n}.
  2. Prove that D_{2n}/\langle r^k \rangle \cong D_{2k}.

  1. By a previous exercise it suffices to note that rr^kr^{-1} = r^k \in \langle r^k \rangle and sr^ks^{-1} = r^{-k} = (r^k)^{-1} \in \langle r^k \rangle.
  2. Define a mapping \overline{\varphi} : \{ r,s \} \rightarrow D_{2n}/\langle r^k \rangle by r \mapsto r \langle r^k \rangle and s \mapsto s \langle s \langle r^k \rangle. Note the following.
    1. Note that 1 \leq k \leq n. Now (r \langle r^k \rangle)^a = r^a \langle r^k \rangle = \langle r^k \rangle if and only if r^a \in \langle r^k \rangle. The least integer a such that (r \langle r^k \rangle)^a = \langle r^k \rangle is thus k.
    2. \overline{\varphi}(s)^2 = s^2 \langle r^k \rangle = \langle r^k \rangle = \overline{1}, so that |\overline{\varphi}(s)| = 2.
    3. \overline{\varphi}(r) \overline{\varphi}(s) = r \langle r^k \rangle \cdot s \langle r^k \rangle = (rs) \langle r^k \rangle = sr^{-1} \langle r^k \rangle = s \langle r^k \rangle \cdot (r \langle r^k \rangle)^{-1} = \overline{\varphi}(s) \overline{\varphi}(r)^{-1}.
    4. Suppose by way of contradiction that \overline{\varphi}(s) \in \langle \overline{\varphi}(r) \rangle; then we have s = r^m for some integer m, a contradiction.

    By a lemma to a previous exercise, \overline{\varphi} induces an injective group homomorphism \varphi : D_{2k} \rightarrow D_{2n}/\langle r^k \rangle. Moreover, \varphi is surjective because (s^ar^b) \langle r^k \rangle = \varphi(s^ar^b).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: