Normalizer of Reflection in Dihedral Group

Theorem
Let $n \in \N$ be a natural number such that $n \ge 3$.

Let $D_n$ be the dihedral group of order $2 n$, given by:


 * $D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$

Let $\map {N_{D_n} } {\set \beta}$ denote the normalizer of the singleton containing the reflection element $\beta$.

Then:
 * $\map {N_{D_n} } {\set \beta} = \begin{cases} \set {e, \beta} & : n \text { odd} \\ \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta} & : n \text { even} \end{cases}$

Proof
By definition, the normalizer of $\set \beta$ is:


 * $\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$

That is:
 * $\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$

First let $g = \beta^k$ for $k \in \set {0, 1}$.

Then:
 * $\beta \beta^k = \beta^k \beta$

Thus:
 * $\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$

Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$.

Suppose $g \alpha = \alpha g$.

Then:

Thus when $n$ is odd, there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.

But when $n$ is even such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.

Hence the result.