Normalizer of Reflection in Dihedral Group
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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:
\(\ds \alpha^j \beta^k \beta\) | \(=\) | \(\ds \beta \alpha^j \beta^k\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^j \beta \beta^k\) | \(=\) | \(\ds \beta \alpha^j \beta^k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^j \beta\) | \(=\) | \(\ds \beta \alpha^j\) | Cancellation Laws | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^j \beta\) | \(=\) | \(\ds \alpha^{n - j} \beta\) | Product of Generating Elements of Dihedral Group | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^j\) | \(=\) | \(\ds \alpha^{n - j}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^2 j\) | \(=\) | \(\ds \alpha^n\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 j\) | \(=\) | \(\ds n\) |
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.
$\blacksquare$