Normalizer of Rotation 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 $\map {N_{D_n} } {\set \alpha}$ denote the normalizer of the singleton containing the rotation element $\alpha$.

Then:

$\map {N_{D_n} } {\set \alpha} = \gen \alpha$

where $\gen \alpha$ is the subgroup generated by $\alpha$.

Proof

By definition, the normalizer of $\set \alpha$ is:

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

That is:

$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$

First let $g = \alpha^k$ for some $k \in \Z$.

Then:

$\alpha \alpha^k = \alpha^k \alpha$

which includes $k = 0$, that is $e$.

Thus:

$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$

Now let $g = \alpha^k \beta$.

Suppose $g \alpha = \alpha g$.

Then:

\(\ds \alpha^k \beta \alpha\)

\(=\)

\(\ds \alpha \alpha^k \beta\)

\(\ds \leadsto \ \ \)

\(\ds \beta \alpha\)

\(=\)

\(\ds \alpha \beta\)

Cancellation Laws

But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$.

Hence the result:

$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$

Hence the result, by definition of generator of subgroup.

$\blacksquare$