I have to say I'm not doing a good job of interpreting my textbook, here, which leaves me rather lost to prove what it asks. But here's a shot...Let the dihedral group $\displaystyle D_n$ be given by elements $\displaystyle a$ of order $\displaystyle n$ and $\displaystyle b$ of order $\displaystyle 2$, where $\displaystyle ba=a^{-1}b$. Show that any subgroup of $\displaystyle <a>$ is normal in $\displaystyle D_n$.

If I read the previous chapter correctly, a dihedral group $\displaystyle D_n=\{a^pb^q\}$, for all integers $\displaystyle p,q$ with $\displaystyle 0\leq p<n$ and $\displaystyle q\in\{0,1\}$. Hopefully this much is correct.

In order for the subgroup $\displaystyle <a>$ to be normal, we need to show that for each element $\displaystyle g\in D_n$, it is true that $\displaystyle g^{-1}a^rg=a^s$ for any integer $\displaystyle r\in[0,n)$ and some other integer $\displaystyle s\in[0,n)$. This clearly holds for all $\displaystyle g=a^pb^0=a^p$, since $\displaystyle (a^p)^{-1}a^ra^p=a^{n-p}a^ra^p=a^r$.

So now we need to show that $\displaystyle (a^pb)^{-1}a^r(a^pb)=a^s$. Now, we know that since $\displaystyle o(b)=2$ then $\displaystyle b^{-1}=b$. That means $\displaystyle (a^pb)^{-1}=ba^{n-p}$. So now we evaluate $\displaystyle ba^{n-p}a^ra^pb=ba^rb$.

Note that since $\displaystyle ba=a^{-1}b=a^{n-1}b$, then $\displaystyle ba^rb=a^{n-1}ba^{r-1}b$. And we can get by repeating this:

$\displaystyle ba^rb=a^{n-1}ba^{r-1}b$

$\displaystyle =a^{2n-2}ba^{r-2}b$

$\displaystyle =a^{3n-3}ba^{r-3}b$

$\displaystyle =a^{4n-4}ba^{r-4}b$

...and so on, up to:

$\displaystyle =a^{rn-r}ba^{r-r}b=a^{rn-r}\in D_n$

Q.E.D. ... ?

Have I done this correctly? If not, how did I err? If so, is there a quicker/easier way to do it?

Thanks for your help, guys.