I have a question that needs attention regarding the dihedral group $D_{n}$. Here is the context of the problem:
Consider the Dihedral group $D_{n}$. Let $\sigma$ be a rotation counterclockwise by $\frac{2\pi}{n}$ and $\tau$ be a reflection.
(1) Show that the subgroup generated by $\sigma\tau$ and $\sigma^{2}\tau$ is all of $D_{n}$.
(2) Find the subgroup generated by $\sigma^{2}$ and $\sigma^{2}\tau$.
For part (1) I am unsure that the language is clear enough that I know what I am being asked to prove. Am I simply "multiplying" (or operating on) $\sigma\tau$ and $\sigma^{2}\tau$ several times until I have $D_{n}$? That seems unlikely given the order of $D_{n}$ is $2n$.
Any help would be appreciated.
EDIT: I can show that $\sigma=\sigma\tau\sigma^{2}\tau$ using the fact that $\sigma^{-1}=\sigma$ and also $\tau^{2}=1$. But it is unclear how I can obtain $\tau$ with any finite operations of $\sigma\tau$ and $\sigma^{2}\tau$.
The only thing I've come up with is $(\sigma^{2}\tau)^{3}=\sigma\sigma\tau\sigma\sigma\tau\sigma\sigma\tau$ and if we use $\sigma^{-1}=\sigma$ this reduces to $\tau\tau\tau=\tau^{2}\tau$. Noting that $\tau^{2}=1$, it then follows that $(\sigma^{2}\tau)^{3}=\tau$, I think. . .
For (1), let $H$ denote the subgroup generated by $\sigma \tau$ and $\sigma^2 \tau$. Note that $\sigma = (\sigma^2 \tau)(\sigma \tau)^{-1} \in H$, hence $\langle \sigma \rangle \leq H$, which shows that $|H| \geq n$, so $|D_n : H| \leq 2$. Since $\sigma \tau$ is also in $H$ but not in $\langle \sigma \rangle$, this means that $|H| > n$, hence $|D_n : H| < 2$, which forces $|D_n : H| = 1$, so $H = D_n$.
[The above of course implies that $\tau \in H$. To see this explicitly, use the fact (shown in the previous paragraph) that $\sigma \in H$. Therefore $\tau = (\sigma^2 \tau)\sigma^2 \in H$.]
For (2), let $K$ denote the subgroup generated by $\sigma^2$ and $\sigma^2 \tau$. Note that $\tau = (\sigma^2)^{-1}(\sigma^2 \tau) \in K$. Also note that for any integer $n$, we have $\sigma^{2n} \in K$.
Observe that $\langle \sigma \rangle$ is a normal subgroup of $D_n$ because it has index 2. Then note that $\langle \sigma^2 \rangle$ is a characteristic subgroup of $\langle \sigma \rangle$ (every subgroup of a finite cyclic group is characteristic). Therefore $\langle \sigma^2 \rangle$ is normal in $D_n$. It follows that $\langle \sigma^2 \rangle \langle \tau \rangle$ is a subgroup of $D_n$. Clearly this subgroup contains $\sigma^2$ and $\sigma^2\tau$, hence it contains $K$, the subgroup generated by these elements. This shows that $K \leq \langle \sigma^2 \rangle \langle \tau \rangle$.
Conversely, an arbitrary element of $\langle \sigma^2 \rangle \langle \tau \rangle$ is of the form $\sigma^{2n}$ or $\sigma^{2n}\tau$ for some integer $n$. Hence (by the observations before the previous paragraph), this element is in $K$. This shows that $\langle \sigma^2 \rangle \langle \tau \rangle \leq K$.
Since we have shown both containments, we conclude that $K = \langle \sigma^2 \rangle \langle \tau \rangle$.