How to prove isomorphism with the Dihedral group

Question

I have a group that I'm trying to prove is isomorphic to the Dihedral group.

I know that it is finite, that it is generated by two elements $\alpha$ and $\beta$ such that: $\alpha^2=\beta^n=1$ and that $\alpha\beta\alpha=\beta^{-1}$. EDIT: also, $\alpha\neq \alpha^2$ and $\beta\neq \beta^2\neq\ldots\neq\beta^n$.

I also know that is has at least $2n$ unique elements.
EDIT: Is this assumption redundant?
EDIT: It is redundant for $n$>2. For $n=2$, $\alpha\neq\beta$ is enough (i.e. $D_2\cong C_2\times C_2$).

Is this enough in order to imply that this group is the Dihedral group with $2n$ elements?

Will appreciate any help :)

Answer 1

Adding my own proof with sufficient conditions:

Let $n>2$ and let $\alpha$ and $\beta$ be two elements in a group such that:

1. $o(\alpha)=2$ and $o(\beta)=n$, where $o(g)$ is the minimal power of $g$ such that $g^{o(g)}=1$.
2. $\alpha\beta=\beta^{-1}\alpha$.

Then $G=\left<\alpha, \beta\right>\cong D_n$, the dihedral group of order $2n$.

Proof.

It is sufficient to show that $G$ has no other relations. We show that $|G|=2n$.
Since $\alpha\beta=\beta^{-1}\alpha$, every element in $G$ can be written in the form $\beta^i\alpha^j$ where $i\in[0,n-1]$ and $j\in[0,1]$.
Assume to the contrary that there exists $1\leq k\leq n-1$ such that $\beta^k\alpha=Id$. This implies that $\alpha=\beta^{n-k}$. Therefore $k=\frac{n}{2}$, since $\alpha^2=1$.
Since $\alpha\beta=\beta^{-1}\alpha$, we have $\beta^{\frac{n}{2}+1}=\beta^{\frac{n}{2}-1}$ which is true only for $n=2$.