How to show that the dihedral group $D_{2\cdot 8}$ is the quotient of the free group on $2$ generators by a certain normal subgroup?

Question

Let $D_{2\cdot 8}$ be given by the group presentation $\langle x,y\mid xy = yx^{-1} , y^2 = e, x^8 = e\rangle$. Let $G = F_{\{x,y\}}$ be the free group on two generators and $N = \langle\{xyx^{-1}y,y^2,x^8\}\rangle$. Then

1. $N$ is a normal subgroup of $G$.
2. The quotient group $G/N$ is isomorphic to $D_{2\cdot 8}$.

For 1. I need to show that $gng^{-1}\in N$ for all $g\in G$, $n\in N$, i.e. that $N$ is closed under conjugation by elements of $G$. But for example I don't see how we could write $yx^8y^{-1}$ as a product of elements of $\{xyx^{-1}y, y^2,x^8\}$ and their inverses.

For 2. I see that the quotient group must consist of $2\cdot8=16$ equivalence classes. But I don't see how we find these explicitly.

I believe it suffices to exhibit a group homomorphism $\varphi$ such that $\ker\varphi = \langle\{xyx^{-1}y,y^2,x^8\}\rangle$ and $\mathrm{Im}\varphi\cong G/N$, by the first isomorphism theorem. But I am not sure how to define this homomorphism exactly.

Answer 1

Hint: Define $\varphi \colon F_{\lbrace x,y \rbrace} \rightarrow D_{2 \cdot 8}$, by sending $x$ and $y$ to the two generators of the dihedral group that you also called $x$ and $y$ in the presentation above. The kernel will then be given by the relations of the dihedral group which is exactly what you want and what you can see from the presentation.

Answer 2

If I remember correctly, $N$ should be defined as the smallest normal subgroup containing $\langle x^8,y^2, xyx^{-1}y\rangle $. That is $N$ is its normalizer; hence normal.

The homomorphism $\varphi$ is just the canonical projection onto the quotient.

The order of the quotient can be established by using the commutativity relation $xy=y^{-1}x$ to write every word in the group in the form $x^ny^m$.

It remains to be seen that the symmetries of a regular octagon satisfy the given relations.