# 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?

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.

• Part 1 is wrong. – Derek Holt Dec 28 '19 at 8:32
• @reuns Dummit and Foote defines $$D_{2n} = \langle r,s\mid r^n=s^2=1, rs = sr^{-1}\rangle.$$ So I don't believe that it should be $xy=y^{-1}x^{-1}$. – Math1000 Dec 28 '19 at 8:37
• The edited presentation does indeed define the dihedral group of order $16$. But it is not true that the subgroup $N$ that you have defined is a normal subgroup of $G$. The normal subgroup $N$ such that $G/N = D_{2,8}$ is the normal closure in $G$ of the group that you have defined, and is a free group on $17$ generators. And then you have $G/N = D_{2,8}$ by definition, so it is very difficult to understand what this question is about! – Derek Holt Dec 28 '19 at 12:38
• @Math1000 ?? You have corrected it now, $y=y^{-1}$ so $xy=y^{-1}x^{-1}=yx^{-1}$ is correct, but $xy=y^{-1}x$ is not as it would give $xy=yx$ – reuns Dec 28 '19 at 21:27

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 $$x^ny^m$$.
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.