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.

  • $\begingroup$ Part 1 is wrong. $\endgroup$ – Derek Holt Dec 28 '19 at 8:32
  • $\begingroup$ @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 $xy=y^{-1}x^{-1}$. $\endgroup$ – Math1000 Dec 28 '19 at 8:37
  • 2
    $\begingroup$ 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! $\endgroup$ – Derek Holt Dec 28 '19 at 12:38
  • $\begingroup$ @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 $xy=yx$ $\endgroup$ – 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 $x^ny^m$.

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


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.


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