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I'd like to show that the linear functions $$ \varphi(z) = z+b, \;\;\; 0\neq b\in \mathbb{C}$$ $$ \psi(z) = -z+c, \;\;\; c\in \mathbb{C}$$ generate, under composition, a group isomorphic to $Dih_\infty$, the infinite dihedral group.

Now $Dih_\infty$ may be presented as follows: $$ \langle r,s \, | \, s^2=1, srs=r^{-1}\rangle.$$

These relations are satisfied by $\varphi$ and $\psi$, since $\psi^2 = 1$, ($1$ stands for the identity function) and $\psi\circ \varphi \circ \psi = \varphi^{-1}$. How do I show no further relations hold ?

An alternative proof is also welcome.

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An idea:

$$\tau(z):=\psi\circ\phi(z)=\psi(z+b)=-(z+b)+c=-z+(c-b)$$

$$\tau^2(z)=\tau(-z+(c-b))=-\left[-z+(c-b)\right]+(c-b)=z$$

And clearly, $\;\phi=\psi^{-1}\tau=\psi\tau\;$ , thus our group is

$$\langle\;\psi\,,\,\tau\;;\;\psi^2=\tau^2=1\;\rangle=C_2*C_2\cong Dih_\infty$$

Of course, in order to be completely formal (or perhaps "too" formal) you may want to show that no finite product of the for $\,a_1b_1a_2b_2\cdot\ldots\cdot\;$ , with $\,a_i,b_i\in\{\psi\,,\,\tau\}\,\,,\,\,a_i\neq b_i\;$ , can be the identity map (trivial element) , but I think this is almost "trivial", since

$$\tau\circ\psi(z)=\tau(-z+c)=-(-z+c)+c-b=z-b\neq z\;,\;\;\text{since}\;\;b\neq 0$$

$$\text{and as already noted above,}\;\;\psi\circ\tau(z)=\psi(-z+c-b)=-(-z+c-b)+c=z+b\neq z$$

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  • $\begingroup$ Thank you for your answer. You are correct in stating that the relations $\psi^2=\tau^1=1$ hold. Still, as you mention, one must show that no further relations hold. How does that follow from the statements $\tau \psi\neq 1$ and $\psi\tau \neq 1$ ? $\endgroup$ – Teddy Jun 11 '13 at 5:58
  • $\begingroup$ Induction, @Teddy. Note that any normal element in that free product $\,C_2*C_2\,$ is a finite product (or concatenation, if you will) of pairs of the form $\,\tau\psi\;,\;\psi\tau\,$ and/or one more element, for example: $$\tau\psi\tau\;,\;\;\psi\tau\psi\tau\psi\;,\;\tau\;,etc.$$ $\endgroup$ – DonAntonio Jun 11 '13 at 6:01
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Without loss of generality, you can assume that $b=1$ and $c=0$.

Let $G$ be the group generated by $\phi$ and $\psi$. You know there is a surjection $f:D_\infty\to G$. Using the relations defining $D_\infty$ one can check easily that every element $g$ of $D_\infty$ is of the form $s^ar^b$ with $0\leq a<2$ and $b\in\mathbb Z$.

If $f(g)$ is the identity, then $f(g)(\sqrt{-1})=\sqrt{-1}$. Show that in that case $a=b=0$.

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