Originally Posted by

**MichaelMath** I have:

$\displaystyle D_n= \langle a,b | a^2 = b^2 =(ab)^2=1 \rangle$

This is a presentation for the dihedral group $\displaystyle D_2$ , better known as the Klein group, the non cyclic

group of order 4

A dihedral group can be generated by two distinct elements of order 2

(a) Show that this definition allows only one infinite group up to isomorphism and determine its centre.

As already noted, the above is __not__ an infinite group but a pretty finite one...

Tonio

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Is this right:

The only non-trivial rotation that commutes with all reflections is the rotation over $\displaystyle \pi$ . So $\displaystyle a$ is a rotation by $\displaystyle 2 \pi/n$ and $\displaystyle b$ is a reflection.

Since $\displaystyle n \geq 3, \, a^{-1} \neq a$. Therefore,

$\displaystyle a^{r+1}b=a(a^rb)=(a^rb)a=a^{r-1}b$

So $\displaystyle a^2=1$ , which is a contradiction; thus, no element of the form $\displaystyle a^rb$ is in the center.

Similarly, if for $\displaystyle 1 \leq s < n, a^sb=ba^s=a^{-s}b$ , then $\displaystyle a^{2s}=1$ , which is possible only if $\displaystyle 2s=n$. Hence, $\displaystyle a^2$ commutes with $\displaystyle b$ iff $\displaystyle n=2s.$

So if $\displaystyle n=2s$ , the the center of $\displaystyle D_n$ is $\displaystyle \{1,a^s \}$; if $\displaystyle n$ is odd the center is $\displaystyle \{ 1 \}$.