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## Isomorphic groups

Groups, Rings, Domains, Modules, etc, Galois theory
Riemann
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### Isomorphic groups

Let $n >2$. Define the group

$$\mathcal{Q}_{2^n} = \langle x, y \mid x^2=y^{2^{n-2}} , y^{2^{n-1}} = 1, x^{-1} yx =y^{-1} \rangle$$

Show that $\mathcal{Q}_{2^n} / \mathcal{Z} \left ( \mathcal{Q}_{2^n} \right ) \simeq \mathcal{D}_{2^{n-1}}$ where $\mathcal{D}$ is the dihedral group.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tolaso J Kos Articles: 2
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### Re: Isomorphic groups

Using  $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of  $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$. Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$. Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n-1}-1\}$. It is easy to prove inductively that $y^tx = xy^{-t}$.

Let $\mathcal{Z} = \mathcal{Z}(\mathcal{Q}_{2^n})$. We prove that $\mathcal{Z}= \{1,y^{2^{n-2}}\}$. Obviously $1 \in \mathcal{Z}$. Furthermore,  $y^{2^{n-2}} \in \mathcal{Z}$ since

$$y^{2^{n-2}}\left(x^ry^s\right) = xy^{-2^{n-2}}x^{r-1}y^s = \cdots = x^ry^{2^{n-2}}y^s =\left(x^ry^s\right)y^{2^{n-2}}$$

If $y^k \in \mathcal{Z}$ (such that $0 \leq k < 2^{n-1}$) then $xy^k = y^kx = xy^{-k}$ hence $y^{2k} = 1$ and therefore $k = 0$ or $k = 2^{n-2}$. If $xy^k \in \mathcal{Z}$ then $xy^{k+1} = yxy^{k} = xy^{k-1}$ hence $y^2 = 1$ which is a contradiction since $n \geq 2$.

Therefore,

$$\mathcal{Q}_{2^n}/\mathcal{Z} = \langle x,y| x^2 = y^{2^{n-2}}=1, yx = xy^{-1} \rangle$$

which is precisely the dihedral group with $2^{n-1}$ elements.
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