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Isomorphic groups

Let n>2 . Define the group

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

Prove that \displaystyle \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.


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.


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

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

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