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The group $Q\subset GL_2(\mathbb{C})$ is generated by $\langle A,B\rangle$

$$ A= \left( \begin{array}{ccc}0 & 1 \\ -1 & 0\end{array} \right) B= \left( \begin{array}{ccc}0 & i \\ i & 0\end{array} \right)$$

I'm asked to prove that Q is a non-abelian group of order 8, and Q is not isomorphic to D4.

  1. Q is not abelian because matrix product is non commutative.

  2. $A$ and $B$ are two elements of $Q$. $A^2= \left( \begin{array}{ccc}-1 & 0 \\ 0 & -1\end{array} \right), A^3= \left( \begin{array}{ccc}0 & -1 \\ 1 & 0\end{array} \right), A^4= Id, B^2= \left( \begin{array}{ccc}-1 & 0 \\ 0 & -1\end{array} \right)=A^2, B^3= \left( \begin{array}{ccc}0 & -i \\ -i & 0\end{array} \right), B^3= Id. AB= \left( \begin{array}{ccc}i & 0 \\ 0 & -i\end{array} \right), BA= \left( \begin{array}{ccc}-i & 0 \\ 0 & i\end{array} \right) .$

  3. Now it looks like $Q$ is at least order 8 since $Q=\{Id, A, A^2, A^3, B, B^3, AB, BA\}$. But how can I prove that there are no more elements in $Q$?

  4. $D_4$ is generated by two elements $a,b$ such that $a^4=e, b^2=e, ba=a^3b$. Then both $a^2$ and $b$ have order two, but there is only one element of order two in $Q$ which is $A^2$. Then $Q$ is not isomorphic to $D_4$

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    $\begingroup$ For the first question you should, for the sake of completeness, show manually that $AB\neq BA$. Matrix product is indeed not commutative in general but the product of two matrices can be commutative in particular (take 2 diagonal matrices for instance) $\endgroup$ – pmichel31415 Apr 21 '16 at 13:18
  • $\begingroup$ @Mandrathax In the second point I calculated both $AB$ and $BA$. But I agree, I should point that out after calculating the products. $\endgroup$ – Cure Apr 21 '16 at 13:20
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    $\begingroup$ You can deduce using $A^4=I$, $A^2=B^2$, $BA = AB^3$ that $|Q| \le 8$. $\endgroup$ – Derek Holt Apr 21 '16 at 13:23
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    $\begingroup$ A possible way to show that $Q$ only contains these 8 elements and fill out Cayley table by multiplying any pair to see that you will not get any new elements and that every element in this table has an inverse: en.wikipedia.org/wiki/Quaternion_group#Cayley_table $\endgroup$ – Martin Sleziak Apr 21 '16 at 14:39
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    $\begingroup$ See also: The order of the Quaternion Group. Another question asking whether $Q_8$ and $D_8$ are isomorphic has an answer using elements of order $2$, too. (These questions were listed among related questions in the sidebar on the right. Sometimes it might be useful to check various lists of similar questions which are generated by SE software.) $\endgroup$ – Martin Sleziak Apr 21 '16 at 14:40
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I think it's import to express the element you computed in terms of A,B and $I_2$ :

$$A^2=B^2=-I_2$$ $$A^3=-A$$ $$B^3=-B$$ $$A^4=B^4=I_2$$ $$AB=-BA$$

The important thing is that you have a kind of "anti-commutativity" in the last line, so you can rewrite any word $AABA...BABBA$ as $\pm A^nB^m$, where $n,m$ are the number of $A$ and $B$ in your word (prove this by induction on the length of the word). Then $A^n=\pm A\mathrm{\ or\ }\pm I_2$, and same for $B$.

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