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The quaternion group has a particular presentation $$ \langle x,y,z: xz=zx,yz=zy,xy=yxz,x^4=y^4=z^2=1\rangle $$

So it must have order $8$, but can you deduce that just from the relations? The relations show that anything can be put in form $x^iy^jz^k$ for $0\leq i,j\leq 3$, and $0\leq k\leq 1$, so there are at most $32$ elements.

How can you eliminate redundancies?

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  • $\begingroup$ what have you tried?? you tried eliminating at least some redundancies?? please write what ever you have tried.. that would be really helpful... $\endgroup$ – user87543 Nov 28 '13 at 5:37
  • $\begingroup$ you forgot $x^2=y^2$.. I am not sure if this relation can be traced out from given relations $\endgroup$ – user87543 Nov 28 '13 at 5:42
  • $\begingroup$ @B.S. : I am sorry, I do not know what Tietze method is.. I have never heard of it.. $\endgroup$ – user87543 Nov 28 '13 at 5:51
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    $\begingroup$ @Nastassaja: GAP tells that your group is (C4 x C2) : C4 of order $32$. $\endgroup$ – Mikasa Nov 28 '13 at 6:21
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    $\begingroup$ So the answer to the question is no, because the group has order 32. $\endgroup$ – Derek Holt Nov 28 '13 at 8:53
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You have already shown that your group $$ P = \langle x,y,z: xz=zx, yz=zy, xy=yxz, x^4=y^4=z^2=1 \rangle $$ has order at most $32$. To prove it has exactly order $32$, one way is to construct a group $G$ of order $32$ and elements $X, Y, Z$ in it which satisfy the relations for $x, y, z$. So $G$ is a homomorphic image of $P$, and $P$ has order at least $32$.

One such $G$ is the semidirect product of the abelian group $$ \langle Y \rangle \times \langle Z \rangle, $$ with $X$ of order $4$ and $Z$ of order $4$, by a cyclic group $\langle X \rangle$ of order $4$, acting via the automorphism (of order $2$) $$ Y^X = Y Z, \qquad Z^X = Z. $$

Now note that the relations you gave to define $P$ are certainly satisfied in $Q_{8}$, taking $x = i$, $y = j$ and $z = -1$. But they do not define $Q_{8}$, as we have just seen. To do that you have to modify them, for instance as $$ P = \langle x,y,z: xy=yxz, x^2=y^2 = z, z^2=1 \rangle, $$ where I have omitted $xz=zx, yz=zy$, which now follow from $x^2=y^2 = z$.

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Hint: Try to convert your representation in to the following representation:

$Q_{8}= \langle a,b\mid a^4=e,a^2=b^2,aba=b \rangle$

This relation shows that every element of $Q_8$ is of the form:

($a^{s} b^{t}$), $0\leq s\leq 3$,$0\leq t \leq 1$

Clearly, $\left |Q_8 \right | \leq 8$.

Next, we consider a subgroup of $GL(2,\mathbb{C})$ generated by two matrices:

$A=\begin{pmatrix} i& 0\\ 0& -i \end{pmatrix},B=\begin{pmatrix} 0 & 1\\ -1& 0 \end{pmatrix}$

It is easy to see that $A^4=I,A^2=B^2,ABA=B$

So the correspondence $a\mapsto A,b\mapsto B$ defines a surjective homomorphism $Q_8 \to \left \langle A,B\right \rangle$.

On the other hand, $\left \langle A,B\right \rangle$ has $8$ elements,namely:

$\left \langle A,B \right \rangle=\left \{\pm \begin{pmatrix} 1& 0\\ 0& 1 \end{pmatrix} ,\pm \begin{pmatrix} i&0 \\ 0& -i \end{pmatrix},\pm \begin{pmatrix} 0& 1\\ -1&0 \end{pmatrix},\pm \begin{pmatrix} 0 &i \\ i& 0 \end{pmatrix}\right \}$

So, $Q_8 \cong\left \langle A,B \right \rangle$ has order $8$

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