Problem: Let $Q_8$ be the group (under ordinary matrix multiplication) generated by the complex matrices $A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$, where $i^2=-1$. Show that $Q_8$ is a nonabelian group of order 8.

Suggestion: I don't know what this group $Q_8$ is made of ( I guess it's got 8 elements), therefore I have no solution for this. However I presume we can use this identity for the monoid part: $A^4=B^4=I$.


First show it has eight elements by writing down $I, A, A^2, A^3, A^4, ...$ and $B, B^2, \ldots$ and $AB, (AB)^2, \ldots$, etc.; that's what's meant by the group "generated by these matrices: all arbitrarily long products of all possible powers of the matrices.

You'll find that every sequence of $A$s and $B$s can be simplified to one of just 8. As an example, $A^2 = -I$ and $$ AB = \begin{bmatrix} i & 0\\ 0 & -i\end{bmatrix}, $$ so $(AB)^2 = -I$, hence $ABAB = A^2$, hence $BAB = A$. So in any sequence of $A$s and $B$s, you can replace a $BAB$ with a single $A$. And you can replace any odd number of $A$s, like $A^7$, with $A$ times an even number of $A$s, which turns into $\pm I$. In the $A^7$ example, we get $A^7 = A \cdot A^6 = A (A^2)^3 = A (-I)^3 = A(-I) = -A$.

(I suspect that one possible set of eight representatives is $I, A, A^2, A^3, B, AB, A^2B, A^3B$, but I'm not certain of this). To prove non-abelian-ness, you just have to find two elements $p$ and $q$ such that $pq \ne qp$.

  • $\begingroup$ Oh right. That comes from the theorem: If $G$ is a group and $\emptyset\ne X\subset G$, then the subgroup $\langle X\rangle$ generated by $X$ consists of all finite products $a_1^{n_1}\cdots a_t^{n_t}, (a_i\in X; n_i\in \mathbb{Z})$. Thanks! $\endgroup$ – Mill Jul 23 '14 at 22:14

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