Let $$A=\begin{pmatrix}i & 0\\ 0 & -i\end{pmatrix}$$ and

$$B=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}.$$

Let $Q=\langle A,B\rangle.$

Prove that $|Q|=8$

The question is given with three hints.

Work out the powers of $A$ and $B$.

I have done this, they both have power $4$.

Find an expression for $B^{-1}AB=A^j$ for a suitable $j$.

I have done this, $j=3$.

Use this to show that any element of $Q$ can be written as $A^iB^j$, but not all of these are distinct.

I am not sure how to go any further.

I have found that $A^2=B^2$ and $A^4=B^4=e$ and the $8$ elements of $Q$ are $$\{e, A,A^2,A^3,B, B^3, AB, A^3B\}$$

which could be written as

$$Q=\{A^1B^4, A^1B^2, A^2B^4, A^4B^4, A^4B^1, A^2B^4,A^2B^1, A^1B^1, A^3B^1\},$$ but I don't think that is what the question wants me to do.

Thank you

Also have another extension to this question. Prove that Q has an automorphism of order 3

  • $\begingroup$ Ok, what does $\;\langle A,B\rangle\;$ mean for you? I thought this was the euclidean inner product for complex square matrices but then I get $\;Q=0\;$ , so it must be something different... $\endgroup$ – DonAntonio Oct 30 '13 at 17:50
  • $\begingroup$ Sorry, <A,B> is the group generated by the matrices A and B. $\endgroup$ – ZZS14 Oct 30 '13 at 17:51
  • $\begingroup$ Oooooh! You meant the group generated by $\;A,B\;$...holy mother of me! $\endgroup$ – DonAntonio Oct 30 '13 at 17:51
  • $\begingroup$ Yes, sorry I didnt realise <> meant something else as well I've only ever seen it used as the group generated by! $\endgroup$ – ZZS14 Oct 30 '13 at 17:53
  • $\begingroup$ Have you already studied a little about the two non-isomorphic and non-abelian groups of order $\;8\;$, the dihedral $\;D_4\;$ and the quaternion group $\;Q_8\;$ ? $\endgroup$ – DonAntonio Oct 30 '13 at 17:54

A general element of the group will look something like $BBABAABABBAB$. However you have $AB=BA^3$, so you can pass the $B$'s to the left through the $A$'s, turning them into $A^3$ as you go. $BBABAABABBAB=BBABAABABBBA^3=BBABAABBA^3BBA^3=BBABAABBBA^9BA^3=BBABAABBBBA^{30}=\cdots$

  • $\begingroup$ I don't really understand this, sorry. $\endgroup$ – ZZS14 Oct 30 '13 at 17:53
  • $\begingroup$ @Emily Since $AB = BA^3$, we can also say that some product like $AABBABAB\mathbf{AB} = AABBABAB\mathbf{BA^3}$ (bold font for emphasis, not any mathematical significance) and further that $AABBAB\mathbf{AB}BA^3 = AABBAB\mathbf{BA^3}BA^3$ and so on, seemingly moving all the $B$'s to the left, only tripling each $A$ every time it's passed through. $\endgroup$ – Arthur Oct 30 '13 at 17:57
  • $\begingroup$ okay I think I get it now sorry $\endgroup$ – ZZS14 Oct 30 '13 at 18:15
  • $\begingroup$ @vadim123 Also have another extension to this question. Prove that Q has an automorphism of order 3 $\endgroup$ – ZZS14 Oct 30 '13 at 20:47
  • $\begingroup$ @Emily, you should ask that as a separate question. $\endgroup$ – vadim123 Oct 30 '13 at 21:13

Well in terms of the last hint in your question, if you want to write it as integral powers of $A$ and $B$ with $A$ first, then you can use the established facts to show $BA=A^3B$, which effectively means you can shift all $B$'s to the right and $A$'s to the left (so the reverse of the hints given in comments before). So you will end up with $A^iB^j$ and then you can just use $A^4=e=B^4$, reducing it to a term with $i,j\leq3$. Now one can just use $A^2=B^2$ and so on to reduce it to one of the 8 elements in the group proving $|Q|=8$.

  • $\begingroup$ Could you explain a little bit more. I am confused how you use A^4=e=B^4 to reduce and from there on really. Thank you $\endgroup$ – ZZS14 Oct 31 '13 at 9:43
  • $\begingroup$ Could you please explain a little further. $\endgroup$ – ZZS14 Nov 3 '13 at 17:08
  • $\begingroup$ sorry I only get back to you now...Well you understand how to get to a term $A^iB^j$ right? Now suppose you have $A^{21}B^{18}=A(A^4)^5B^2(B^4)^4$. Now use $A^4=e=B^4$: $A(e)^5B^2(e)^4=AB^2$ which happens to be one of the elements in $Q$ - the group. $\endgroup$ – Christiaan Hattingh Nov 16 '13 at 19:19

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