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pretty much got stuck with the following question (it has several parts):

a). Show that $D_8$ isn't isomorphic to $Q_8$

b). Let $K$ be a subgroup of $GL_2(\mathbb C)$ so that $$K=\left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\right \rangle.$$ Show $K$ is non-abelian order 8 group and is isomorphic exactly to either $D_8$ or $Q_8$. Show this by building specific isomorphism.

Thank you for any assistance!

P.S. Can someone perhaps expand more on the notation of groups such as $K$? I know it means smaller subgroup containing both elements, but are there any more properties, etc'?

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  • $\begingroup$ for a) look at the order of the members of the groups. for b) do the same. The notation $\langle A,B \rangle$ means the group generated by those elements. $\endgroup$ – Tyler Apr 20 '14 at 13:51
  • $\begingroup$ How do I do the second part? Couldn't solve it... $\endgroup$ – charlie Apr 20 '14 at 14:25
  • $\begingroup$ You can also characterise $K$ as the set of all products $\{A_1^{e_1}A_2^{e_2}\ldots A_n^{e_n} : n\in \mathbb N,\; e_i = \pm 1 \}$ where the $A_i$ are equal to either of your two generating matrices. To construct an explicit isomorphism, it's easiest to see what relations you have between your matrices - how do they behave when you raise them to various powers, what does the effect of conjugating one by another have? You should see similarities with one of your groups of order 8... $\endgroup$ – ah11950 Apr 20 '14 at 14:41
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Hints:

$$\begin{align*}D_8&=\langle s,t\;;\;s^2=t^4=1\;,\;sts=t^3\}=\{1,s,t,t^2,t^3,st,st^2,st^3\}\;,\;\text{with usual relations}\\ Q_8&=\{a,b\;;\;a^4=1\,,\,a^2=b^2\,,\,aba=b\}=\{1,-1,i,j,k,-i,-j,-k:\}\;,\;\;\text{w.u.r.}\end{align*}$$

Check, for example, that

$$s^2=1\;,\;\;(st)^2=(sts)t=t^3t=1$$

Now, how many elements of order two are there in $\;Q_8\;$ ? This solves (a)

For (b): put

$$a=\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}\;,\;\;b=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$

and now verify that

$$a^4=1\,,\,b^2=1\,,\,bab=a^3\;\ldots$$

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  • $\begingroup$ Thank you! You are very helpful all of the time, much gratitude. $\endgroup$ – charlie Apr 20 '14 at 15:59
  • $\begingroup$ When generally approaching group generated by set such as in this case, what's the approach of knowing what's the entire group? Is there a way other than trial and error for finding out all of the element in $K$? $\endgroup$ – charlie Apr 20 '14 at 16:05
  • $\begingroup$ @charlie , in general that can be a tough problem. In this case we have nice, easy-to-work-with complex matrices $\;2\times 2\;$, so calculating their first few powers and products is rather easy. $\endgroup$ – DonAntonio Apr 20 '14 at 16:20

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