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'?
Best Answer
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
$$a^4=1\,,\,b^2=1\,,\,bab=a^3\;\ldots$$