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Find all the matrices in the subgroup $N(\langle A \rangle)$ of $\operatorname{GL}(2,\mathbb{R})$ where $|A|=4$. Is $N(\langle A \rangle)$ abelian? Is $N(\langle A \rangle)$ finite?

$\langle A \rangle$ is the cyclic subgroup of $GL(2,\mathbb{R})$ generated by $A$ and $N(\langle A \rangle)$ its normalizer.

My solution, $\langle A \rangle = \{A^0,A, A^2, A^3\}$ so $N(\langle A \rangle)$ can only be $A^0$ since $BA^0B^{-1}=A^0=I$ for any B in $\operatorname{GL}(2,\mathbb{R})$, therefore it is abelian and has order $1$.

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So $N(H)$ denotes the normalizer of $H$? Certainly the center of $GL(2,\mathbb R)$ is going to belong to it. – JSchlather Jan 29 '13 at 20:13
I understand that $\,\langle A\rangle\,$ may be a sloppy way to denote the subgroup of $\,GL(2,\Bbb R)\,$ generated by all the matrices of determinant $\,4\,$ (if this is really what you meant, of course...) , but what's $\,N(\langle A\rangle)\,$? Tie subgroup's normalizer? – DonAntonio Jan 29 '13 at 20:15
@JacobSchlather $\langle A \rangle$ is the cyclic subgroup of GL(2,$\mathbb{R}$) generated by A – bobdylan Jan 29 '13 at 20:17
Linear groups tag if exists is a better tag than Symmetric groups. – Babak S. Jan 29 '13 at 20:18
@DonAntonio $\langle A \rangle$ is the cyclic subgroup of GL(2,$\mathbb{R}$) generated by A – bobdylan Jan 29 '13 at 20:19
up vote 5 down vote accepted

You didn't say what $A$ is! But any matrix of order 4 in ${\rm GL}_2({\mathbb R})$ has minimal polynomial $x^2+1$, so all elements of order 4 are conjugate, and I am going to assume (without loss of generality!) that $A = \left( \begin{array}{rr}0&1\\-1&0\end{array}\right)$. An element $B$ in the normalizer of $\langle A \rangle$ must conjugate $A$ to either $A$ or to $A^{-1} = \left( \begin{array}{rr}0&-1\\1&0\end{array}\right)$. So we have either $BA=AB$ or $BA = A^{-1}B$. Now just let $B=\left( \begin{array}{rr}a&b\\c&d\end{array}\right)$ and solve the equations. You get $d=a$, $c=-b$ for the first condition (which gives the centralizer of $A$) and $d=-a$, $c=b$ for the elements that invert $A$.

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