Let $G=GL\left ( 2,\mathbb{R} \right )$, the group of 2x2 matrices over $\mathbb{R}$ with non-zero determinant. Let H be the subgroup of matrices of determinant $\pm 1$. If $a,b \in G$ and $aH=bH$, what can be said about $det\left ( a \right )$ and $det\left ( b \right )$?


Following from the hypothesis: $H=a^{-1}bH$ implies $a^{-1}b \in H$.


$det\left ( a^{-1}b \right )=det\left ( a^{-1} \right )det\left ( b \right ) =\frac{1}{\pm 1}det\left ( b \right ) =\left ( \pm 1 \right )det\left ( b \right )=det\left ( a \right )det\left ( b \right )$

So, have established that $det\left ( a \right )=det\left ( a^{-1} \right )=\pm 1$

From here on, it feels as though it is a game of tail chasing. I've tried various manipulation but to no avail. Any hint is appreciated.


1 Answer 1


From $a^{-1}b\in H$, what you can deduce is that $\det\bigl(a^{-1}b\bigr)=\pm1$. In other words, $\det(a)=\pm\det(b)$. Actually, this is an equivalence; $aH=bH$ if and only if $\det(a)=\pm\det(b)$.

  • $\begingroup$ Very nice way to put it.Thank you. $\endgroup$ Jul 1, 2017 at 6:15

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