MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be 'ax+b' topological group i.e subgroup of $GL(2,\mathbb R)$ containing $2\times 2 $ matrices of type $\{\left(\begin{matrix} a&b\\\0&1\end{matrix}\right): a\neq 0\; a,b\in \mathbb R\}$. Further let $A$ be an element of order $2$ in G. Is is true that only continuous functions /homeomorphisms $f$ on $G$ satisfying $f(AX)=f(X)A$ for all $X\in G$ are $f(x)=Bx^{-1}$, where $B$ is any element of $G$.

share|cite|improve this question
There is only one element of order 2 ($a=-1$, $b=0$). – Florian Sep 19 '11 at 17:38
With $a=-1$ and $b$ arbitrary! They are the unique elements of order $2$. – Valerio Capraro Sep 19 '11 at 18:13
I've tried to make the title more specific. Apologies for the multiple edits... – j.c. Sep 19 '11 at 19:18
Your problem, modulo mistakes, is equivalent to finding continuous $\alpha:\mathbb{R}^*\times\mathbb{R}\to\mathbb{R}^*$ and $\beta:\mathbb{R}^*\times\mathbb{R}\to\mathbb{R}$ such that $\alpha(-x,-y+b)=-\alpha(x,y)$ & $\beta(-x,-y+b)=b\cdot\alpha(x,y)+\beta(x,y)$ for every $x\in\mathbb{R}^*$ and $y\in\mathbb{R}$. Voting to close as too localized, unless the question is more motivated. – Qfwfq Sep 19 '11 at 19:35
I think by "functions" the OP means $f : G \rightarrow G$. – Homology Sep 19 '11 at 20:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.