Hello,
I would like to know whether the only non trivial involutary group isomorphism $f:(\mathbb{C}^{*},\times)\to(\mathbb{C}^{*},\times)$ is the complex conjugation or not. Thank you in advance.
$\begingroup$
$\endgroup$
1
asked Jul 21, 2011 at 20:20
-
2$\begingroup$ See ccc's answer here for a complete list of all (measurable) endomorphisms. You can easily check which ones of those are involutive: math.stackexchange.com/questions/52319 $\endgroup$– Theo BuehlerCommented Jul 21, 2011 at 21:21
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
$z\mapsto \frac{1}{z}$
answered Jul 21, 2011 at 20:32
-
$\begingroup$ Indeed...Maybe it's time for me to go to bed :/ $\endgroup$ Commented Jul 21, 2011 at 20:41
$\begingroup$
$\endgroup$
How about $$ z\mapsto \frac{z}{|z|^2}?$$
answered Jul 21, 2011 at 20:33