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Suppose $G$ denotes the multiplicative group {$-1,1$} and $S=${$z\in \mathbb C:|z|=1$}. Let $G$ act on $S$ by complex multiplication.Then the cardinality of the orbit of $i$ is

$a)1$

$b)2$

$c)5 $

$d)\infty$

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    MSE shouldn't be used as alternative to reading through course material. I'm sure orbit is one of the first things defined after introducing notion of group action. – 2017-01-03

1 Answers 1

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It's two. The orbit of $i$ is the set $G{i}=\{x\in S : x=g\cdot i$, with $ g\in \{-1,1\}\}$. Hence $G{i}=\{i,-i\}$.

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    what is orbit?i donot have any knowledge about it..plz expain it – 2017-01-03
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    Why are you asking this question if you don't know what an orbit is? – 2017-01-03
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    If you have a group G acting on a set X . The orbit of an element $x\in X$ is the set $Gx=\{y\in X : y=g\cdot x, g\in G\}$ – 2017-01-03
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    actual i found this problem from question paper.. – 2017-01-03
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    Please first look up, what a group operation and an orbit is. Then your question should answer itself in a few seconds! – 2017-01-03