Let $K$ be a field. Consider the action of the multiplicative group $K^* := K-\{0 \}$ on the vector space $K^n$ given by scaling.
a) Describe the orbits of this action.
b) Describe the stabilizer groups for this action.
I did solve the question but since this is my first course in abstract algebra, I just wanted some feedbacks from you guys and see if I approached this question correctly. Below is what I have. Thanks!
a) Orbits are the column vectors described by the following: {1st entry 1, rest 0's}, {2nd entry 1, rest 0's},..., {nth entry 1, rest 0's}.
b) {1} is the stabilizer since for every $v \in K^n$, $1 \cdot v = v$.
Not quite. The orbit of an element $v \in K^n$ is the set of all $k \cdot v$, with $k \in K^*$. We write the orbit as $K^*v = \{k \cdot v | k \in K^* \}$.
Thus the orbit of a non-zero vector $v \in K^n$ is just the line it spans, minus the origin. If $v$ is the origin, then its orbit is, of course, itself, since $k \cdot 0 = 0$.
The stabilizer of an element $v \in K^n$ is the set of group elements that fix $v$. Since $1$ is the only element that fixes $v$, you are correct here.
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As an aside, this action on a vector space is really important: If you identify the orbits of this action and throws away the origin, you get what is called the projectivization of $K^n$. We write this as $\mathbb{P}K^n.$