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This is a problem in my homework that I am a bit confused over.

Let $G$ act on a set $X$. Given $A\subseteq X$ we can define $gA=\{ga : a \in A\}$. Thus, $gA$ is a subset of $X$. Check that this is an action of $G$ on the powerset $P(X)$ of $X$ (the set of all subsets of $X$). What is the stabilizer of the empty set $\varnothing$ under this action? What is the stabilizer of a one-element set $\{x\}$, for $x \in X$?

As far as the first part goes, showing $gh(x)=g(hx)$ is fairly easy. I'm not sure how to show that $ea=a$. Any advice?

What I do not understand is how to find the stabilizer of the empty set. Would it be the entire $G$ trivially since you can't technically perform a group action on nothing?

As far as the stabilizer of a one element set $\{x\}$, I know that it consists of elements of $G$ s.t. $gx=x$; however, the only guaranteed one in my mind is e since I don't know how the elements of G interact with elements of $X$. Any recommendations?

anon
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Richard Cao
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    Checking $eA=A$ for all sets $A\subseteq X$ should be obvious. What is $eA$, by definition? "You can't technically perform a group action on nothing" is not a sensible way of saying "because $g{}={}$ for all $g\in G$." And lastly, "elements of $G$ such that $gx=x$" is by definition the what? – anon Apr 14 '17 at 02:52
  • So, to respond to all your points: 1. from a general perspective, for the group action, is the operation the same as the group operation in G? In that case then the identity of G when applied with the group operation of G to an element x of X should return X correct? 2. I really do not know how else to think of this. 3. By definition of the center of G. I'm looking for the center, but can I be more specific than "the stabilizer is the center of G?" – Richard Cao Apr 14 '17 at 02:59
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    (1) $A$ is a subset of $X$. What is $eA$ according to the definition of $G$'s action on subsets of $X$? (2) I just told you how to think about it. Observe $g{}={}$ for all $g\in G$, so every $g$ fixes the empty set. (3) No, that's not the center of $G$. You've determined that $g{x}={x}$ (the condition for $g$ being in the stabilizer of ${x}$) is equivalent to $gx=x$, and the set of all $g$ such that $gx=x$ is the ___? – anon Apr 14 '17 at 03:09
  • Sorry, I think I misunderstood your question the first time around. As for 1, $eA=${$ea: a∈A$} and for 3, the definition is the stabilizer group of x. – Richard Cao Apr 14 '17 at 03:13
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    Yes. [If you want a proof why $g{}={}$, suppose $y\in g{}$, then there exists an $x\in{}$ such that $y=gx$, but there does not exist any $x$ in ${}$, a contradiction, hence there is no $y$ in $g{}$, i.e. $g{}={}$.] – anon Apr 14 '17 at 03:13
  • Thank you very much! I think I had just a question in general: how do we know that $ea=a$? I'm not sure why I keep getting stumped at this point, but isn't e for the group G only guaranteed to return g for $eg$ for an element g of G? Why is it the case that $ea=a$? Also, in this case, is the operation on the action the same as the group operation? For the stabilizer of ${x}$, can we get more specific than simply just writing out {$g∈G:gx=x$}? – Richard Cao Apr 14 '17 at 03:17
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    Look up the definition of "group action." It has $ea=a$ for all $a\in X$ bulit into the definition. Therefore, $eA={ea:a\in A}={a:a\in A}=A$. I don't know what you mean by "is the operation on the action the same as the group operation." If you mean is $G\times X\to X$ (the group action) the same as $G\times G\to G$ (the group operation), then no of course not, $X$ and $G$ can be very different sets after all. Note that in (3), the question is to find the stabilizer of ${x}$, and the answer is the stabilizer of $x$. It should be obvious you cannot be more specific than that. – anon Apr 14 '17 at 03:23
  • I guess for part 1 I was hung up on the part where it asks to check that this was a group action. In order to achieve that, all I have to do is to show that $gh(x)=g(hx)$, and $eA=A$; however, I guess just writing $eA={ea:a∈A}={a:a∈A}=A$ is sufficient since the problem tells you G acts on X? – Richard Cao Apr 14 '17 at 03:30

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