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I have been reading up a bit on group actions and whether they are faithful, free and transitive. A question I found states:

Consider the following group actions

1) The symmetric group $S_n$ acting on an n-element set.

2) The orthogonal group $O(n)$ acting on $\mathbb{R}^n$.

3) The orthogonal group $O(n)$ acting on the $(n-1)$-sphere $S^{n-1}$.

Which of these actions is faithful, transitive or free and what are the group orbits?

An explicit action is not specified, but I assumed that we could take the 'obvious' action: 1) I took $S_n$ as permuting the elements of the set and for 2) and 3) I took $O(n)$ as rotations. Is this correct, or can groups only act on a set in one way anyway?

Going with these actions, I thought that 1) would be faithful as the only permutation which leaves a set unchanged is the identity permutation; transitive as you can always find a bijection to send one permutation to another; and free as the only bijection which leaves a permutation unchanged is the identity permutation. The orbit would be all possible permutations of $n$ elements. My problem with this though is that my reasonings for faithful and free seem identical, so I don't think my train of thinking is correct.

For 2) and 3), I also don't see how what they are acting on seems to affect the characteristics of the group action. What kind of things do I need to think about?

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2 Answers

up vote 2 down vote accepted

Let us remember the definition of faithful, transitive and free:

The action of a group $G$ on a set $X$ is called:

Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.

Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.

Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.

Now we can determine your three group actions:

The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.

The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.

The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.

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Thanks for your detailed answer. It really helped that you restated the definitions of free/transitive/faithful, as the notes I was reading stated them slightly differently and I didn't fully understand them until now. –  Andrew Ledesma May 30 '13 at 4:07
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I think your assumption about the "obvious" action is spot on. Note that, in general, one should specify the action - there are often different actions. (The branch of representation theory, in some very simplified sense, exists to answer the question "how many ways can a group act on a vector space").

You're right that 1) if faithful and your reason is fine. But when you say it's transitive because "you can always find a bijection to send one permutation to another", note that the action of $S_n$ does not "send one permutation to another", but rather, sends one element of your $n$-element set to another. That said, yes, the action is transitive.

A transitive action is rarely free - in particular, in this case, it's not (except when $n\leq 2$): Given any $p$ in your $n$ element set, there are plenty of elements of $S_n$ which fix $p$ and just move around the other points. In fact, the isotropy subgroup at $p$ is isomorphic to $S_{n-1}$.

For your statement about orbits, again, the orbit wont consist of a subset of $S_n$, but instead of a subset of your $n$ element set. To answer this, try proving that an action is transitive iff the orbit through any point is the whole space being acted on.

For 2) and 3), here's a hypothetical situation highlighting how the set you're acting on affects things. (If you'd like to keep a concrete example in mind, apply what I'm about to say to $G = \mathbb{Z}/2\mathbb{Z} = \{1,-1\}$ under multiplication acting on the set $X = \{-1,0,1\}$ by multiplication.)

Suppose $G$ acts on $X$ and you want to check if this action is free - that is, that the only element of $G$ which fixes any point of $X$ is the identity in $G$. One way to prove this is to look at each element of $X$ one at a time and ask "What $g\in G$ fixes this element?" As long as the answer is always "only $g = e$", the action is free.

Suppose you check every point but one and you keep finding the answer is "only $g= e$" so the action is looking free. Then you check that last point ($0\in X$ in the concrete example) and find "every element of g fixes this point!" Well, then the action is not free, woops.

But, since $G$ fixes this "bad" point, it must also preserve the other points: if $y\in X$ is not the bad point, then $gy$ isn't the bad point either, for any choice of $g\in G$. Thus, $G$ also acts on a subset of $X$ and the action on this subset is free. So, a nonfree action, when restricted appropriately, can become free.

In the case of restricting the $O(n)$ action from $\mathbb{R}^n$ to $S^{n-1}$, nothing so drastic happens, but there still is at least two changes - the action on $\mathbb{R}^n$ has a fixed point and is nontransitive, while the action on $S^{n-1}$ doesn't have a fixed point and is transitive.

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Thank-you for taking the time to write a very detailed and helpful answer, it improved my understanding quite a bit. –  Andrew Ledesma May 30 '13 at 4:10
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