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The two sets attached to a point of an action, why orbits partition and stabilisers are subgroups, and the theorem $|G| = |Gx| \cdot |G_x|$ read in both directions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the orbit and the stabiliser of a point, say why the orbits partition the set and why every stabiliser is a subgroup, apply the orbit-stabiliser theorem to find whichever of the three numbers is missing, prove the theorem as a bijection between an orbit and a set of cosets, and use it to count the rotations of a solid without listing any of them.
Lagrange's theorem said a subgroup tiles its group with cosets of equal size, so the order of the subgroup divides the order of the group. The proof of this lesson's theorem is that proof with a new label on the tiles: each coset is now the set of group elements that send one point to one place.
For $G$ acting on $X$ and $x \in X$, the orbit is $Gx = \{g \cdot x : g \in G\}$, a subset of $X$; the stabiliser is $G_x = \{g \in G : g \cdot x = x\}$, a subgroup of $G$. The index $[G : G_x]$ counts the cosets of the stabiliser. An action with a single orbit is transitive.
Fix a point $x$ of the set. Two things are attached to it, and they live in different places: the orbit $Gx$ is a subset of $X$, and the stabiliser $G_x$ is a subgroup of $G$.
Orbits partition. Writing $x \sim y$ when some $g$ carries $x$ to $y$ gives an equivalence relation — the identity makes it reflexive, inverses make it symmetric and products make it transitive. So $X$ falls into disjoint orbits, and every counting argument in this unit begins by adding up their sizes.
Stabilisers are subgroups. $e$ fixes $x$; if $g$ and $h$ fix $x$ then $gh$ does; and $g \cdot x = x$ gives $x = g^{-1} \cdot x$.
The theorem.
$$|Gx| = [G : G_x], \qquad \text{so} \qquad |G| = |Gx| \cdot |G_x|$$
for a finite group. The proof is a bijection: send the coset $gG_x$ to the point $g \cdot x$. It is well defined and injective for the same one-line reason — $gG_x = hG_x$ exactly when $h^{-1}g$ fixes $x$ — and surjective because that is what an orbit is.
Why it is used so much. It converts between a count of things and a count of symmetries. Knowing the group, it bounds how large an orbit can be; knowing an orbit and a stabiliser, it gives the order of a group nobody has listed. Both readings are used constantly, and the second is how the order of a rotation group is found without naming a single rotation.
Conjugate points, conjugate stabilisers. If $y = g \cdot x$ then $G_y = gG_xg^{-1}$. So points in one orbit have stabilisers that are different subgroups but the same size — which is consistent with the theorem, since the orbit is the same.
A second course in algebra counts before it constructs. An orbit size divides the order of the group, a Sylow count is pinned between a divisibility and a congruence, and a degree multiplies up a tower — and each of those numbers rules out structures nobody has to go looking for.
Another way: picture
Stand at one point of the set and push it around with every element of the group. Where it lands is the orbit. Some pushes do nothing at all, and those form the stabiliser. Now pair up the group elements by where they send the point: elements landing in the same place form one coset, and there is exactly one coset per landing site. Counting the sites and counting the elements per site multiplies back to the whole group.
Another way: steps
To use orbit-stabiliser: 1. Choose the set the group acts on, and pick a convenient point. 2. Find its orbit — often all of $X$, when the action is transitive. 3. Find its stabiliser, usually by asking what a symmetry fixing that point can still do. 4. Multiply for the order of the group, or divide for whichever of the two is unknown. 5. Sanity-check: both factors must divide the order.
Downwards: the group is known. Then the theorem restricts what can happen. A group of order $15$ acting on a set of $4$ points has orbits whose sizes divide $15$, so each orbit has $1, 3, 5$ or $15$ points; but $15$ and $5$ will not fit in a set of four, and $3 + 1$ or $1 + 1 + 1 + 1$ are the only options. That is already enough to show the action cannot be transitive.
The same reading proves the useful fact that a group of order $n$ has no faithful action on fewer than a handful of points unless $n$ is small: the image sits inside $S_k$, so $n$ divides $k!$.
Upwards: the group is unknown. Then the theorem computes. The order of the rotation group of a cube is $24$ because the six faces form one orbit and four rotations hold a face still. No rotation had to be described, and the same two observations handle every regular solid.
This reading also gives the cleanest count of $|S_n|$: $S_n$ acts transitively on $n$ letters, the stabiliser of one letter is a copy of $S_{n-1}$, so $|S_n| = n \cdot |S_{n-1}|$ and induction finishes it.
A warning about the choice of point. Orbits vary. Acting on the corners and the centre of a square gives an orbit of size $4$ and an orbit of size $1$, with stabilisers of orders $2$ and $8$. The theorem holds for each point separately; there is no single orbit size for the action.
Treating the orbit as a subgroup. It is a subset of the set being acted on, and lives somewhere the group does not. Only the stabiliser is a subgroup.
Adding instead of multiplying. $|G| = |Gx| \cdot |G_x|$. The additive version would make an orbit of size $6$ inside a group of order $24$ have a stabiliser of order $18$, which is not even a divisor.
Assuming every orbit has the same size. They need not, and the ones of size $1$ are usually the interesting ones — they become the centre in the class equation.
Assuming stabilisers of different points are equal. They are conjugate when the points share an orbit, and unrelated otherwise.
Forgetting to check transitivity before using $|Gx| = |X|$. The orbit is all of $X$ only when the action is transitive, and that is an assumption to be checked rather than assumed.
The group has $8$ elements. A corner can be carried to any corner, so its orbit has $4$ points.
Transitive on corners.
Orbit-stabiliser gives a stabiliser of order $8 / 4 = 2$, and indeed the identity and the diagonal reflection through that corner are the only symmetries fixing it.
Divide.
The centre of the square is fixed by everything: orbit $1$, stabiliser $8$, and $1 \times 8 = 8$ again. Different point, different split, same product.
Every point multiplies back to the order.
$S_n$ acts on $\{1, \ldots, n\}$ transitively, so the orbit of the letter $1$ has $n$ points.
One orbit.
Its stabiliser is the set of permutations fixing $1$, which is a copy of $S_{n-1}$.
The stabiliser is a smaller symmetric group.
So $|S_n| = n \cdot |S_{n-1}|$, and with $|S_1| = 1$ this gives $n!$ by induction — without ever writing a permutation down.
A recursion from the theorem.
Any edge can be carried to any other, so the twelve edges form a single orbit.
Transitive.
The rotations holding one edge in place are the identity and the half turn about the axis through the middles of that edge and the opposite one.
A stabiliser of order $2$.
So the group has $12 \times 2 = 24$ elements — the same answer the faces gave, as it must be, because it is the same group counted a different way.
The rotations of a cube act on its faces, its vertices, its edges and its four space diagonals. The orbit sizes are given. Fill in the size of one stabiliser in each case, and the product of the two.
| Size of the orbit | Size of a stabiliser | Their product | |
|---|---|---|---|
| The faces | 6 | ||
| The vertices | 8 | ||
| The edges | 12 | ||
| The space diagonals | 4 |
Consider $\mathbb{Z}_6$ acting on itself by addition, at any element. The group has $6$ elements and the stabiliser of that point has $1$. How many points are in its orbit?
Answer:
Select every statement that is true of a finite group $G$ acting on a set $X$.
This task has no paper form; do it on a device.
Match each action to the collection its orbits turn out to be.
| The conjugacy classes | The left cosets of $H$ | The families of conjugate subgroups | One orbit, holding all four corners | The elements of the centre, one each | |
|---|---|---|---|---|---|
| $G$ acting on itself by conjugation | |||||
| A subgroup $H$ acting on $G$ by $h \cdot g = gh^{-1}$ | |||||
| $G$ acting on its subgroups by conjugation | |||||
| The rotations of a square acting on its corners |
Build the proof of the orbit-stabiliser theorem: $|Gx| = [G : G_x]$.
This task has no paper form; do it on a device.
How many rotations carry a regular tetrahedron onto itself?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The rotations of a cube act on its faces, its vertices, its edges and its four space diagonals. The orbit sizes are given. Fill in the size of one stabiliser in each case, and the product of the two.
| Size of the orbit | Size of a stabiliser | Their product | |
|---|---|---|---|
| The faces | 6 | ||
| The vertices | 8 | ||
| The edges | 12 | ||
| The space diagonals | 4 |
You can find an orbit and a stabiliser and multiply them back to the order of the group. Say in your own words why the orbit is not a subgroup and the stabiliser is. Next: averaging fixed points, which counts the orbits themselves.
9. Your turn: the rotations of a cube acting on its twelve edges, step 3