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A group as a way of moving a set: the two axioms, the homomorphism into $\operatorname{Sym}(X)$ they amount to, kernels and faithfulness, and the four actions a group has on things built out of itself.
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 check the two axioms of a group action, convert freely between an action and a homomorphism into a symmetric group, name the orbit, stabiliser and kernel of an action, decide whether an action is faithful or transitive, recognise the four standard actions a group has on itself and its subgroups, and prove Cayley's theorem by building an action from nothing.
The first course built $S_n$ and showed that its elements compose, invert and have parity. It also proved Lagrange's theorem by tiling a group with cosets. This lesson joins the two: a group is allowed to permute a set, the tiles become orbits, and Lagrange's counting argument turns into a tool that works on any set at all rather than only on the group.
An action of $G$ on a set $X$ writes $g \cdot x$ for a point of $X$, with $e \cdot x = x$ and $h \cdot (g \cdot x) = (hg) \cdot x$. The orbit of $x$ is $Gx = \{g \cdot x : g \in G\}$; its stabiliser $G_x = \{g : g \cdot x = x\}$ is a subgroup. The kernel of the action is $\bigcap_x G_x$, the elements that move nothing; the action is faithful when the kernel is trivial and transitive when there is only one orbit.
An action of a group $G$ on a set $X$ is a rule $G \times X \to X$, written $g \cdot x$, satisfying exactly two conditions:
$$e \cdot x = x, \qquad h \cdot (g \cdot x) = (hg) \cdot x.$$
Nothing else is demanded. In particular $X$ need have no relation to $G$: the eight symmetries of a square act on four corners, on four edges, on two diagonals and on one centre, all at once.
The same thing said twice. Fix $g$ and let $x$ vary. The map $\sigma_g(x) = g \cdot x$ has $\sigma_{g^{-1}}$ as a two-sided inverse, so it is a bijection of $X$; and the second axiom says exactly $\sigma_{hg} = \sigma_h \circ \sigma_g$. So an action of $G$ on $X$ is a homomorphism
$$\varphi : G \to \operatorname{Sym}(X),$$
and conversely any such homomorphism defines an action. The two descriptions are the same information, and each is convenient somewhere: the action is what one computes with, the homomorphism is what has a kernel.
The kernel. $\ker \varphi$ is the set of group elements that move no point at all. The action is faithful when this is trivial, and then $G$ is literally a group of permutations of $X$. When it is not, the set simply cannot see part of the group: an abelian group conjugating itself moves nothing, and the kernel is everything.
The four standard actions. A group always has a set to act on, because it has itself. Left multiplication gives Cayley's theorem; conjugation gives the class equation; the cosets of a subgroup give a homomorphism into a small symmetric group; and conjugation on Sylow subgroups gives the Sylow counts. The whole of this unit is those four, counted.
A group is not only a set with an operation; it is a way of moving something. Reading $G$ as permutations of a set turns questions about $G$ into questions about orbits, and the answers come back as arithmetic: conjugation makes the class equation, multiplication makes Cayley's theorem, and cosets make the rest.
Another way: picture
Think of $X$ as a board and the elements of $G$ as the moves available. The axioms say doing nothing is a move, and that two moves in succession are again one of the moves on the list. What a point can reach is its orbit; what leaves a point alone is its stabiliser; and a move that leaves the whole board unchanged is invisible from the board, which is what the kernel collects.
Another way: steps
To read a situation as an action: 1. Name the set $X$ the group is moving. This is the step with the choice in it. 2. Check the two axioms — usually one line each. 3. Ask which elements fix everything: that is the kernel, and whether it is trivial says whether the action is faithful. 4. Ask whether one point can reach all of $X$: that is transitivity. 5. Then count. Orbits and stabilisers are where the theorems come from.
Every application in this unit follows the same two-line pattern: invent a set the group moves, then count orbits and stabilisers. The mathematics is in the first line.
On itself, by left multiplication. $X = G$, $g \cdot x = gx$. Faithful, transitive, and it embeds $G$ in $S_{|G|}$ — Cayley's theorem.
On itself, by conjugation. $X = G$, $g \cdot x = gxg^{-1}$. The orbits are conjugacy classes, the stabilisers are centralisers, and the kernel is the centre. Orbit-stabiliser applied here is the class equation, and everything the next lessons prove about $p$-groups comes out of it.
On the cosets of a subgroup. $X = G/H$ (as a set of cosets), $g \cdot xH = gxH$. Transitive, with $|X| = [G : H]$, so $G$ maps into $S_{[G:H]}$. The kernel is the largest normal subgroup of $G$ inside $H$, which is how one proves that a group of order $24$ cannot be simple: it would embed in $S_3$.
On a geometric object. The symmetries of a cube act on $6$ faces, $8$ vertices, $12$ edges and $4$ space diagonals. Counting the last of these transitively is what identifies that group as $S_4$.
A useful habit: when a problem resists, ask what set the group could be made to move. A question about the order of a group is often a question about a set whose size is already known.
Thinking the set has to be the group. It usually is not. Most of the power comes from acting on something small — cosets, Sylow subgroups, faces of a cube — because a small set means a small symmetric group and a strong restriction.
Writing $g \cdot (h \cdot x) = (gh) \cdot x$ and not checking the order. With the convention above it is $h \cdot (g \cdot x) = (hg) \cdot x$. Getting it backwards gives a right action, which is a perfectly good object and a different one; mixing the two in one argument produces sign-like errors that are hard to find.
Confusing the stabiliser with the kernel. The stabiliser depends on the point; the kernel is the intersection of all of them. A reflection stabilises two corners of a square and is not in the kernel.
Assuming an action is faithful. Conjugation by an abelian group moves nothing at all. When an action has a kernel, the useful statement is about $G/\ker$, not about $G$.
Expecting one orbit. Transitivity is a real condition, not part of the definition. Conjugation is almost never transitive — the identity is always alone in its class.
The group has eight elements; the set has four corners. The axioms hold because composing symmetries is composing the maps they induce on corners.
An action of a group on a set unrelated to it.
Every corner reaches every corner, so the action is transitive with one orbit of size $4$. The corner at the top right is fixed by the identity and by the diagonal reflection through it, so its stabiliser has $2$ elements.
Orbit $4$, stabiliser $2$.
The kernel is trivial: a symmetry fixing all four corners is the identity. So the eight symmetries embed in $S_4$, which has $24$ elements.
Faithful, so the group is a permutation group on four letters.
$X = G$ and $g \cdot x = gxg^{-1}$. The axioms hold: $exe^{-1} = x$, and $h(gxg^{-1})h^{-1} = (hg)x(hg)^{-1}$.
An action of the group on itself.
The orbit of $x$ is its conjugacy class and the stabiliser is its centraliser $C(x)$. An element is alone in its orbit exactly when it commutes with everything.
Orbits are classes, stabilisers are centralisers.
The kernel is $\{g : gxg^{-1} = x \text{ for all } x\} = Z(G)$. So this action is faithful exactly when the centre is trivial, and useless for an abelian group.
The kernel is the centre.
The rule is $\sigma \cdot \{a, b\} = \{\sigma(a), \sigma(b)\}$, which is an action because applying two permutations in turn is applying their product.
Check the axioms.
Any pair can be carried to any other, so the action is transitive and the single orbit has $6$ points.
One orbit.
Orbit-stabiliser then gives a stabiliser of order $24 / 6 = 4$: the permutations preserving the split of $\{1,2,3,4\}$ into that pair and its complement. And the kernel is trivial, since a permutation fixing every pair fixes every letter — so $S_4$ also embeds in $S_6$.
A group has four standard actions on things built out of itself. Match each to the theorem it produces.
| Cayley's theorem: every group is a permutation group | The class equation | A homomorphism into the symmetric group on the index | The number of Sylow subgroups divides the order | Every subgroup of an abelian group is normal | |
|---|---|---|---|---|---|
| $G$ acting on itself by left multiplication | |||||
| $G$ acting on itself by conjugation | |||||
| $G$ acting on the left cosets of a subgroup | |||||
| $G$ acting on its Sylow subgroups by conjugation |
Select every statement that is true of a group $G$ acting on a set $X$.
This task has no paper form; do it on a device.
Put in order the steps that turn an action of $G$ on $X$ into a homomorphism $G \to \operatorname{Sym}(X)$.
Number the steps in order (write the number in the box):
Is this action faithful for every group it is defined on: $G$ acting on itself by conjugation?
A group with $5$ elements acts on itself by left multiplication. How many elements does the kernel of that action have?
Answer:
Build the proof of Cayley's theorem: every group of order $n$ is isomorphic to a subgroup of $S_n$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Is this action faithful for every group it is defined on: $S_n$ acting on the $n$ letters?
You can check that a rule is an action, read it as a homomorphism into a symmetric group, and find its kernel. Say in your own words why choosing which set the group acts on is where the mathematics is. Next: orbits and stabilisers, and the counting theorem that relates them.
9. Your turn: $S_4$ acting on the six pairs of letters, step 3