Back to the on-screen lesson ·
Conjugation read as an action: classes are orbits, centralisers are stabilisers, the centre is the fixed set — and adding the class sizes forces a group of prime power order to have a centre.
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 conjugacy classes and centralisers of a small group, write down its class equation and check that the sizes add correctly, say why every class size divides the order, prove that a group of prime power order has a non-trivial centre, list the candidates for normal subgroups from the class sizes alone, and deduce that every group of order $p^{2}$ is abelian.
The first course used conjugation to define normality: $N$ is normal when $gNg^{-1} = N$. It also observed that conjugate permutations have the same cycle shape, because conjugating is renaming the letters. This lesson turns conjugation into an action and counts its orbits, and both of those old facts come back as statements about that action.
The conjugacy class of $x$ is $\{gxg^{-1} : g \in G\}$. Its centraliser $C(x) = \{g : gx = xg\}$ is the subgroup of elements commuting with $x$. The centre $Z(G) = \{x : gx = xg \text{ for all } g\}$ is the set of elements whose class has one member; it is a normal subgroup, and $Z(G) = G$ exactly when $G$ is abelian.
Let $G$ act on itself by $g \cdot x = gxg^{-1}$. The axioms hold — $exe^{-1} = x$ and $h(gxg^{-1})h^{-1} = (hg)x(hg)^{-1}$ — so everything proved about actions applies.
The dictionary. Orbits are conjugacy classes. Stabilisers are centralisers $C(x)$. The fixed points, meaning elements whose orbit is a single point, are exactly the centre $Z(G)$. Orbit-stabiliser says the class of $x$ has $[G : C(x)]$ elements.
The class equation. The classes partition $G$. Splitting them into the singletons and the rest:
$$|G| = |Z(G)| + \sum_{i} [G : C(x_i)],$$
with one term for each class of size greater than one, and $x_i$ a representative of it. Every term on the right divides $|G|$, and every term of the sum is greater than $1$.
Why this is worth so much. It is a partition of $|G|$ into divisors of $|G|$, with strong constraints on which. That is a very tight arithmetic condition, and it decides structural questions that look nothing like counting:
That last argument is worth noticing: a normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $|G|$. Listing the class sizes therefore lists the candidates for normal subgroups, and often there are none.
Another way: picture
Lay the group out as a set and colour each element by its class. The centre is the collection of classes painted one element at a time; everything else comes in blocks whose sizes divide the order and exceed one. The class equation is just the statement that the colours cover the group without overlapping — and almost every theorem below comes from the fact that there are very few ways to write a number as such a sum.
Another way: steps
To use the class equation: 1. Write $|G|$ and list the divisors. 2. Work out the class sizes, either directly or as indices of centralisers. 3. Check they add to $|G|$, with the $1$s accounting for the centre. 4. To bound the centre, read the equation modulo a prime dividing the order. 5. To hunt for normal subgroups, look for collections of classes including the identity whose sizes add to a divisor of $|G|$.
A normal subgroup is closed under conjugation, so it is a union of conjugacy classes. It contains the identity, so one of those classes is the singleton $\{e\}$. And its order divides $|G|$. Those three facts together turn the search for normal subgroups into an arithmetic puzzle with a short list of candidates.
$A_5$, of order $60$. Its classes have sizes $1, 15, 20, 12, 12$. A normal subgroup's order is $1$ plus some sub-collection of $15, 20, 12, 12$, and must divide $60$. Running through them: $1, 16, 21, 13, 13, 36, 28, 28, 33, 25, 48, 40, 45, 53, 60$ — of these only $1$ and $60$ divide $60$. So $A_5$ is simple, and the proof is four lines of addition.
$S_4$, of order $24$. Classes $1, 6, 3, 8, 6$. Candidates: $1$; $1 + 3 = 4$; $1 + 3 + 8 = 12$; $1+6+3+8+6=24$; and the rest fail to divide. The three that work are exactly the trivial subgroup, $V_4$, $A_4$ and $S_4$ — which is the complete list of normal subgroups of $S_4$, found without exhibiting a single one.
A caution. The condition is necessary, not sufficient: a union of classes whose size divides the order need not be a subgroup. What the method does is narrow an infinite-looking search to a handful of cases that can then be checked by hand.
Treating a conjugacy class as a subgroup. It is not one, except for $\{e\}$. The transpositions of $S_3$ form a class; their products do not lie in it.
Confusing the centraliser $C(x)$ with the centre $Z(G)$. The first is attached to one element and is usually large; the second is the intersection of all of them and is often trivial.
Thinking the class sizes must be equal. If they were, the equation would say nothing. Their inequality is the content.
Forgetting that the sum omits the singletons. $|Z(G)|$ collects all the classes of size one; writing them again in the sum double-counts them.
Extending the $p$-group argument too far. The centre is non-trivial needs $|G|$ to be a power of a single prime. $S_3$ has order $6$ and a trivial centre.
The six elements split by cycle shape: the identity, the three transpositions, the two three-cycles.
Conjugate means same shape.
So $6 = 1 + 3 + 2$. The centre contributes the $1$ alone, so $Z(S_3)$ is trivial.
The class equation.
A normal subgroup is $1$ plus a sub-collection of $\{3, 2\}$ dividing $6$: that gives $1, 3, 6$, and $3$ is $A_3$. So $A_3$ is the only proper non-trivial normal subgroup.
The normal subgroups, read off.
Non-central class sizes divide $8$ and exceed $1$, so each is $2$, $4$ or $8$ — all even.
Each class size is an index.
The class equation gives $8 = |Z(G)| + (\text{a sum of even numbers})$, so $|Z(G)|$ is even.
Read modulo $2$.
So $|Z(G)| \in \{2, 4, 8\}$. For the symmetries of a square and for the quaternion group it is $2$; for the abelian groups it is $8$; it is never $1$.
Never trivial.
The eight elements are the identity, the half turn, the two quarter turns, the two diagonal reflections and the two axis reflections.
List by type.
The identity and the half turn commute with everything; each remaining pair forms one class of size $2$.
Classes of sizes $1, 1, 2, 2, 2$.
So $8 = 2 + 2 + 2 + 2$ with $|Z| = 2$, and the candidates for normal subgroups are $2$, $2 + 2 = 4$ three ways, and $8$ — which is the correct list.
In $S_4$, two permutations are conjugate exactly when they have the same cycle shape. The identity's row is given. Fill in how many permutations have each shape, the order of the centraliser of one of them, and the total.
| Permutations of this shape | Order of the centraliser of one | |
|---|---|---|
| Shape $e$ | 1 | 24 |
| Shape $(a\,b)$ | ||
| Shape $(a\,b)(c\,d)$ | ||
| Shape $(a\,b\,c)$ | ||
| Shape $(a\,b\,c\,d)$ | ||
| Total number of permutations | — |
How many conjugacy classes does $\mathbb{Z}_{12}$ have?
Answer:
Select every statement that is true of the conjugacy classes of a finite group $G$.
This task has no paper form; do it on a device.
The class equation is $|G| = |Z(G)| + \sum [G : C(x_i)]$. Match each ingredient to what it is, read as an action.
| The conjugacy classes | Its centraliser $C(x)$ | The centre $Z(G)$ | The index $[G : C(x)]$ | The order of the group | |
|---|---|---|---|---|---|
| The orbits of the action | |||||
| The stabiliser of an element | |||||
| The points fixed by every element of $G$ | |||||
| The size of one orbit |
Build the proof that a group of order $p^{n}$, with $p$ prime and $n \ge 1$, has a centre with more than one element.
This task has no paper form; do it on a device.
$G$ has $5^{2}$ elements, and $5$ is prime. How many elements does its centre have?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In $S_4$, two permutations are conjugate exactly when they have the same cycle shape. The identity's row is given. Fill in how many permutations have each shape, the order of the centraliser of one of them, and the total.
| Permutations of this shape | Order of the centraliser of one | |
|---|---|---|
| Shape $e$ | 1 | 24 |
| Shape $(a\,b)$ | ||
| Shape $(a\,b)(c\,d)$ | ||
| Shape $(a\,b\,c)$ | ||
| Shape $(a\,b\,c\,d)$ | ||
| Total number of permutations | — |
You can write the class equation of a small group and use it to bound the centre. Say in your own words why a normal subgroup has to be a union of conjugacy classes. Next: the Sylow theorems, which are the same counting pushed as far as it will go.
9. Your turn: the class equation of the symmetries of a square, step 3