Conjugacy Classes, Symmetric and Alternating Groups — Studies in Algebra and Group Theory

24 Conjugacy Classes, Symmetric and Alternating Groups

29 October 2025 · Denis Auroux

Conjugation and the class equation

Key observation for classifying finite groups: acts on itself by conjugation: acts by . We’ve seen that this does define a group homomorphism , so it is indeed an action.

The orbits of this action are the conjugacy classes in , and the stabilizer of an element is the subgroup of elements which commute with . This is called the centralizer of , . Note , the center of , is the kernel of the action (i.e. the subgroup of elements which act trivially). So: the action is trivial when is abelian; it is faithful iff .

How does this help?

The conjugacy classes form a partition of , so
For each conjugacy class, divides .
Moreover for the identity element, and iff .

is called the class equation of the group .

This is extremely useful. For example:

Theorem 24.1. If for prime, then must be abelian.

Proof.

Conjugacy classes have order , and . Thus the number of conjugacy classes such that , i.e. of central elements of , must be a multiple of . Hence .
is a subgroup of , so divides : it’s or . If then is abelian!
Now assume , and let . Then commutes with itself and with , so , hence . But is a subgroup of , so . This implies , i.e. commutes with all elements of , i.e. : contradiction. So , is abelian.

∎

(Hence the only groups of order up to isomorphism are and .)

Proposition 24.2. There are exactly 5 groups of order 8 up to isomorphism.

We know the 3 abelian ones: , , . We know = the symmetries of the square. Finally: the quaternion group with (multiplication by flips signs).

Two ways to show there are only two nonabelian groups of order 8:

“By hand” (see the HW hint): suppose and is not abelian.
- Step 1: a group where every element has must be abelian, so there must be an element of order 4 (order 8 would make ).
- Step 2: the order 4 subgroup generated by is normal. Work out the possibilities for multiplication by an element such that . (Need: ? ?)
Using conjugacy and the class equation:
- Step 1: class equation , , has order 2, 4, or 8. abelian. 4 is impossible by the same argument as for above. So .
- Step 2: if then , but . So , and . Hence the class equation is Then work out the possibilities!

Conjugacy classes in the symmetric group

A -cycle , where are distinct elements of , is a permutation mapping and all other elements to themselves.
Two cycles are disjoint if the subsets of elements they cycle are disjoint. Disjoint cycles commute.

Proposition 24.3. Any permutation can be expressed as a product of disjoint cycles, uniquely up to reordering the factors (disjoint cycles commute, so the order doesn’t matter).

Algorithm: look at the successive images of 1 under ; this gives a subset of elements that are cyclically permuted by . Then consider the elements not in this subset, and repeat.

In other terms: the various cycles are the restrictions of to the orbits of on .

Example 24.4. Here = the successive images of 1 under until it returns to 1; : the same for the other elements not in the previous cycles.

Proposition 24.5. Let be a -cycle and any permutation; then .

Proof. Calculate: (under , then , then ), so the action on is as claimed. Other elements: . ∎

Corollary 24.6. All -cycles are conjugate in .

More generally, are conjugate iff they have the same cycle lengths in their disjoint cycle decompositions.

Hence, conjugacy classes in correspond to partitions of , i.e. ways to write as a sum of positive integers (up to reordering the terms).

Example 24.7 ().

partition	description	size of conj. class
	identity (only “1-cycles”)	1
	transpositions	3
	3-cycles	2

Example 24.8 ().

partition	description	size of conj. class
	id	1
	transposition	6
	2 transpositions	3
	3-cycle	8
	4-cycle	6

The class equation of is .

This helps us find normal subgroups of : is normal iff . So a normal subgroup is a union of conjugacy classes! Also, it must include id, and divides . Here: apart from and , the only candidates are

: . This is indeed a normal subgroup ().
: . This is the alternating group .

Example 24.9 ().

partition	description	size of conj. class
	id	1
	transposition	10
	2 transpositions	15
	3-cycle	20
	3-cycle + transposition	20
	4-cycle	30
	5-cycle	24

Class equation: .

Search for normal subgroups (besides and ): the only options are

: . This is not a subgroup (not closed under composition): .
: id, , 5-cycles, and either 3-cycles or (3-cycle)(transposition) (2 possibilities for the 20). Only the first option (3-cycles) works, and gives . (The other isn’t closed under composition.)

The alternating group

Recall we’ve defined the sign homomorphism by using that transpositions generate ; we still need to check this is independent of how we express as a product of transpositions. I mentioned: , where (but then… check it’s a homomorphism?).

Now we can do better: take a vector space with basis ; then to each we associate an element of : the linear map such that . This gives an injective homomorphism (with image the subgroup of “permutation matrices”).

Now, has finite order (since does), hence is a root of unity, hence . We can define : this is clearly well defined and a homomorphism.

Concretely, to compute the sign: acts on by , and the sign is , counting the transpositions needed to switch these back in order; so this agrees with the other definition.

Notes not reached in this lecture, covered at the start of {{LINK:L25}}

Note (Observe). A -cycle has sign (since ).

So if has cycle lengths (including the 1’s), i.e. corresponds to the partition , then

Definition 24.10. , the alternating group (a normal subgroup of index 2 in ).

Proposition 24.11. If is a conjugacy class, then either

is odd, ; or
(a) is a conjugacy class in ; or (b) splits into 2 conjugacy classes in .

Case 2a vs. 2b: take and its centralizer . Is or not? If yes, then the conjugates of by odd permutations are different from the conjugates by even permutations, and these form two conjugacy classes in . If not, then all conjugates of in are conjugates by elements of .

Example 24.12 ().

3-cycles still form a single conjugacy class in (because ); also for the ’s ().
5-cycles split into 2 conjugacy classes in .

So the class equation of is .

We can now look for normal subgroups of . We can’t reach a divisor of 60 in any nontrivial way, hence the only ones are and .

Hence:

Proposition 24.13. is simple.