IB Groups, Rings and Modules

1Groups

1.4 Conjugacy, centralizers and normalizers

We have seen that every group acts on itself by multiplying on the left. A group

G can also act on itself in a different way, by conjugation:

g ∗ g

= gg

−1

Let

G → Sym

(

) be the associated permutation representation. We know,

by definition, that

(

) is a bijection from

as sets. However, here

not an arbitrary set, but is a group. A natural question to ask is whether

(

)

is a homomorphism or not. Indeed, we have

φ(g)(g

· g

) = gg

−1

= (gg

−1

)(gg

−1

) = φ(g)(g

)φ(g)(g

(

) is a homomorphism from

. Since

(

) is bijective (as in any

group action), it is in fact an isomorphism.

Thus, for any group

, there are many isomorphisms from

to itself, one

for every g ∈ G, and can be obtained from a group action of G on itself.

We can, of course, take the collection of all isomorphisms of

, and form a

new group out of it.

Definition (Automorphism group). The automorphism group of G is

Aut(G) = {f : G → G : f is a group isomorphism}.

This is a group under composition, with the identity map as the identity.

This is a subgroup of

Sym

(

), and the homomorphism

G → Sym

(

) by

conjugation lands in Aut(G).

This is pretty fun — we can use this to cook up some more groups, by taking

a group and looking at its automorphism group.

We can also take a group, take its automorphism group, and then take its

automorphism group again, and do it again, and see if this process stabilizes, or

becomes periodic, or something. This is left as an exercise for the reader.

Definition (Conjugacy class). The conjugacy class of g ∈ G is

ccl

(g) = {hgh

−1

: h ∈ G},

i.e. the orbit of g ∈ G under the conjugation action.

Definition (Centralizer). The centralizer of g ∈ G is

(g) = {h ∈ G : hgh

−1

= g},

i.e. the stabilizer of

under the conjugation action. This is alternatively the set

of all h ∈ G that commute with g.

Definition (Center). The center of a group G is

Z(G) = {h ∈ G : hgh

−1

= g for all g ∈ G} =

g∈G

(g) = ker(φ).

These are the elements of the group that commute with everything else.

By the orbit-stabilizer theorem, for each

x ∈ G

, we obtain a bijection

ccl(x) ↔ G/C

(x).

Proposition. Let G be a finite group. Then

|ccl(x)| = |G : C

(x)| = |G|/|C

(x)|.

In particular, the size of each conjugacy class divides the order of the group.

Another useful notion is the normalizer.

Definition (Normalizer). Let H ≤ G. The normalizer of H in G is

(H) = {g ∈ G : g

−1

Hg = H}.

Note that we certainly have

H ≤ N

(

). Even better,

HCN

(

), essentially

by definition. This is in fact the biggest subgroup of G in which H is normal.

We are now going to look at conjugacy classes of

. Now we recall from IA

Groups that permutations in

are conjugate if and only if they have the same

cycle type when written as a product of disjoint cycles. We can think of the

cycle types as partitions of

. For example, the partition 2

1 of 5 corresponds

to the conjugacy class of (1 2)(3 4)(5). So the conjugacy classes of

are exactly

the partitions of n.

We will use this fact in the proof of the following theorem:

Theorem. The alternating groups

are simple for

n ≥

5 (also for

= 1

3).

The cases in brackets follow from a direct check since

∼

{e}

and

∼

, all of which are simple. We can also check manually that

has

non-trivial normal subgroups, and hence not simple.

Recall we also proved that

is simple in IA Groups by brute force — we

listed all its conjugacy classes, and see they cannot be put together to make a

normal subgroup. This obviously cannot be easily generalized to higher values

of n. Hence we need to prove this with a different approach.

Proof. We start with the following claim:

Claim. A

is generated by 3-cycles.

As any element of

is a product of evenly-many transpositions, it suffices

to show that every product of two transpositions is also a product of 3-cycles.

There are three possible cases: let a, b, c, d be distinct. Then

(i) (a b)(a b) = e.

(ii) (a b)(b c) = (a b c).

(iii) (a b)(c d) = (a c b)(a c d).

So we have shown that every possible product of two transpositions is a product

of three-cycles.

Claim. Let H C A

. If H contains a 3-cycle, then we H = A

We show that if

contains a 3-cycle, then every 3-cycle is in

. Then we

are done since

is generated by 3-cycles. For concreteness, suppose we know

(a b c) ∈ H, and we want to show (1 2 3) ∈ H.

Since they have the same cycle type, so we have

σ ∈ S

such that (

a b c

) =

(1 2 3)

−1

. If

is even, i.e.

σ ∈ A

, then we have that (1 2 3)

∈ σ

−1

Hσ

by the normality of H and we are trivially done.

is odd, replace it by

¯σ

σ ·

(4 5). Here is where we use the fact that

n ≥ 5 (we will use it again later). Then we have

¯σ(1 2 3)¯σ

−1

= σ(4 5)(1 2 3)(4 5)σ

−1

= σ(1 2 3)σ

−1

= (a b c),

using the fact that (1 2 3) and (4 5) commute. Now

¯σ

is even. So (1 2 3)

∈ H

above.

What we’ve got so far is that if

H C A

contains any 3-cycle, then it is

Finally, we have to show that every normal subgroup must contain at least one

3-cycle.

Claim. Let H C A

be non-trivial. Then H contains a 3-cycle.

We separate this into many cases

(i)

Suppose

contains an element which can be written in disjoint cycle

notation

σ = (1 2 3 ···r)τ,

for

r ≥

4. We now let

= (1 2 3)

∈ A

. Then by normality of

, we know

−1

σδ ∈ H

. Then

−1

σδ ∈ H

. Also, we notice that

does not contain

3. So it commutes with

, and also trivially with (1 2 3

··· r

). We

can expand this mess to obtain

−1

σδ = (r ··· 2 1)(1 3 2)(1 2 3 ··· r)(1 2 3) = (2 3 r),

which is a 3-cycle. So done.

The same argument goes through if

= (

··· a

)

for any

, ··· , a

(ii)

Suppose

contains an element consisting of at least two 3-cycles in disjoint

cycle notation, say

σ = (1 2 3)(4 5 6)τ

We now let δ = (1 2 4), and again calculate

−1

σδ = (1 3 2)(4 6 5)(1 4 2)(1 2 3)(4 5 6)(1 2 4) = (1 2 4 3 6).

This is a 5-cycle, which is necessarily in

. By the previous case, we get a

3-cycle in H too, and hence H = A

(iii)

Suppose

contains

= (1 2 3)

, with

a product of 2-cycles (if

contains

anything longer, then it would fit in one of the previous two cases). Then

= (1 2 3)

= (1 3 2) is a three-cycle.

(iv)

Suppose

contains

= (1 2)(3 4)

, where

is a product of 2-cycles. We

first let δ = (1 2 3) and calculate

u = σ

−1

σδ = (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3) = (1 4)(2 3),

which is again in

. We landed in the same case, but instead of two

transpositions times a mess, we just have two transpositions, which is nicer.

Now let

v = (1 5 2)u(1 2 5) = (1 3)(4 5) ∈ H.

Note that we used

n ≥

5 again. We have yet again landed in the same case.

Notice however, that these are not the same transpositions. We multiply

uv = (1 4)(2 3)(1 3)(4 5) = (1 2 3 4 5) ∈ H.

This is then covered by the first case, and we are done.

So done. Phew.