1Groups

IB Groups, Rings and Modules



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
1
= gg
1
g
1
.
Let
φ
:
G Sym
(
G
) be the associated permutation representation. We know,
by definition, that
φ
(
g
) is a bijection from
G
to
G
as sets. However, here
G
is
not an arbitrary set, but is a group. A natural question to ask is whether
φ
(
g
)
is a homomorphism or not. Indeed, we have
φ(g)(g
1
· g
2
) = gg
1
g
2
g
1
= (gg
1
g
1
)(gg
2
g
1
) = φ(g)(g
1
)φ(g)(g
2
).
So
φ
(
g
) is a homomorphism from
G
to
G
. Since
φ
(
g
) is bijective (as in any
group action), it is in fact an isomorphism.
Thus, for any group
G
, there are many isomorphisms from
G
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
G
, 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
(
G
), and the homomorphism
φ
:
G Sym
(
G
) 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
(g) = {hgh
1
: h G},
i.e. the orbit of g G under the conjugation action.
Definition (Centralizer). The centralizer of g G is
C
G
(g) = {h G : hgh
1
= g},
i.e. the stabilizer of
g
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} =
\
gG
C
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
G
(x).
Proposition. Let G be a finite group. Then
|ccl(x)| = |G : C
G
(x)| = |G|/|C
G
(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
N
G
(H) = {g G : g
1
Hg = H}.
Note that we certainly have
H N
G
(
H
). Even better,
HCN
G
(
H
), 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
S
n
. Now we recall from IA
Groups that permutations in
S
n
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
n
. For example, the partition 2
,
2
,
1 of 5 corresponds
to the conjugacy class of (1 2)(3 4)(5). So the conjugacy classes of
S
n
are exactly
the partitions of n.
We will use this fact in the proof of the following theorem:
Theorem. The alternating groups
A
n
are simple for
n
5 (also for
n
= 1
,
2
,
3).
The cases in brackets follow from a direct check since
A
1
=
A
2
=
{e}
and
A
3
=
C
3
, all of which are simple. We can also check manually that
A
4
has
non-trivial normal subgroups, and hence not simple.
Recall we also proved that
A
5
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
n
is generated by 3-cycles.
As any element of
A
n
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
n
. If H contains a 3-cycle, then we H = A
n
.
We show that if
H
contains a 3-cycle, then every 3-cycle is in
H
. Then we
are done since
A
n
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
n
such that (
a b c
) =
σ
(1 2 3)
σ
1
. If
σ
is even, i.e.
σ A
n
, then we have that (1 2 3)
σ
1
Hσ
=
H
,
by the normality of H and we are trivially done.
If
σ
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
as
above.
What we’ve got so far is that if
H C A
n
contains any 3-cycle, then it is
A
n
.
Finally, we have to show that every normal subgroup must contain at least one
3-cycle.
Claim. Let H C A
n
be non-trivial. Then H contains a 3-cycle.
We separate this into many cases
(i)
Suppose
H
contains an element which can be written in disjoint cycle
notation
σ = (1 2 3 ···r)τ,
for
r
4. We now let
δ
= (1 2 3)
A
n
. Then by normality of
H
, we know
δ
1
σδ H
. Then
σ
1
δ
1
σδ H
. Also, we notice that
τ
does not contain
1
,
2
,
3. So it commutes with
δ
, and also trivially with (1 2 3
··· r
). We
can expand this mess to obtain
σ
1
δ
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
σ
= (
a
1
a
2
··· a
r
)
τ
for any
a
1
, ··· , a
n
.
(ii)
Suppose
H
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
σ
1
δ
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
H
. By the previous case, we get a
3-cycle in H too, and hence H = A
n
.
(iii)
Suppose
H
contains
σ
= (1 2 3)
τ
, with
τ
a product of 2-cycles (if
τ
contains
anything longer, then it would fit in one of the previous two cases). Then
σ
2
= (1 2 3)
2
= (1 3 2) is a three-cycle.
(iv)
Suppose
H
contains
σ
= (1 2)(3 4)
τ
, where
τ
is a product of 2-cycles. We
first let δ = (1 2 3) and calculate
u = σ
1
δ
1
σδ = (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3) = (1 4)(2 3),
which is again in
u
. 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.