Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Structure (mathematical logic) wikipedia , lookup

Birkhoff's representation theorem wikipedia , lookup

Fundamental group wikipedia , lookup

Covering space wikipedia , lookup

Tensor product of modules wikipedia , lookup

Corecursion wikipedia , lookup

Oscillator representation wikipedia , lookup

Transcript
```Groups acting on sets
1
Definition and examples
Most of the groups we encounter are related to some other structure. For
example, groups arising in geometry or physics are often symmetry groups
of a geometric object (such as Dn ) or transformation groups of space (such
as SO3 ). The idea underlying this relationship is that of a group action:
Definition 1.1. Let G be a group and X a set. Then an action of G on X
is a function F : G × X → X, where we write F (g, x) = g · x, satisfying:
1. For all g1 , g2 ∈ G and x ∈ X, g1 · (g2 · x) = (g1 g2 ) · x.
2. For all x ∈ X, 1 · x = x.
When the action F is understood, we say that X is a G-set.
Note that a group action is not the same thing as a binary structure. In
a binary structure, we combine two elements of X to get a third element of
X (we combine two apples and get an apple). In a group action, we combine
an element of G with an element of X to get an element of X (we combine
an apple and an orange and get another orange).
Example 1.2.
1. The trivial action: g · x = x for all g ∈ G and x ∈ X.
2. Sn acts on {1, . . . , n} via σ · k = σ(k) (here we use the definition of
multiplication in Sn as function composition). More generally, if X is
any set, SX acts on X, by the same formula: given σ ∈ SX and x ∈ X,
define σ · x = σ(x). To see that this is an action as we have defined it,
note that, given σ1 , σ2 ∈ SX and x ∈ X,
σ1 · (σ2 · x) = σ1 · (σ2 (x)) = σ1 (σ2 (x)) = (σ1 ◦ σ2 )(x) = (σ1 ◦ σ2 ) · x,
since the group operation on SX is function composition. Clearly
IdX ·x = IdX (x) = x for all x ∈ X. Thus SX acts on X. Note that
1
SX acts on many other objects associated to X, such as the power set
P(X), the set of all subsets of X, by the formula that, for all σ ∈ SX
and A ⊆ X,
σ · A = σ(A) = {σ(a) : a ∈ A}.
Since #(σ·A) = #(A), if A is finite, SX also acts on the subset of P(X)
consisting of all subsets of X with 2 elements, or with 3 elements, or
with k elements for any fixed k.
3. The group R∗ acts on the vector space Rn by scalar multiplication:
given t ∈ R∗ and v ∈ Rn , let t · v = tv ∈ Rn be scalar multiplication.
That this is an action follows from familiar properties of scalar multiplication: t1 (t2 v) = (t1 t2 )v and 1v = v, for all t1 , t2 ∈ R∗ and v ∈ Rn .
(Of course, these properties hold for t1 , t2 ∈ R as well, but R is not a
group under multiplication. Also, scalar multiplication has additional
properties having to do with addition of scalars or vectors.)
4. GLn (R) acts on Rn by the usual rule A · v = Av is the multiplication
of the matrix A on the vector v, and is the same thing as F (v), where
F : Rn → Rn is the linear function corresponding to A. Similarly, On
and SOn act on Rn . In addition, On and SOn act on the (n−1)-sphere
S n−1 of radius 1 defined by
S n−1 = {v ∈ Rn : kvk = 1}.
Note that S 1 = U (1) is the unit circle in R2 and S 2 is the unit sphere
in R3 . An (n − 1)-sphere of radius r is defined similarly.
5. Let Pn be a regular n-gon in R2 , n ≥ 3. For example, we could take
Pn to be centered at the origin and to have vertices
pk = (cos(2πk/n), sin(2πk/n)),
k = 0, 1, . . . , n − 1.
The dihedral group Dn acts Pn and on the set of vertices {p0 , . . . , pn−1 },
as well as on the set of edges {p0 p1 , p1 p2 , . . . , pn−1 p0 }. With the
above notation, it is easy to see (as we have described in various homework problems) that
Dn = {A ∈ O2 : A(Pn ) = Pn }.
6. In a partial analogy with the previous example, let S be a regular solid
(or Platonic solid) in R3 , known to Euclid, Plato, and before Plato to
2
the Pythagoreans. We will not give a precise definition. We view S
as centered at the origin. Here, unlike the case of R2 where there is
a regular n-gon for every n ≥ 3, there are just 5 regular solids. A
regular solid is an example of a polyhedron, which has vertices, edges
and faces. If we tabulate this information, we have the following list of
the regular solids (where v is the number of vertices, e is the number
of edges, and f is the number of faces):
name
tetrahedron
cube
octahedron
dodecahedron
icosahedron
v
4
8
6
20
12
e
6
12
12
30
30
f
4
6
8
12
20
n
3
4
3
5
3
Here n is the number of edges of a face, which is a regular n-gon. This
can be determined by the above data, since each edge meets exactly
two faces, and thus 2e = nf . For example, the faces of a dodecahedron
are regular pentagons. Note Euler’s formula, which in this case says
that
v − e + f = 2.
To every regular solid S there is an associated dual solid S ∨ , where
the number of vertices of S ∨ is equal to the number of faces of S, and
vice versa. Here the tetrahedron is its own dual, while the dual of
the cube is the octahedron and the dual of the dodecahedron is the
icosahedron.
Given a regular solid S, we define its symmetry group G(S) by
G(S) = {A ∈ SO3 : A(S) = S}.
Then G(S) acts on S, and on the sets of vertices, edges, or faces of S.
It is not hard to show that G(S) = G(S ∨ ). Note that (unlike the case
of Dn where we allow elements of O2 ) we only consider elements of
SO3 . The group G(S) is always finite, and we shall say a little more
7. The remaining two examples are more directly connected with group
theory. If G is a group, then G acts on itself by left multiplication:
g · x = gx. The axioms of a group action just become the fact that
multiplication in G is associative (g1 (g2 x) = (g1 g2 )x) and the definition
3
of the identity (1x = x for all x ∈ G). More generally, if H ≤ G is a
subgroup, not necessarily normal, then G acts on the set of left cosets
G/H via: g · (xH) = (gx)H. The argument that this is indeed an
action is similar to the case of left multiplication.
8. G acts on itself by conjugation ig : ig (x) = gxg −1 . (We write it this
way instead of as g · x to avoid confusion with the left multiplication
action.) To see that this is an action, note that, for all g1 , g2 ∈ G,
ig1 (ig2 (x)) = ig1 (g2 xg2−1 ) = g1 (g2 xg2−1 )g1−1 = (g1 g2 )x(g2−1 g1−1 )
= (g1 g2 )x(g1 g2 )−1 = ig1 g2 (x),
where we have used the familiar fact that (g1 g2 )−1 = g2−1 g1−1 . Since
clearly
i1 (x) = 1x1−1 = 1x1 = x,
conjugation does give an action of G on itself. This action is the trivial
action ⇐⇒ gxg −1 = x for all g, x ∈ G ⇐⇒ gx = xg for all g, x ∈ G
⇐⇒ G is abelian.
One principle that we have seen implicitly in some of the above examples
is the following:
Proposition 1.3. If X is a G-set and f : G0 → G is a homomorphism, then
X becomes a G0 -set via g 0 · x = f (g 0 ) · x. In particular, if H ≤ G, then a
G-set X is also an H-set via the inclusion homomorphism of H in G.
Proof. Given g10 , g20 ∈ G0 ,
g10 · (g20 · x) = g10 · (f (g20 ) · x) = f (g10 ) · (f (g20 ) · x) = ((f (g10 )f (g20 )) · x
= f (g10 g20 ) · x = (g10 g20 ) · x,
using the fact that f is a homomorphism. Also, if 10 is the identity in G0 ,
then f (10 ) = 1 is the identity in G, and thus, for all x ∈ X,
10 · x = f (10 ) · x = 1 · x = x.
It follows that the formula g 0 · x = f (g 0 ) · x defines an action of G0 on X.
Example 1.4. If G is a group, then SG acts on G and on the set P(G) of
all subsets of G. Thus, so does Aut G, the group of automorphisms of G (i.e.
isomorphisms from G to G), which is a subgroup of G under composition.
Note that Aut G also acts on the set of all subgroups of G, which is a
subset of P(G), whereas SG does not act on this set (because a bijection
from G to itself will not in general take a subgroup to a subgroup).
4
Using Proposition 1.3, we can give a partial generalization of Cayley’s
theorem. Recall that, for the action of G on itself by left multiplication, we
define a bijection `g : G → G by:
`g (x) = gx.
More generally, let G act on a set X, and define `g : X → X by the formula
`g (x) = g · x.
Lemma 1.5. With notation as above,
(i) For all g1 , g2 ∈ G, `g1 ◦ `g2 = `g1 g2 .
(ii) `1 = IdX .
(iii) For all g ∈ G, `g is a bijection from X to X, i.e. `g ∈ SX for all
g ∈ G, and the inverse of `g is `g−1 .
Proof. (i) We must check that, for all x ∈ X, `g1 ◦ `g2 (x) = `g1 g2 (x). By
definition,
`g1 ◦ `g2 (x) = `g1 (`g2 (x)) = `g1 (g2 · x)
= g1 · (g2 · x) = (g1 g2 ) · x = `g1 g2 (x).
(ii) Clearly, for all x ∈ X, `1 (x) = 1 · x = x, and thus `1 = IdX .
(iii) It is enough to prove that (`g )−1 = `g−1 , i.e. that `g ◦ `g−1 = `g−1 ◦ `g =
IdX . Using (i) and (ii),
`g ◦ `g−1 = `gg−1 = `1 = IdX ,
and similarly `g−1 ◦ `g = IdX .
Note in particular that, if y = g · x, then x = g −1 · y.
Corollary 1.6. If X is a G-set, then the function f : G → SX defined by
f (g) = `g is a homomorphism from G to SX .
Proof. By (iii) above, `g ∈ SX . The equation `g1 ◦ `g2 = `g1 g2 says that
f (g1 g2 ) = f (g1 ) ◦ f (g2 ), in other words that f is a homomorphism.
Remark 1.7. For the left multiplication action of G on itself, the homomorphism f : G → SG is easily seen to be injective; this is the content of
Cayley’s theorem. In general, though, f need not be injective. For example,
if G acts on X by the trivial action g · x = x for all g ∈ G, then `g = IdX
for all g ∈ G, so that f is the trivial homomorphism.
5
We can also reverse the construction of the corollary: Given a homomorphism f : G → SX , since SX acts on X, X becomes a G-set by Proposition 1.3. Finally, the two constructions just described (passing from an
action of G on X to a homomorphism G → SX , and passing from a homomorphism G → SX to an action of G on X) are inverse constructions. Thus,
the concept of a G-set X is equivalent to the concept of a homomorphism
G → SX .
Example 1.8. If G is a group, the Aut G is a subgroup of SG , and thus
Aut G acts on G. We have the conjugation homomorphism f : G → Aut G
defined by f (g) = ig , where as usual ig (x) = gxg −1 . The composition G →
Aut G → SG is the same thing as the action of G on itself by conjugation.
Note that G acts on the set of all subgroups of G by conjugation, since
Aut G does (Example 1.4).
Definition 1.9. If X is a G-set, then a G-subset Y of X is a subset Y ⊆ X
such that, for all g ∈ G and y ∈ Y , g · y ∈ Y . A G-subset is itself a G-set.
Definition 1.10. If X1 and X2 are G-sets, an isomorphism f from X1 to
X2 of G-sets, or briefly a G-isomorphism is a bijection f : X1 → X2 such
that f (g · x) = g · f (x) for all g ∈ G and x ∈ X. In this case we say that
X1 and X2 are are isomorphic as G-sets or G-isomorphic, and write this as
X1 ∼
=G X2 . Clearly IdX is an isomorphism of G-sets. If f : X1 → X2 is
an isomorphism of G-sets, then so is f −1 . Likewise the composition of two
isomorphisms is again an isomorphism. Thus, as with the usual definition
of isomorphism, the relation ∼
=G is reflexive, symmetric, and transitive.
2
Orbits and isotropy subgroups
Definition 2.1. If X is a G-set and x ∈ X, the orbit of X (under G) is the
set G · x = {g · x : g ∈ G}. Thus G · x ⊆ X. Clearly G · x is a G-subset of
X and is the smallest G-subset of X containing x.
Example 2.2. In the case of the action of Sn on {1, . . . , n}, given σ ∈ Sn ,
we have previously defined the orbits Oσ (i) of σ. The link with the current
definition is as follows: the orbits of σ in the previous sense are the orbits of
hσi acting on {1, . . . , n} as a subgroup of Sn . In other words, Oσ (i) = hσi · i.
In fact, both sides are equal to
{σ a (i) : a ∈ Z}.
6
We defined the orbits Oσ (i) by an equivalence relation, and so it is
natural to try to do the same thing for the orbit G · x.
Proposition 2.3. Let G act on a set X, and define x ∼G y ⇐⇒ there
exists a g ∈ G such that g · x = y. Then ∼G is an equivalence relation, and
the equivalence class containing x is the orbit G · x. Thus, two orbits of G
are either disjoint or identical.
Proof. Reflexive: x = 1 · x, hence x ∼G x. Symmetric: if x ∼G y, then by
definition there exists a g ∈ G such that g · x = y. We have seen that, ion
this case, g −1 · y = x. Thus y ∼G x. Transitive: Say that x ∼G y and that
y ∼G z. Then there exists a g1 ∈ G such that g1 · x = y and there exists a
g2 ∈ G such that g2 · y = z. Thus z = g2 · y = g2 · (g1 · x) = (g1 g2 ) · x, and
so x ∼G z.
The remaining statements, that the equivalence class containing x is the
orbit G · x and that two orbits of G are either disjoint or identical, are then
clear by definition and from general properties of equivalence classes.
Definition 2.4. If X is a G-set and if G · x = X for one (or equivalently
all) x ∈ X, we say that G acts transitively on X.
Example 2.5. (1) Sn acts transitively on {1, . . . , n}. This just says that,
for all k ∈ {1, . . . , n}, there exists a σ ∈ Sn such that σ(1) = k, hence
Sn · 1 = {1, . . . , n} and there is just one orbit. Likewise, it is easy to see
that An acts transitively on Sn for n ≥ 3, but not for n = 2. But if σ ∈ Sn ,
then the subgroup hσi acts transitively on {1, . . . , n} ⇐⇒ there is just one
orbit of σ and it has n elements ⇐⇒ σ is an n-cycle.
(2) GLn (R) acts on Rn . There are two orbits: {0} and Rn − {0}. SOn
acts on Rn . For n ≥ 2, the orbits are {0} and the (n − 1)-spheres of
radius r > 0 centered at the origin. In particular, for n ≥ 2, SOn acts
transitively on S n−1 . Because this action is transitive, the geometry of
S n−1 is homogeneous, i.e. it looks the same at every point.
(3) The dihedral group Dn acts transitively on {p0 , . . . , pn−1 }, the set of
vertices of Pn .
(4) The group G acts on itself by left multiplication. This action is transitive,
since for example the orbit G · 1 is clearly G.
(5) The group G acts on itself by conjugation. The orbit of x ∈ G is
the conjugacy class of x, the subset C(x) of G consisting of all elements
conjugate to x. Thus by definition
C(x) = {gxg −1 : g ∈ G}.
7
For example, C(1) = {1}, so that the conjugation action is never transitive
as long as G is not the trivial group.
Definition 2.6. If X is a G-set and x ∈ X, the isotropy subgroup Gx is
the set {g ∈ G : g · x = x}.
Lemma 2.7. Gx is a subgroup of G.
Proof. Closure: if g1 , g2 ∈ Gx , then
(g1 g2 ) · x = g1 · (g2 · x) = g1 · x = x.
Identity: as 1·x = x, 1 ∈ Gx for every x. Inverses: if g ∈ Gx , then g·x = x by
defintion. As we have seen, g −1 · x = x, hence g −1 ∈ Gx . Thus Gx ≤ G.
Definition 2.8. If X is a G-set, then the fixed set X G is the set {x ∈
X : g · x = x for all g ∈ G}. It is the largest G-subset of X for which the
G-action is trivial. Clearly x ∈ X G ⇐⇒ Gx = G ⇐⇒ G · x = {x} ⇐⇒
the orbit G · x contains exactly one element.
Proposition 2.9. (i) If X is a G-set, x ∈ X, and y = g · x ∈ G · x, then
Gy = gGx g −1 . In other words, the isotropy groups of x and y are conjugate
by g.
(ii) If X is a G-set and x ∈ X, then there is an isomorphism of G-sets from
G · x to G/Gx , where G acts on the set of left cosets of Gx in the usual way
(by left multiplication of cosets).
Proof. (i) Given h ∈ G, h · y = y ⇐⇒ h · (g · x) = g · x ⇐⇒ (hg) · x = g · x
⇐⇒ g −1 · ((hg) · x) = x ⇐⇒ (g −1 hg) · x = x ⇐⇒ g −1 hg ∈ Gx ⇐⇒
h ∈ gGx g −1 . Thus Gy = gGx g −1 .
(ii) It is simplest to define a function F : G/Gx → G · x and prove first that
it is a bijection and then that it is an isomorphism of G-sets. Given a coset
gGx , define F (gGx ) = g · x (note that this is indeed an element of G · x). We
must show that F is well-defined, i.e. independent of the representative g of
the coset gGx . Any other element of gGx is of the form gh with h ∈ Gx ,
and hence
(gh) · x = g · (h · x) = g · x,
since by the definition of Gx h · x = x. Hence F is well-defined and it is
surjective by the definition of G · x. Next, we claim that F is injective.
Suppose that F (g1 Gx ) = F (g2 Gx ). Then by definition g1 · x = g2 · x, so that
x = g1−1 · (g1 · x) = g1−1 · (g2 · x) = (g1−1 g2 ) · x.
8
Thus g1−1 g2 ∈ Gx , and so g1 Gx = g2 Gx . Thus F is injective, hence a
bijection. Finally, we must check that F is an isomorphism of G-sets. This
follows since, for all g ∈ G and cosets hGx ∈ G/Gx , by the definition of the
action of G on G/Gx ,
F (g · (hGx )) = F ((gh)Gx ) = (gh) · x = g · (h · x) = g · F (hGx ).
Thus F is an isomorphism of G-sets.
Corollary 2.10. Suppose that G is finite. Let X be a G-set and let x ∈ X.
Then
#(G) = #(Gx ) · #(G · x).
Equivalently,
(G : Gx ) = #(G)/#(Gx ) = #(G · x).
Hence the order of an orbit of G in X divides the order of G. In particular,
if G acts transitively on X, then #(G) = #(Gx ) · #(X), or equivalently
#(X) = (G : Gx ).
Example 2.11. (1) Sn acts transitively on {1, . . . , n} and the isotropy subgroup of n is
Hn = {σ ∈ Sn : σ(n) = n} ∼
= Sn−1 .
Thus Sn /Hn is Sn -isomorphic to {1, . . . , n}. Note that #(Sn ) = n!, #(Hn ) =
#(Sn−1 ) = (n − 1)!, and #({1, . . . , n}) = n = n!/(n − 1)! = #(Sn )/#(Hn ).
(2) SOn acts transitively on S n−1 and the isotropy subgroup of en is easily
seen to be SOn−1 . Thus SOn /SOn−1 is SOn -isomorphic to S n−1 . In topology, this is an important relationship between the (n − 1)-sphere and the
group SOn .
(3) Dn acts transitively on the set of vertices {p0 , . . . , pn−1 } of Pn and the
isotropy subgroup of p0 , say, is the reflection in p0 . This gives another
argument that #(Dn ) = 2n.
(4) Let S be the dodecahedron and G(S) the group of symmetries of S. By
experimenting with a model of S, it is plausible that G(S) acts transitively
on the 12 faces of S and that the isotropy group of a face has order 5,
corresponding to the possible rotations of a pentagon. Thus we expect that
#(G(S)) = 60. We would get a similar conclusion by looking at the action
of G(S) on the 20 vertices, where the order of the isotropy group is 3, or on
the 30 edges, where the order of the isotropy group is 2.
9
(5) If H is a subgroup of G, then G acts transitively on G/H since the orbit
G · H = G/H. The isotropy subgroup of H is by definition
{g ∈ G : gH = H} = H.
The isotropy subgroup of xH is
{g ∈ G : gxH = xH} = xHx−1 ,
since gxH = xH ⇐⇒ x−1 gx ∈ H ⇐⇒ g ∈ xHx−1 .
(6) If G acts on itself by conjugation, then the fixed set GG is the center
Z(G), the orbit of x ∈ G is, as we have seen, the conjugacy class C(x) =
{gxg −1 : g ∈ G}, and the isotropy group of x is the centralizer of x, namely
the subgroup
ZG (x) = {g ∈ G : gxg −1 = x}.
3
The class equation and applications
Suppose X is a finite G-set with orbits O1 = G·x1 , . . . , Ok = G·xk (for some
choices of x1 , . . . , xk ∈ X). Then, since every element of X is in exactly one
G-orbit,
k
X
#(X) =
#(Oi ).
i=1
We rewrite this by grouping together the one-element orbits into X G , as
X
#(X) = #(X G ) +
#(Oi ),
#(Oi )>1
where the second sum is over all of the orbits which have more than one
element. Note that, if G is finite and if #(Oi ) > 1, then #(Oi ) is a nontrivial
divisor of #(G), i.e. #(Oi ) divides #(G) and #(Oi ) 6= 1.
In particular, if G is a finite group and we let G act on itself by conjugation, then the fixed set GG is just the center of G and the orbits G · xi are
the conjugacy classes
C(xi ) = {gxi g −1 : g ∈ G}.
The isotropy subgroup of xi ∈ G for the G-action of conjugation is the
centralizer ZG (xi ), and #(C(xi )) = (G : ZG (xi )), the index of the centralizer
of xi in the group G. Thus we get the class equation:
X
X
#(G) = #(Z(G)) +
#(C(xi )) = #(Z(G)) +
(G : ZG (xi )),
i
i
10
where the sum is over the distinct conjugacy classes C(xi ) which have more
than one element (i.e. for which xi ∈
/ Z(G)).
Example 3.1. In Sn , the conjugacy classes are described by the “shapes”
of a product of disjoint cycles. In other words, given σ ∈ Sn , there exist
integers k1 , . . . , kr ≥ 2 with k1 + · · · + kr ≤ n such that σ is a product
of disjoint cycles of lengths k1 , . . . , kr , and every two such products with
the same k1 , . . . , kr are conjugate. Here 1 is the empty product. Thus, for
example in S3 , there are three conjugacy classes:
C(1) = {1}, C((1, 2)) = {(1, 2), (1, 3), (2, 3)}, C((1, 2, 3)) = {(1, 2, 3), (1, 3, 2)}.
Note that the sum of all the elements is #(S3 ) = 6, and that the order of
each conjugacy class divides 6. The situation in S4 is already much more
complicated. Here, C(1) = {1} has just one element. The conjugacy
class of
a transposition is the set of all transpositions, and there are 42 = 6 of these.
The conjugacy class of a product of two disjoint transpositions is the set of
of two disjoint transpositions, and we have seen that there are
1 4
=
3
of
these.
The conjugacy class of a 3-cycle is the set of all 3-cycles,
2 2
and there are (4 · 3 · 2)/3 = 8 of these. The conjugacy class of a 4-cycle is
the set of all 4-cycles, and there are (4 · 3 · 2)/4 = 6 of these. Thus, the order
of each conjugacy class divides 24 = #(S4 ), and the total number is
1 + 6 + 3 + 8 + 6 = 24.
Returning to the case of a general action of a finite group G on a finite
set X, we have:
Proposition 3.2. If #(G) = pr , where p is a prime, and if X is a finite
G-set, then
#(X) ≡ #(X G ) (mod p).
P
Proof. In the formula #(X) = #(X G ) + #(Oi )>1 #(Oi ) above, all of the
terms #(Oi ) for #(Oi ) > 1 are nontrivial divisors of pr , hence are of the
form ps for 1 ≤ s ≤ r. Thus, if #(Oi ) > 1, then #(Oi ) ≡ 0 (mod p) and
hence #(X) ≡ #(X G ) (mod p).
Corollary 3.3. Let p be a prime number. If #(G) = pr with r ≥ 1, then
Z(G) 6= {1}. In particular, if #(G) > p, then G is not simple (and hence,
by induction, solvable).
11
Proof. In this case, we let X = G with the conjugation action. Then the
fixed set X G is just the center of G, so that
#(Z(G)) ≡ #(G) = pr ≡ 0
(mod p).
Thus Z(G) is a subgroup of G with order divisible by p, and hence Z(G) 6=
{1}. We have seen that Z(G)CG. Thus, either Z(G) is a proper (nontrivial)
subgroup of G, or Z(G) = G. In this last case, G is abelian and every
subgroup of G is normal, so that G is simple ⇐⇒ there are no proper
nontrivial subgroups of G ⇐⇒ G is isomorphic to Z/pZ (by a homework
problem).
Corollary 3.4. Let p be a prime number. If #(G) = p2 , then G is abelian.
Proof. By the preceding corollary, the center Z(G) is a subgroup of G with
#(Z(G)) = p or #(Z(G)) = p2 . If #(Z(G)) = p2 , then G = Z(G) and
hence G is abelian. So let us assume that #(Z(G)) = p. We will show
that in this case G is also abelian, a contradiction since then Z(G) = G. If
#(Z(G)) = p, then Z(G) C G and G/Z(G) is a group of order p. But every
group of order p is cyclic, so in particular G/Z(G) is a cyclic group. By a
homework problem, G is then abelian.
Remark 3.5. In fact, it is easy to show that, if #(G) = p2 , then either
G∼
= Z/p2 Z is cyclic or G ∼
= (Z/pZ) × (Z/pZ).
12
```