GROUP ACTIONS AND THE SYLOW THEOREMS
MICHAEL PENKAVA
1. Group Actions
Definition 1.1. Let G be a group and X be a set. Then G is said to act on X if
there is a map G × X → X, (g, x) 7→ g.x, satisfying
(1) (gh).x = g.(h.x).
(2) e.x = x.
A set X equipped with an action of the group G is called a G-set.
There are other notations for group actions. It is common to see g ⋆ x instead of
g.x. When there is no ambiguity, the notation gx instead of g.x is commonly used.
Example 1.2. If G is a group, there are two natural actions of G on itself. The
first action is multiplication, given by g.x = gx for g, x ∈ G. The second action is
conjugation, given by g.x = gxg −1 .
Exercise 1.3. Verify that multiplication and conjugation satisfy the axioms of a
group action.
Proposition 1.4. If G acts on X, then the map α : G → SX , given by αg (x) = g.x
is a morphism of G to the group SX of permutations of X. Conversely, if α : G →
SX is a morphism of groups, then the map G × X → X, given by g.x = αg (x) is a
group action.
Exercise 1.5. Prove the above proposition.
A group action, as we have defined it, is also called a left group action. A right
group action of G on X is defined similarly, as a map X × G → X, given by
(x, g) 7→ x.g, satisfying x.(gh) = (x.g).h and x.e = x.
Proposition 1.6. A left action of G on X induces a right action by x.g = g −1 .x.
Similarly, a right action determines a left action by g.x = x.g −1 .
Example 1.7. Prove the above proposition.
Definition 1.8. If G acts on X and x ∈ X, then the G-orbit of x or just the orbit
of x is the set Ox = {g.x|g ∈ G}. This set is often denoted by Gx. If there is is
only one orbit, then G is said to act transitively on X.
The set StabG (x) = {g ∈ G|g.x = x} is called the G-stabilizer of x, or just the
stabilizer of x. This set is often denoted as Gx , or Stab(x).
Proposition 1.9. If G acts on X and x ∈ X, then G acts on Ox . This restricted
action is transitive.
Theorem 1.10 (Fundamental Theorem of Transitive Group Actions). Suppose
that G acts transitively on X and x ∈ X. Then the map G/ Stab(x) → X given by
ḡ 7→ g.x is a well defined bijection. Thus |X| = [G : Stab(x)]. If G is finite, then
|X| = |G|/| Stab(x)|.
1
2
MICHAEL PENKAVA
Proof. To show the map is well defined, suppose that g ′ ∈ ḡ. Then g ′ = gh where
h ∈ Stab(x). Thus g ′ .x = (gh).x = g.(h.x) = g.x. This shows the map ḡ → X
is well defined. Let y ∈ X. Then y = g.x for some g ∈ G, since the action is
transitive, which shows the map is surjective. Finally, suppose that ā and b̄ have
the same image. Then a.x = b.x, so
(b−1 a).x = b−1 .(a.x) = b−1 .(b.x) = (b−1 b).x = e.x = x.
This shows that b−1 a ∈ Stab(x), so b ∈ ā. It follows that ā = b̄. This shows that
the map is injective. The other statements follow from the fact that |G/ Stab(x)| =
[G : Stab(x)] = |/| Stab(x)|, the latter equality holding when |G| < ∞.
Definition 1.11. A finite group with order pn for a prime p and positive integer
n is called a p-group
Definition 1.12. If G acts on X, then denote the subset of X consisting of all
points in X which are fixed by every element in G by X0 . This set is also often
denoted as Fix(G).
Proposition 1.13. Let G be a finite p-group acting on a finite set X. Then
|X0 | = |X| mod p.
Proof. Let S be the set of orbits which are not singletons. Then X is a disjoint
union of X0 and the collection S. If Ox ∈ S, then |Ox | = [G : Stab(x)] is divisible
by p. Thus p divides the order of the union of the elements in S. It follows that
|X| = |X0 | mod p.
Definition 1.14. If G is a group and a ∈ G, then the centralizer of a in G, denoted
by CG (a) or C(a), is the set C(a) = {g ∈ G|gag −1 = a}. The center of G, denoted
by Z(G) is the set of elements of G which commute with every element in G.
Proposition 1.15. If a ∈ G, then C(a) is a subgroup of G.
Theorem 1.16 (Class Equation). Let G be a finite group. Let S be a set of representatives for the conjugacy classes of G which do not consist of a single element.
Then
∑
G = |Z(G)| +
[G : C(x)].
x∈S
Corollary 1.17. Let G be a finite p-group. Then the center of G is nontrivial.
Corollary 1.18. Let G be a group of order p2 . Then G is abelian.
Proof. Since Z(G) is nontrivial, either |Z(G)| = p or |Z(G)| = p2 . If the latter case
holds, then G is abelian, so assume the former. Then |G/Z(G)| = p, so G/Z(G) is
isomorphic to Zp . However, by a theorem from Abstract Algebra I, if G/Z(G) is
cyclic, then G is abelian. This shows that it is impossible for Z(G) to have order
p.
Proposition 1.19. Let X be a set. Then the symmetric group Sn acts on X n by
σ.(x1 , . . . , xn ) = (xσ−1 (1) , . . . xσ−1 (n) ).
Proof. ( Let σ, τ ∈ Sn , and x = (x1 , . . . , xn ) ∈ X n . Let y = (y1 , . . . , yn ) = τ.x, so
that yi = xτ −1 (i) . Then σ.(τ.x) = z, where zi = yσ−1 (i) = xτ −1 (σ−1 (i)) = x(στ )−1 (i) .
But this means that z = (στ ).x, which shows that σ.(τ.x) = (στ ).x. Since it is
immediate to see that e.x = x, we see that Sn act on X n .
GROUP ACTIONS AND THE SYLOW THEOREMS
3
Corollary 1.20. Let X be a set. Then there is an action of Zn on X n which
is completely determined by the requirement that if 1.x = (x2 , . . . , xn , x1 ), where
x = (x1 , . . . xn ).
Proof. Let σ = (n, n − 1, . . . , 1). Then σ is an n-cycle, so o(σ) = n and ⟨σ⟩ ∼
= Zn .
We have σ −1 = (1, 2, . . . , n), so σ.x = (xσ−1 (1) , . . . , xσ−1 (n) ) = (x2 , . . . , xn , x1 ). Theorem 1.21 (Cauchy’s Theorem). Let G be a finite group and p be a prime
dividing the order of G. Then G has an element of order p.
Proof. Zp acts on X = {x ∈ Gp |x1 x2 · · · xp = e} by 1.(x1 , . . . xp ) = (x2 , . . . , xp , x1 ).
Now x ∈ X0 precisely when x1 = · · · = xp . The map Gp−1 → X, given by
−1
(x1 , . . . xp−1 ) 7→ (x1 , . . . xp−1 , x−1
p−1 · · · x1 ), is bijective, so |X| = 0 mod p. It
follows that |X0 | = 0 mod p. Now (e, . . . , e) ∈ X0 , so there must be at least p
elements in X0 . Let x be an element in X0 which is not (e, . . . , e). Then xp1 = e,
and x1 ̸= e, so o(x1 ) = p. Thus G has an element of order p.
Proposition 1.22. Let H ≤ G. Then G acts on G/H by
xḡ = xg.
Proposition 1.23. Let K, H ≤ G and consider the restriction of the action of G
on G/H to K. Then ḡ ∈ Fix(K) iff g −1 Kg ⊆ H.
Proof. We have
ḡ ∈ Fix(K) ⇐⇒ kg = ḡ for all k ∈ K
⇐⇒ kg = ḡ for all k ∈ K
⇐⇒ g −1 kg = g −1 ḡ for all k ∈ K
⇐⇒ g −1 kg = ē for all k ∈ K
⇐⇒ g −1 kg ∈ H for all k ∈ K
⇐⇒ g −1 Kg ⊆ H
Proposition 1.24. Let H ≤ G, and π : G → G/H be the projection π(g) = ḡ. If
K ≤ G, then π −1 (π(K)) = KH. In particular, if H ≤ K, then π(K) = K/H and
π −1 (K/H) = K.
Proof. We have
g ∈ π −1 (π(K)) ⇐⇒ π(g) ∈ π(K)
⇐⇒ ḡ = k̄ for some k ∈ K
⇐⇒ g = kh for some k ∈ K
⇐⇒ g ∈ KH,
which shows the first statement. If H ≤ K, then KH = K. Moreover, the definition
of K/H and π(K) coincide.
Proposition 1.25. Let H be a p-subgroup of G for some prime p. Then
[G : H] = [NG (H) : H]
mod p.
4
MICHAEL PENKAVA
Proof. Let H act on G/H. Then ḡ ∈ Stab(H) precisely when g −1 Hg ⊆ H, in other
words, when g ∈ NG (H). Thus we have
[NG (H) : H] = o(NG (H)/H) = o(G/H) = [G : H]
because H is a p-group.
mod p,
Theorem 1.26 (First Sylow Theorem). Suppose that pm+1 |o(G) and o(H) = pm .
Then there is some K ≤ G such that H ▹ K and o(K) = pm+1 . In particular, if pn
is the largest power of p dividing o(G), then there is a subgroup of G with order pn .
Proof. Since o(G/H) is divisible by p, so is o(NG (H)/H). Since NG (H)/H is a
group whose order is divisible by p, it has a subgroup K̄ of order p. Let K =
π −1 (K̄), where π : NG (H) → NG (H)/H is the projection. But then o(K) =
po(H) = pm+1 . By construction, K ≤ NG (H) and H ≤ K, so H ▹ K.
Definition 1.27. If pn is the largest power of p which divides o(G), then a subgroup
of order pn is called a p-Sylow subgroup of G.
Theorem 1.28 (Second Sylow Theorem). The p-Sylow subgroups of G are all
conjugate.
Proof. Let H, K be two p-Sylow subgroups of G, and consider the action of K on
G/H by kḡ = kg. Since p does not divide NG (H)/H, it follows that there must
be some element ḡ which is fixed under this action. But then g −1 Kg ⊆ H. Since
H and K have the same order, it follows that g −1 Kg = H. Thus H and K are
conjugate.
Theorem 1.29 (Third Sylow Theorem). Let k be the number of p-Sylow subgroups
of G. Then k|o(G) and k = 1 mod p.
Proof. Since the set X of p-Sylow subgroups of G are exactly the set of subgroups
of G of order pn , where pn is the largest power of p dividing o(G), they are an orbit
under the action of G on its subgroups by conjugation. This implies that k|o(G).
Let H be a p-Sylow subgroup of G. Then H also acts on X by conjugation. We
no longer can assume that there is a single orbit. Instead, consider X0 = Stab(H).
Since conjugation of H by elements in H leaves H fixed, we know that H ∈ Stab(H).
Suppose that K ∈ Stab(H). Then conjugation of K by any element in H fixes K,
which implies that H ≤ NG (H). But then both H and K are subgroups of NG (K),
and they must be p-Sylow subgroups of NG (K). Therefore H and K must be
conjugate as subgroups of NG (K). Since K ▹ NG (K), it is only conjugate to itself.
Thus H = K. This shows Stab(H) = {H}. Thus 1 = |X0 | = |X| mod p.
GROUP ACTIONS AND THE SYLOW THEOREMS
5
2. Applications of the Sylow Theorems
Example 2.1. Suppose that o(G) = 6. Let us compute the number of 2-Sylow and
3-Sylow subgroups of G. Let k be the number of 3-Sylow subgroups, so k|6, and thus
k is 1, 2, 3 or 6. We can eliminate the multiples of 3, leaving 1 and 2. Since 2 ̸= 1
mod 3, we must have k = 1. In particular, if H is a 3-Sylow subgroup, it must be
normal in G. Note, we could have seen this by using the fact that [G : H] = 1.
Next, let k be the number of 2-Sylow subgroups. Eliminating the multiples of 2,
we have k is either 1, or 3. Since both 3 and 1 are equal to 1 mod 2, we cannot
determine which is correct. Let K be a 2-Sylow subgroup of G.
Since H is normal, and H ∩ K can only consist of the identity, and 32̇ = 6,
we know that G = H o K. What are the possible actions of Z2 on Z3 ? Well,
Aut(Z3 ) = Z2 , so there are exactly 2 morphisms from Z2 to Aut(Z3 ), the trivial
morphism, and the identity morphism.
The trivial morphism always determines the direct product structure on H × K,
which is Z3 × Z3 = Z6 . In this case, there is exactly one subgroup of order 2.
The identity morphism determines the group structure D3 . In this case there
are 3 subgroups of order 2. Thus both possibilities given by the Sylow theorems do
occur!
Example 2.2. Let G be a group of order 4. By Cauchy’s Theorem, it has a
subgroup H of order 2. Now, if G has an element of order 4, it is isomorphic to
Z4 , so assume otherwise. Then there must be an element not in H of order 2. Let
K be the subgroup generated by this element. One can use the index to show that
H and K are normal, or one can use the Sylow theorem, which says that H must
be normal in a subgroup of order 22 and similarly for K. Now, H ∩ K = {e}, so we
must have G = H × K, because both H and K are normal, and o(H)o(K) = o(G).
Theorem 2.3. Let H, K ≤ G. Then
|HK| =
|H| |K|
.
|H ∩ K|
As a consequence,
|H ∩ K| ≥
|H| |K|
.
|G|
Proof. Let H × K act on HK by (h, k) ∗ x = hxk −1 . To check that this is a group
action, note that
(h1 , k1 ) ∗ (h2 , k2 ) ∗ x = h1 (h2 xk2−1 )k1−1 = (h1 h2 )x(k1 k2 )−1
= (h1 h2 , k1 k2 ) ∗ x = ((h1 , k1 )(h2 , k2 )) ∗ x.
(e, e) ∗ x = exe−1 = x.
Note that e = ee ∈ HK, and (h, k −1 ) ∗ e = hk, so the action is transitive. Moreover
Stab(e) = {(h, k) ∈ H × K|hk −1 = e}
= {(h, h)|h ∈ H ∩ K},
So | Stab(e)| = |H ∩ K|. By the fundamental theorem of group actions, we obtain
|H×K|
|K|
that |HK| = Stab(e)
= |H|
|H∩K| . Now use the fact that |HK| ≤ |G| and rearrange
the inequality |G| ≥
|H| |K|
|H∩K|
to obtain the second statement!
6
MICHAEL PENKAVA
Example 2.4. Let G be a group of order 8. If G has an element of order 8, then
it is isomorphic to Z8 , so assume otherwise. We know that G has an element of
order 2, and thus a subgroup of order 2, which is normal in a subgroup of order 4.
In particular, G has a subgroup H of order 4, which is normal in G, either by the
index theorem or the Sylow theory.
If every element of G has order 2, then G must be abelian. In this case, we must
have H ∼
= Z2 × Z2 , and there is an element not in H, which must give a normal
subgroup K or order 2, Thus G = H × K = Z32 .
Thus we can assume that G has an element of order 4, which generates a subgroup
H of order 4, which is normal in G. If there is an element of order 2 which is not in
H, let K be the subgroup it generates. We have H ∩ K = {e}, and H is normal in
G, so G = H o K. Now Aut(H) = Z∗4 = {1, 3} ∼
= Z2 . Thus there are precisely two
morphisms from K to Aut(H), the trivial morphism and the identity morphism.
For the trivial morphism, the group is H × K = Z4 × Z2 , and for the identity
morphism, we obtain the group D4 .
Finally, suppose that H ∼
= Z4 , and there is no element of order 2 not in H. Then
there must be some element of order 4 not in H and it generates a subgroup K. By
the theorem above, we know that |H ∩ K| ≥ 484̇ = 2, so we must have |H ∩ K| = 2.
Since H is normal in G, we can form the semidirect product H × K with product
(h, k)(h′ , k ′ ) = (hkh′ k −1 , kk ′ ), and the map ϕ(h, k) = hk is a morphism from
H o K → G, whose image is HK. By the theorem, o(HK) = 4·4
2 = 8, so this
morphism is surjective. Moreover, the kernel of this map is ker(ϕ) = {(h, h−1 )|h ∈
H ∩ K}. Moreover ker(ϕ) ≤ Z(H o K). To see this, note that
(h, h−1 )(h′ , k) = (hh−1 h′ h, h−1 k) = (h′ h, h−1 k)
= (hh′ , kh−1 ) = (h′ khk −1 , kh) = (h′ , k)(h, h−1 ).
In any semidirect product H o K, where H ∼
=K∼
= Z4 , note that since Aut(H) ∼
=
Z2 , there are precisely two morphisms K → Aut(H). The trivial morphism determines the direct product Z4 × Z4 ), and the subgroup ker(ϕ) corresponds to the
subgroup ⟨(2, 2)⟩. The resulting quotient group has two elements of order 2. In the
quotient group Z4 × Z4 /⟨(2, 2)⟩, the elements (2, 0) and (1, 1) are distinct elements
of order 2. Therefore, if the group G has only one element of order 2, it must be
determined by the nontrivial morphism Z4 → Aut(Z4 ).
Let H = ⟨ρ⟩ and K = ⟨σ⟩. Then the nontrivial morphism determines the action
given by σ ∗ ρ = ρ3 , or if we identify ρ and σ with their images in H o K, we
have the commutation relation σρ = ρ3 σ, which gives a generators and relations
description H o K = ⟨ρ, σ|ρ4 = σ 4 = e, σρ = ρσ⟩. The element ρ2 σ 2 has order
two and lies in the center of H o K, and G = H o K/⟨ρ2 σ 2 ⟩. Then the group G
is isomorphic to the octonion group O8 = {1, −1, i, −i, j, −j, k, −k}, which is the
group of invertible integer quaternions, under the bijection given by
ē → 1,
ρ2 → −1,
ρ̄ → i,
σ̄ → j,
ρσ → k
Example 2.5. Let G be a group of order 12. Then the number of 3 Sylow subgroups
is either 4 or 1, and the number of 2-Sylow subgroups is either 3 or 1. Therefore,
we don’t immediately know if any of the Sylow subgroups is normal in G.
On the other hand, suppose that there really are 4 3-Sylow subgroups. These
subgroups could not have any element in common except the identity, so they would
GROUP ACTIONS AND THE SYLOW THEOREMS
7
account for 8 distinct elements of order 3 in G. On the other hand, if there are 3
2-Sylow subgroups. If H and K are distinct 2-Sylow subgroups, then o(H ∩ K) ≥
4 · 4/12 > 1, so the intersection must consist of 2 elements, since they are distinct.
Thus H ∪ K must consist of 6 elements. Since we cannot have 14 elements in G,
it follows that either the 3-Sylow or the 2-Sylow subgroup must be normal in G.
Since the 2-Sylow subgroups are conjugate, they must either all be isomorphic
to Z4 or all be isomorphic to Z2 × Z2 . Let us suppose that there are 3 2-Sylow
subgroups, and they are isomorphic to Z4 . Then notice that if H and K are distinct
2-Sylow subgroups, they intersect in a subgroup of order 2, and each contains exactly
one element of order 2, which means that there can only be one element of order
2 in G, because, by the first Sylow theorem, any element of order 2 is contained
in a 2-Sylow subgroup, and the intersection of any two such subgroups contains the
same element of order 2. Thus we now can count the orders of all elements in G.
There are 2 elements of order 3, because the 3-Sylow subgroup is normal in G, 6
elements of order 4, 1 element of order 2, and 1 element of order 1. Therefore, the
two elements we have not accounted for must have order 6. This means we do have
a normal subgroup of G of order 6, but since this subgroup contains an element of
order 2, we cannot express G as a semidirect product Z6 o Z2 .
Instead, we proceed as follows. Let H be the 3-Sylow subgroup, which is normal
in G, and let K be a 2-Sylow subgroup. Note that Aut(H) ∼
= Z2 . Assume K ∼
= Z4 ,
then we note that there are precisely two distinct morphisms K → Aut(H). The
trivial morphism gives the group Z3 × Z4 ∼
= Z12 . To understand the group given by
the nontrivial morphism, let H = ⟨ρ⟩ and K = ⟨σ⟩, so
G = ⟨ρ, σ|ρ3 = σ 4 = e, σρ = ρ2 σ⟩.
Next, suppose that the 3-Sylow subgroup H is normal, but the 2-Sylow subgroups
are isomorphic to Z2 × Z2 . Recall that Aut(Z2 × Z2 ) ∼
= S3 . In fact, if a, b and c are
the elements of order in Z2 × Z2 , then the automorphism group acts just like the
group of permutations of these elements. Let H = ⟨a, b, ab, e⟩ be a 2-Sylow subgroup.
Then a morphism H → Z2 = Aut(H) is either trivial, or has kernel consisting of 2
elements. All 3 nontrivial morphisms K → Aut(H) can be obtained from any one
of them by composition with an automorphism of K, so up to isomorphism, there
are only two semidirect product structures on Z3 o (Z2 × Z2 ). The trivial structure
gives the direct product Z3 × Z2 × Z2 = Z2 × Z6 .
For the other structure, we can write H = ⟨ρ⟩, K = ⟨a, b⟩, with a2 = b2 = e and
ab = ba, and we can assume that a ⋆ ρ = ρ2 , b ⋆ ρ = ρ, in other words aρ = ρ2 a
and bρ = ρb. In this case, we see that the subgroup M = ⟨ρ, a⟩ is isomorphic to
D3 , and the subgroup N = ⟨b⟩ is in the center of G and is isomorphic to Z3 . Since
[G : M ] = 2, both M and N are normal and G ∼
= M ×N ∼
= D3 × Z2 . Since
D6 ∼
D
×
Z
,
this
group
is
just
D
.
= 3
2
6
Now, let us assume that the 2-Sylow subgroup K is normal, so G = K o H.
First, let us suppose that K ∼
= Z4 , so Aut(K) = Z2 . But then the only morphism
Z3 → Z2 is the trivial one, so G = K × H ∼
= Z4 × Z3 .
Next, assume that K ∼
Z
×
Z
,
so
that
Aut(K) = S3 . Since S3 contains a
= 2
2
subgroup isomorphic to Z3 , there is a nontrivial morphism H → Aut(K). In fact,
there are two nontrivial such morphisms, but they differ by an automorphism of H,
so they give the same semidirect product structure. We have G = ⟨a, b, c|a2 = b2 =
c3 = e, ab = ba, ca = bc, cb = abc⟩.
8
MICHAEL PENKAVA
Theorem 2.6. Suppose that α : K → Aut(H) is a morphism, and θ ∈ Aut(K).
Let α′ = α ◦ θ. Then the map H oα′ K → H oα K, given by (h, k) 7→ (h, θ(k)) is
an isomorphism.
Proof. Denote the map H oα′ K → H oα K by ϕ. We compute
ϕ((h, k)(h′ , k ′ )) = ϕ(hαk′ (h′ ), kk ′ ) = (hαθ(k) (h′ ), θ(kk ′ ))
= (hαθ(k) (h′ ), θ(k)θ(k ′ ))(h, θ(k))(h, θ(k ′ )) = ϕ(h, k)ϕ(h′ , k ′ ).
It is easy to see that ϕ is a bijection, so it is an isomorphism.
The theorem above says that if two morphisms K → Aut(H) differ by an automorphism of K, then they induce isomorphic semidirect products.
Theorem 2.7. Suppose that α : K → Aut(H) is a morphism, and θ ∈ Aut(H).
Define α′ by αk′ = θ ◦ αk ◦ θ−1 . Then the map H oα K → H oα′ K, given by
(h, k) 7→ (θ(h), k) is an isomorphism.
Proof. Denote the map H oα K → H oα′ K by ϕ. Then
ϕ((h, k)(h′ , k ′ )) = ϕ(hαk (h′ ), kk ′ ) = (θ(hαk (h′ )), kk ′ )
= (θ(h)θ(αk (h′ )), kk ′ ) = (θ(h)θ(αk (θ−1 (θ(h′ )), kk ′ )
= (θ(h)αk′ (θ(h′ )), kk ′ ) = (θ(h), k)(θ(h′ ), k ′ ) = ϕ(h, k)ϕ(h′ , k ′ ).
Since ϕ is clearly bijective, it is an isomorphism.
Let p and q be primes, and suppose that q < p. Suppose that G is a group of
order pq. Let H be a p-Sylow subgroup and K be a q-Sylow subgroup. By the third
Sylow theorem, H must be normal in G, so G = H o K. If q ̸ |(p − 1) then K is also
normal in G, so G = H × K, by a theorem decomposing a group as direct product
of subgroups. Thus, we obtain only one possible group structure Zp × Zq = Zpq .
Now, consider the case that q|(p−1). Now Aut(H) = Aut(Zp ) = Zp∗ = Zp−1 , and
therefore, there is a unique subgroup M of order q in Aut(H), and this subgroup
consists of all elements in Aut(H) whose order divides q. Let K = ⟨σ⟩, and suppose
that K is written multiplicatively, so that K = {σ m |1 ≤ m ≤ q}.
For each element x ∈ M , there is a morphism αx : K → Aut(H), which is
uniquely determined by the condition ασx = x, and every morphism α : K →
Aut(K) must be of this form. Therefore there are exactly q such morphisms.
Now, let us study these morphisms. First, there is the trivial morphism α1 ,
which maps σ to the identity of Aut(H). In this case, the semidirect product is
just the direct product, which gives the group Zpq .
The other q − 1 elements of the subgroup M of Aut(H) are all generators of
the subgroup M . Moreover, if we choose x to be one of these elements, then every
element in M is of the form xm for some 1 ≤ m ≤ q, and we have ασx m = xm .
Moreover, if 1 ≤ m < q, then the map θm : K → K, given by θm (σ) = σ m , is an
isomorphism, so is an automorphism of K. We compute that αθxm (σ) = ασx m = xm .
Therefore if y = xm , we have αy = αx ◦ θm , which means that the semidirect
products determined by αx and αy are isomorphic.
Finally, we want to give a presentation of G in terms of generators and relations.
Express H = ⟨ρ⟩, written multiplicatively, so H = {ρm |1 ≤ m ≤ p}. Now, the
automorphism x is of the form x(ρ) = ρk for some k, and since xq is the identity,
q
we have ρ = xq (ρ) = (x(ρ))q = ρk , we must have k q = 1 (mod p). There are
GROUP ACTIONS AND THE SYLOW THEOREMS
9
exactly q solutions to this equation (mod p), and if we choose any solution k
except k = 1, we can express all q of the solutions as powers of k. If we denote the
elements (ρi , σ j ) in H o K as ρi σ j , then we have
σρ = (e, σ)(ρ, e) = (eασx (ρ), σ ė) = (ρk , σ) = ρk σ.
Thus we can express G in terms of generators and relations as
G = ⟨ρ, σ|ρp = σ q = e, σρ = ρk σ⟩.
We summarize the results in the following theorem.
Theorem 2.8. Let G be a group of order pq, where p and q are primes with q < p.
Then G ∼
= Zp o Zq . Moreover, if q ̸ |(p − 1), then G ∼
= Zpq . If q|(p − 1), then there
is some 1 < k < p − 1 such that k q = 1 mod p, and if we choose any such k, then
G is isomorphic to the group given in terms of generators and relations by
⟨ρσ|ρp = σ q , σρ = ρk σ⟩.
Example 2.9. Let G be a group of order 21. Since 21 = 7 · 3 and 3|(7 − 1), there
are exactly two such groups, up to isomorphism. The first possibility is G ∼
= Z21 .
Let us study the other possibility. First, we need to find a k ̸= 1 such that k 3 = 1
(mod 7). Evidently, k = 2 works, but we could also have chosen k = 22 = 4. This
means that have a group G given by
G = ⟨ρ, σ|ρ7 = σ 3 = e, σρ = ρ2 σ⟩.
Department of Mathematics, University of Wisconsin-Eau Claire, Eau Claire, WI
54729 USA
E-mail address: [email protected]