Theorem 11.9. A group of orderp2, wherepis prime, is cyclic or is isomorphic toZ/pZ×
Z/pZ.
Proof. LetGbe a group of orderp2and assume thatGis not cyclic. According to
theorem9.7, the centre ofGhas order at leastp. So take an elementα ∈Z(G),
11.2. GROUPS OF SMALL ORDER 181
is not cyclic, it follows that|β| = ptoo. And because α ∈ Z(G), αβ = βα. Therefore the mapping
Z/pZ×Z/pZ→G
given by
(a, b)7→αaβb , a, b∈Z/pZ.
is a well-defined homomorphism. This homomorphism is injective. But both groups have orderp2, so it is in fact an isomorphism. ThusGis either cyclic or
isomorphic to(Z/pZ)2.
For example a group of order9is either cyclic or isomorphic to(Z/3Z)2. The next result deals with groups of order2p.
Theorem 11.10. A group of order2p, wherep≥3is prime, is isomorphic toZ/2pZor toDp.
Proof. Let Gbe a group of order2p. Its Sylowp-subgroup has order pand is therefore cyclic (see corollary10.5). Similarly, the Sylow2-subgroup is cyclic of order2. So letαbe an element ofGof orderp, andβof order2. Nownp ≡1
(mod p) and np|[G : ⟨α⟩] = 2. This implies that np = 1and that ⟨α⟩ is a
normal subgroup. Therefore
βαβ =αk,
for somek,0< k < p. Conjugating again withβ, we get
α=β2αβ2 =βαkβ = (αk)k=αk2 .
Thereforek2 ≡1 (mod p), which means thatk=±1.
Thus there are two cases: first,
βαβ =α ,
which says thatαandβ commute. But then αβ has order2p(see exercise 5.7) andGis cyclic.
In the second case
βαβ =α−1.
This is the defining relation forDp (equation (7.1)). SoGis isomorphic toDp.
In particular, forp= 3this says again that a group of order6is either cyclic or isomorphic toS3. And forp= 5, it says that a group of order10is either cyclic
or isomorphic toD5.
It is time to sort out the groups of order8.
Theorem 11.11. LetGbe a group of order8. ThenGis isomorphic toD4,Q,Z/8Z,
Z/4Z×Z/2Zor(Z/2Z)3.
Proof. Suppose thatGis not cyclic. Then its non-trivial elements have order2or
4. It is not hard to see that if all of these have order2, then
G∼=(Z/2Z)3.
So let α ∈ G be an element of order 4. Since the index of⟨α⟩is 2, ⟨α⟩ is a normal subgroup. Pick an elementβ ̸∈ ⟨α⟩. Then
βαβ−1 =αk,
wherek=±1. Ifk = 1thenαandβcommute andGis abelian. There are two possibilities: either|β| = 2or|β|= 4. In the first case, arguing as in the proof of theorem11.9we see that
G∼=Z/4Z×Z/2Z.
In the second case, we must have that β2 = α2. Therefore |αβ| = 2. So
replacingβbyαβ we are back to the first case. Now suppose that
11.2. GROUPS OF SMALL ORDER 183
ThusGis not abelian. If |β| = 2, then the relation tells us thatG ∼= D4 (see
equation (7.1)). This leaves us with the case|β|= 4. According to theorem9.7,
Z(G)is not trivial. It is not hard to see that |Z(G)|= 2,
and that ifγ ∈Ghas order4, then
Z(G)⊂ ⟨γ⟩.
So letϵgenerateZ(G). Thenα2 =ϵ,β2 =ϵ, and therefore
(βα)2 = (βα)(α−1β) = β2 =ϵ .
It follows that|βα|= 4. Furthermore,
βα =α−1β =α3β =ϵ(αβ).
Comparing this with the description ofQin exercise 4.5, we see thatG∼=Q. The following result is the key to classifying groups of order12.
Theorem 11.12. LetGbe a group of order12. ThenGhas a normal subgroup of order
3orG∼=A4.
Proof. Write12 = 22·3. According to theorem11.5,n3 ≡1 (mod 3). Accord-
ing to corollary 11.7,n3 divides4. So n3 = 1or 4. Suppose that n3 ̸= 1, in
other words thatGdoes not have a normal subgroup of order3. LetH be one of the subgroups of order3. ThenGacts on the set of 4left cosets,G/H, by multiplication on the left. This defines a homomorphism
σ :G→S4 .
What is the kernel ofσ? Well,σ(α) = 1means that for allβ ∈G,
Equivalently
β−1αβH =H
or
β−1αβ ∈H
for allβ ∈G. But then
α∈βHβ−1
for allβ ∈ G. By assumption,H has four distinct conjugates, and the intersec- tion of two distinct subgroups of order3is trivial. Therefore α = 1 and σ is injective. SoGis isomorphic to a subgroup ofS4of order12, and the only such
subgroup isA4.
With this result it is not hard to classify groups of order12. It turns out that up to isomorphism there are5groups: Z/12Z,Z/3Z×V,D6,A4andG12(see
exercise 4.17).
Example 11.13. LetGbe a group of order15 = 3·5. Thenn3 ≡1 (mod 3)
and n3|5. It follows thatn3 = 1. Similarly, n5 ≡ 1 (mod 5)and n5|3. So
n5 = 1 too. Thus Ghas only one Sylow 3-subgroup and only one Sylow 5-
subgroup and both are normal. Letαbe an element of order3andβ of order
5. Applying exercise 10.8, we see that
αβ =βα .
But then by exercise 5.7,
|αβ|=|α||β|= 15.