Question

derive all groups of order 12 using sylow theorems. please dont use any generalizations show all work and theorems used. all 5 of them. dont use the semi direct product, dont copy and paste anyt answers on google. show all work and cohesive strong arguments using the sylow theorems, counting, and the extended cayley theorem if possible.

Answer 1

will appeal to an isomorphism property of semidirect products: for any

semidirect product H oϕ K and automorphism f : K → K, H oϕ K ∼= H oϕ◦f K. This says

that precomposing an action of K on H by automorphisms (that’s ϕ) with an automorphism

of K produces an isomorphic semidirect product of H and K.

Proof. Let |G| = 12 = 22

· 3. A 2-Sylow subgroup has order 4 and a 3-Sylow subgroup has

order 3. We will start by showing G has a normal 2-Sylow or 3-Sylow subgroup: n2 = 1 or

n3 = 1. From the Sylow theorems,

n2 | 3, n2 ≡ 1 mod 2, n3 | 4, n3 ≡ 1 mod 3.

Therefore n2 = 1 or 3 and n3 = 1 or 4.

To show n2 = 1 or n3 = 1, assume n3 6= 1. Then n3 = 4. Let’s count elements of order 3.

Since each 3-Sylow subgroup has prime order 3, two different 3-Sylow subgroups intersect

trivially. Each of the four 3-Sylow subgroups of G contains two elements of order 3 and

these are not in any other 3-Sylow subgroup, so the number of elements in G of order 3 is

2n2 = 8. This leaves us with 12 − 8 = 4 elements in G not of order 3. A 2-Sylow subgroup

has order 4 and contains no elements of order 3, so one 2-Sylow subgroup must account for

the remaining 4 elements of G. Thus n2 = 1 if n3 6= 1.

Next we show G is a semidirect product of a 2-Sylow and 3-Sylow subgroup. Let P be a

2-Sylow subgroup and Q be a 3-Sylow subgroup of G. Since P and Q have relatively prime

orders, P ∩ Q = {1} and the set P Q = {xy : x ∈ P, y ∈ Q} has size |P||Q|/|P ∩ Q| =

12 = |G|, so G = P Q. Since P or Q is normal in G, G is a semidirect product of P and Q:

G ∼= P o Q if P C G and G ∼= Q o P if Q C G.

1 Groups of order 4 are isomorphic to Z/(4)

or (Z/(2))2

, and groups of order 3 are isomorphic to Z/(3), so G is a semidirect product of

the form

Z/(4) o Z/(3), (Z/(2))2 o Z/(3), Z/(3) o Z/(4), Z/(3) o (Z/(2))2

.

We will determine all these semidirect products, up to isomorphism, by working out all the

ways Z/(4) and (Z/(2))2

can act by automorphisms on Z/(3) and all the ways Z/(3) can

act by automorphisms on Z/(4) and (Z/(2))2

.

First we list the automorphisms of the Sylow subgroups: Aut(Z/(4)) ∼= (Z/(4))× =

{±1 mod 4}, Aut((Z/(2))2

) ∼= GL2(Z/(2)), and Aut(Z/(3)) ∼= (Z/(3))× = {±1 mod 3}.

Case 1: n2 = 1, P ∼= Z/(4).

The 2-Sylow subgroup is normal, so the 3-Sylow subgroup acts on it. Our group is a

semidirect product Z/(4) o Z/(3), for which the action of the second group on the first is

through a homomorphism ϕ: Z/(3) → (Z/(4))×. The domain has order 3 and the target

has order 2, so this homomorphism is trivial, and thus the semidirect product must be

trivial: it’s the direct product

Z/(4) × Z/(3),

which is cyclic of order 12 (generator (1, 1)).

Case 2: n2 = 1, P ∼= Z/(2) × Z/(2).

We need to understand all homomorphisms ϕ: Z/(3) → GL2(Z/(2)). The trivial homo-

morphism leads to the direct product

(Z/(2))2 × Z/(3).

What about nontrivial homomorphisms ϕ: Z/(3) → GL2(Z/(2))? Inside GL2(Z/(2)), which

has order 6 (it’s isomorphic to S3), there is one subgroup of order 3: {(

1 0

0 1 ),(

0 1

1 1 ),(

1 1

1 0 )}. A

nontrivial homomorphism ϕ: Z/(3) → GL2(Z/(2)) is determined by where it sends 1 mod 3,

which must go to a solution of A3 = I2; then ϕ(k mod 3) = Ak

in general. For ϕ to be

nontrivial, A needs to have order 3, and there are two choices for that. The two matrices of

order 3 in GL2(Z/(2)) are inverses. Call one of them A, making the other A−1

. The resulting

homomorphisms Z/(3) → GL2(Z/(2)) are ϕ(k mod 3) = Ak and ψ(k mod 3) = A−k

, which

are related to each other by composition with inversion, but watch out: inversion is not

an automorphism of GL2(Z/(2)). It is an automorphism of Z/(3), where it’s negation. So

precomposing ϕ with negation on Z/(3) turns ϕ into ψ: ψ = ϕ ◦ f, where f(x) = −x

on Z/(3). Therefore the two nontrivial homomorphisms Z/(3) → GL2(Z/(2)) are linked

through precomposition with an automorphism of Z/(3), and therefore ϕ and ψ define

isomorphic semidirect products. This means that up to isomorphism, there is one nontrivial

semidirect product

(Z/(2))2 o Z/(3).

That is, we have shown that up to isomorphism there is only one group of order 12 with n2 =

1 and 2-Sylow subgroup isomorphic to Z/(2) × Z/(2). The group A4 fits this description:

its normal 2-Sylow subgroup is {(1),(12)(34),(13)(24),(14)(23)}, which is not cyclic.

Now assume n2 6= 1, so n2 = 3 and n3 = 1. Since n2 > 1, the group is nonabelian, so it’s

a nontrivial semidirect product (a direct product of abelian groups is abelian).

Case 3: n2 = 3, n3 = 1, and P ∼= Z/(4).

Our group looks like Z/(3) o Z/(4), built from a nontrivial homomorphism ϕ: Z/(4) →

Aut(Z/(3)) = (Z/(3))× There is only one choice of ϕ: it has to send 1 mod 4 to −1 mod

3, which determines everything else: ϕ(c mod 4) = (−1)c mod 3. Therefore there is one

nontrivial semidirect product Z/(3) o Z/(4). Explicitly, this group is the set Z/(3) × Z/(4)

with group law

(a, b)(c, d) = (a + (−1)b

c, b + d).

Case 4: n2 = 3, n3 = 1, and P ∼= Z/(2) × Z/(2).

The group is Z/(3) o (Z/(2))2

for a nontrivial homomorphism ϕ: (Z/(2))2 → (Z/(3))×.

The group (Z/(2))2 has a pair of generators (1, 0) and (0, 1), and ϕ(a, b) = ϕ(1, 0)aϕ(0, 1)b

,

where ϕ(1, 0) and ϕ(0, 1) are ±1. Conversely, this formula for ϕ defines a homomorphism

since a and b are in Z/(2) and exponents on ±1 only matter mod 2. For ϕ to be nontrivial

means ϕ(1, 0) and ϕ(0, 1) are not both 1, so there are three choices of ϕ: (Z/(2))2 →

(Z/(3))×:

ϕ(a, b) = (−1)a

, ϕ(a, b) = (−1)b

, ϕ(a, b) = (−1)a

(−1)b = (−1)a+b

.

This does not mean there are corresponding semidirect products Z/(3) oϕ (Z/(2))2 are

nonisomorphic. In fact, the above three choices of ϕ lead to isomorphic semidirect products:

precomposing the first ϕ with the matrix ( 0 1

1 0 ) produces the second ϕ, and precomposing the

first ϕ with the matrix ( 1 1

0 1 ) produces the third ϕ. Therefore the three nontrivial semidirect

products Z/(3)o(Z/(2))2 are isomorphic, so all groups of order 12 with n2 = 3 (equivalently,

all nonabelian groups of order 12 with n3 = 1) and 2-Sylow subgroup isomorphic to (Z/(2))2

are isomorphic. One such group is D6, with normal 3-Sylow subgroup {1, r2

, r4}.

If we meet a group of order 12, we can decide which of the 5 groups it is isomorphic to

by the following procedure:

• Is it abelian? If so, it’s isomorphic to Z/(4)×Z(3) ∼= Z/(12) or Z/(2)×Z/(2)×Z/(3),

which are distinguished by the structure of the 2-Sylow subgroup.

• Is it nonabelian with n2 = 1? If so, then it’s isomorphic to A4.

• Is it nonabelian with n2 > 1? If so, then it’s isomorphic to D6 if its 2-Sylow

subgroups are noncyclic and it’s isomorphic to the nontrivial semidirect product

Z/(3) o Z/(4) if its 2-Sylow subgroup is cyclic.

For example, here are four nonabelian groups of order 12:

Z/(2) × S3, PSL2(F3), Aff(Z/(6)), Aff(F4).

The group Z/(2) × S3 has n2 > 1 and its 2-Sylow subgroups are not cyclic, so Z/(2) × S3

∼=

D6. The group PSL2(F3) is nonabelian with n2 = 1, so PSL2(F3) ∼= A4. The group

Aff(Z/(6)) has n2 > 1 and its 2-Sylow subgroups are not cyclic, so Aff(Z/(6)) ∼= D6.

Finally, Aff(F4) is nonabelian with a normal 2-Sylow subgroup, so Aff(F4) ∼= A4.

Another way to distinguish the three nonabelian groups of order 12 is to count elements

of order 2 in them: D6 has 7 elements of order 2 (6 reflections and r

3

), A4 has 3 elements

of order 2 (the permutations of type (2, 2)), and Z/(3) o Z/(4) has one element of order 2

(it is (0, 2)).

In abstract algebra textbooks (not group theory textbooks), Z/(3) o Z/(4) is usually

written as T but it is almost never given a name to accompany the label. Should it be

called the “obscure group of order 12”? Actually, this group belongs to a standard family

of finite groups: the dicyclic groups, also called the binary dihedral groups. They are

nonabelian with order 4n (n ≥ 2) and each contains a unique element of order 2. The one

of order 8 is Q8, and more generally the one of order 2m is the generalized quaternion group Q2^m