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

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