D6 - The Inside Story

Shirleen Stibbe s.stibbe@open.ac.uk

Notation

D6 = 〈 r, s : r6 = s2 = e; sr = r5s = r 1s 〉

Standard form

rm, rms, (m = 0,1,...,5),

IB2 Orders of elements

IB4 Normal Subgroups

−

Order

1 2 2 3 6

N1 N2 N3 N4 N5

= = = = =

Note:

rm = r[2πm/6],

rms = q[πm/6] (assuming s = q[0]),

Type

Element

identity reflection rotation rotation rotation

e rms, (m = 0,1,...,5) r3 r 2, r 4 r, r5

3

{e, r } 2 4 {e, r , r } 2 4 2 4 {e, r , r , s, r s, r s} 2 4 3 5 {e, r , r , rs, r s, r s} 2 3 4 5 {e, r, r , r , r , r }

≅ ≅ ≅ ≅ ≅

srm = r ms −

Number

1 6 1 2 2

C2 C3 D3 D3 C6

Note: I’ve use Cn rather than Zn for all cyclic subgroups of order n, since they’re usually groups of rotations.

r2 is rotation through 2π/3; this explains why N3 and N4 are isomorphic to D3

IB4 Non-normal subgroups

Order 2:

K m = { e , r ms }

Order 4:

H 1 = { e , r 3, s , r 3s } ≅ C2 × C2 ≅ D2 H2 = {e, r3, rs, r4s} ≅ C2 × C2 ≅ D2 H3 = {e, r3, r2s, r5s} ≅ C2 × C2 ≅ D2

GR2 Internal Direct Product (Th'm 1.1, HB p24)

a) D6 = N1 N3 b) N1∩ N3 = {e} c) N1 and N3 are normal in D6

≅ C 2,

m = 0, 1, 2, 3, 4, 5

So D6 ≅ N1 × N3 ≅ C2 × D3 Note: N1 and N4 also satisfy the conditions for Theorem 1.1 3

5

2

4

2

4

3

5

GR4 Conjugacy

Conjugacy classes: {e}, {r }, {r, r }, {r , r }, {s, r s, r s}, {rs, r s, r s} [Elements split as: rotations: through the same angle; reflections: axes through (a) vertices, (b) edges.] Class equation: 12 = 1 + 1 + 2 + 2 + 3 + 3

GR4 Centre

Centre of D6: Z(D6) = {e, r } (= N1) 3 Note: r = r[π] commutes with everything

GR4 Correspondence Theorem (& Proofs & Strategies 3.9)

3

Quotient Group D6/Nm

D6/N1: eN1 rN 1 r 2N 1 sN 1 rsN1 r 2s N 1 D6/N1

= = = = = =

3

r N1 r 4N 1 r 5N 1 r 3s N 1 r 4s N 1 r5 sN 1

3

{e, r } 4 {r, r } 2 5 {r , r } 3 {s, r s} 4 {rs, r s} 2 5 {r s, r s}

= = = = = = 2

Nm contained in

Subgroups of D6/Nm

N5 normal in D6

N5/N1 = {eN1, rN1, r N1} Normal

H 1, H 2, H 3 all non-normal in D6

2

= {eN1 , rN1, r N1 , sN1, rsN1, r sN1} ≅ D3

Expect: D6/N1 has 1 normal and 3 non-normal subgroups

2

H1/N1 ={eN1, sN1} Non-normal H2/N1 = {eN1, rsN1} Non-normal 2

H3/N1 = {eN1, r sN1} Non-normal

http://www.shirleenstibbe.co.uk/m336

GR4 Correspondence Theorem (cont)

Quotient Group D6/Nm

D6/N2: eN2 rN 2 sN 2 rsN2

2

= r N2 3 = r N2 2 = r sN 2 3 = r sN 2

= = = =

r 4N 2 r 5N 2 r 4s N 2 r 5s N 2

2

4

= {e, r , r } 3 5 = {r, r , r } 2 4 = {s, r s, r s} 3 5 = {rs, r s, r s}

Nm contained in

Subgroups of D6/Nm

N 3, N 4, N 5

N3/N2 = {eN2, sN2}

all normal in D6

Normal

Expect: D6/N2 has 3 normal subgroups

Normal

N5/N2 = {eN2, rN2}

D6/N2 = { eN2 , rN2, sN2, rsN2 } ≅ D2 ≅ V ≅ C2 x C2 D6/N3 = {eN3, rN3} ≅ C2

Expect: D6/N2 has 3 N4/N2 = {eN2, rsN2}

Normal Not contained in any proper subgroups

D6/N3 has order 2, so no proper subgroups

Expect: D6/N3 has no proper subgroups D6/N4 = {eN4, rN4} ≅ C2

Not contained in any proper subgroups

D6/N4 has order 2, so no proper subgroups

Expect: No s

Expect: D6/N4 has no proper subgroups D6/N5 = {eN5, sN5} ≅ C2

Not contained in any proper subgroups

D6/N5 has order 2, so no proper subgroups

Expect: D6/N5 has no proper subgroups Note: The table above uses information about the subgroups of D6 that contain Ni to derive information about

the subgroups of D6/Ni. You can also go the other way; i.e. if you know about the subgroups of D6/Ni, you can find the corresponding subgroups of D6 which contain Ni. (see Sylow and the Correspondence Theorem below). GR5 Sylow structure

2

|D6| = 12 = 2 × 3 Let np be the number of Sylow p-subgroups.

n3 ≡ 1 mod 3 and n3|4 so n3 = 1 or 4 n2 ≡ 1 mod 2 and n2|3 so n2 = 1 or 3 Let N be a Sylow 3-subgroup. Then |N| = 3, a prime, so N ≅ C3. There are only 2 elements of order 3 in D6, i.e. enough for only 1 Sylow 3-subgroup, so n3 = 1, and N is unique, hence normal in D6. Let H be a Sylow 2-subgroup. Then |H| = 4, so H ≅ C4 or H ≅ C2 × C2. There are no elements of order 4 in D6, so H ≅ C2 × C2, and H contains 3 elements of order 2. Since there are 7 elements of order 2 in D6 we must have n2 > 1, so n2 = 3.

D6 has a normal Sylow 3-subgroup, and 3 conjugate Sylow 2-subgroups, all isomorphic to C 2 × C 2. Sylow 3–subgroup:

N 2 = { e , r 2, r 4}

Sylow 2–subgroups:

H 1 = { e , r 3, s , r 3s } ≅ C2 × C2 H2 = {e, r3, rs, r4s} ≅ C2 × C2 H 3 = { e , r 3, r 2s , r 5s } ≅ C 2 × C 2

Note:

≅ C3

Hi ∩ Hj = {e, r3} i.e. the Sylow 2–subgroups are not disjoint (order is not prime).

http://www.shirleenstibbe.co.uk/m336

Sylow and the Correspondence Theorem D6

H

D6/N

N

H/N

By the Sylow theorems, D6 has a normal subgroup, N, of order 3. We may therefore form the quotient group, D6 /N, of order 12/3 = 4. There are only 2 groups of order 4, so D6 /N ≅ C4 or C2 × C2. In either case, D6 /N has a normal subgroup of order 2. Therefore, by the Correspondence Theorem, D6 has a normal subgroup, H, which contains N and has order 2 × |N| = 6. [Note: H = D6 ∪ hD6, h∈ H, h ∉ N]

12

Note: This argument shows that, if you know something about the subgroups of a quotient group of a group 6

4 2

3

e

G, then you can use the Correspondence Theorem to find out about the subgroups of G itself. We did not use any information about D6 apart from its order and the order of its normal Sylow subgroup.

N

GR6 Conjugacy Action and its consequences

Let X = {H1, H2, H3}, the Sylow 2−subgroups. The conjugacy group action is defined as: ϕ : D6 → Γ(X) –1 g → ϕg where ϕg(Hi) = g ∧ Hi = gH ig By the Sylow theorems, the Hi are all conjugate, i.e. all in the same orbit under the group action. So |Orb(Hi)| = 3 By the Orbit–Stabiliser theorem, |Orb(Hi)| × |Stab(Hi)| = | D6| = 12. So |Stab(Hi)| = 4 If h ∈ Hi, then hHih–1 = Hi (since Hi is closed), so h ∈ Stab(Hi), and Hi ⊆ Stab(Hi). Since |Hi| = |Stab(Hi)| = 4, we must have Stab(Hi) = Hi.

GR6 Derived Homomorphism

| X | = 3, so that Γ(X) ≅ S3, so the derived homomorphism is:

ψ : D6

This is the composition: D6 → Γ(X) → S3 where, if ϕg(Hi) (= g ∧ Hi) = Hj, then i → j in S3.

S3

→ 3

3

Reminder: H1 = {e, r , s, r s},

H2 = {e, r3, rs, r4s}, H3 = {e, r3, r2s, r5s} Group Action Table

g ∧ Hi = gHig–1

g

H1 H1 H3 H2 H1 H3 H2 H1 H3 H2 H1 H3 H2

e r r2 r3 r4 r5 s rs r 2s r 3s r 4s r 5s

Ker(ψ ) =

Stab ( H ) = H i

H i ∈X

i

H2 H2 H1 H3 H2 H1 H3 H3 H2 H1 H3 H2 H1

H3 H3 H2 H1 H3 H2 H1 H2 H1 H3 H2 H1 H3

Cycle ( ∈ S3)

= {e, r 3 } = N1, (shaded in the table),

e (132) (123) e (132) (123) (23) (13) (12) (23) (13) (12) Im(ψ ) = S3

H i ∈X

First isomorphism theorem:

D6 / Ker(ψ ) ≅ Im(ψ ) i.e. D6 / N1 ≅ S3 cf D6 / N1 ≅ D3 (≅ S3) under Correspondence Theorem

http://www.shirleenstibbe.co.uk/m336