D6 - the inside story

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

D6 - The Inside Story Shirleen Stibbe [email protected]

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

Standard form rm, rms, (m = 0,1,...,5), rm = r[2πm/6], rms = q[πm/6] (assuming s = q[0]), srm = r−ms

IB2 Orders of elements

Order Type Element Number

1 identity e 1 2 reflection rms, (m = 0,1,...,5) 6 2 rotation r3 1 3 rotation r2, r4 2 6 rotation r, r5 2

IB4 Normal Subgroups

N1 = {e, r3} ≅ C2 N2 = {e, r2, r4} ≅ C3 N3 = {e, r2, r4, s, r2s, r4s} ≅ D3 N4 = {e, r2, r4, rs, r3s, r5s} ≅ D3 N5 = {e, r, r2, r3, r4, r5} ≅ C6 Note: r2 is rotation through 2π/3; this explains why N3 and N4 are isomorphic to D3

IB4 Non-normal subgroups

Order 2: Km = {e, rms} ≅ C2, m = 0, 1, 2, 3, 4, 5 Order 4: H1 = {e, r3, s, r3s} ≅ 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 So D6 ≅ N1 × N3 ≅ C2 × D3 Note: N1 and N4 also satisfy the conditions for Theorem 1.1

GR4 Conjugacy

Conjugacy classes: {e}, {r3}, {r, r5}, {r2, r4}, {s, r2s, r4s}, {rs, r3s, r5s} [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, r3} (= N1) Note: r3 = r[π] commutes with everything

GR4 Correspondence Theorem (& Proofs & Strategies 3.9)

Quotient Group D6/Nm Nm contained in Subgroups of D6/Nm

D6/N1: eN1 = r3N1 = {e, r3} rN1 = r4N1 = {r, r4} r2N1 = r5N1 = {r2, r5} sN1 = r3sN1 = {s, r3s} rsN1 = r4sN1 = {rs, r4s} r2sN1 = r5 sN1 = {r2s, r5s}

D6/N1 = {eN1 , rN1, r2N1 , sN1, rsN1, r2sN1} ≅ D3

N5 normal in D6 H1, H2, H3 all non-normal in D6 Expect: D6/N1 has 1 normal and 3 non-normal subgroups

N5/N1 = {eN1, rN1, r2N1} Normal

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

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

H3/N1 = {eN1, r2sN1} Non-normal

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

groups of rotations.


GR4 Correspondence Theorem (cont)

Quotient Group D6/Nm Nm contained in Subgroups of D6/Nm

D6/N2: eN2 = r2N2 = r4N2 = {e, r2, r4} rN2 = r3N2 = r5N2 = {r, r3, r5} sN2 = r2sN2 = r4sN2 = {s, r2s, r4s} rsN2 = r3sN2 = r5sN2 = {rs, r3s, r5s}

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

N3, N4, N5 all normal in D6 Expect: D6/N2 has 3 normal subgroups N3, N4, N5 - all normal in D6 Expect: D6/N2 has 3 normal subgroups

N3/N2 = {eN2, sN2} Normal N4/N2 = {eN2, rsN2} Normal N5/N2 = {eN2, rN2} Normal

D6/N3 = {eN3, rN3} ≅ C2

Not contained in any proper subgroups Expect: D6/N3 has no proper subgroups

D6/N3 has order 2, so no proper subgroups

D6/N4 = {eN4, rN4} ≅ C2

Not contained in any proper subgroups Expect: No subgroups Expect: D6/N4 has no proper subgroups


D6/N5 = {eN5, sN5} ≅ C2

Not contained in any 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

|D6| = 12 = 22 × 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 C2 × C2. Sylow 3–subgroup: N2 = {e, r2, r4} ≅ C3 Sylow 2–subgroups: H1 = {e, r3, s, r3s} ≅ C2 × C2 H2 = {e, r3, rs, r4s} ≅ C2 × C2 H3 = {e, r3, r2s, r5s} ≅ C2 × C2

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


Sylow and the Correspondence Theorem

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]

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

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) g → ϕg where ϕg(Hi) = g ∧ Hi = gH ig–1 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 → S3 This is the composition: D6 → Γ(X) → S3 where, if ϕg(Hi) (= g ∧ Hi) = Hj, then i → j in S3.

Reminder: H1 = {e, r3, s, r3s}, H2 = {e, r3, rs, r4s}, H3 = {e, r3, r2s, r5s}

Group Action Table

g ∧ Hi = gHig–1

g H1 H2 H3 Cycle ( ∈ S3) e H1 H2 H3 e r H3 H1 H2 (132) r2 H2 H3 H1 (123) r3 H1 H2 H3 e r4 H3 H1 H2 (132) r5 H2 H3 H1 (123) s H1 H3 H2 (23) rs H3 H2 H1 (13) r2s H2 H1 H3 (12) r3s H1 H3 H2 (23) r4s H3 H2 H1 (13) r5s H2 H1 H3 (12)

Ker(ψ ) = },{)(Stab 3reHHXH

iXH

iii

==∈∈ = N1, (shaded in the table), Im(ψ ) = S3

First isomorphism theorem:

D6 / Ker(ψ ) ≅ Im(ψ )

i.e. D6 / N1 ≅ S3

cf D6 / N1 ≅ D3 (≅ S3) under Correspondence Theorem

N

12

6

3 2

4

D6

H D6/N

N e

H/N

D6 - the inside story

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