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people:fer:401ws:fall2018:daily_topics

Math 401 - 01 Daily Topics - part 2 (Fall 2018)


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Week 7Topics
10/01/2018Test 1
10/02/2018Lagrange's corollary 1
Orbit-Stabilizer theorem
Examples: cube, truncated icosahedron (soccer ball)
10/03/2018Corollaries 2-5 to Lagrange's theorem
Addendum to cor 3: moreover, there is a unique group of order p, up to isomorphism.
Thm. 7.2
Example 6, p.144
Corollary: if p is the smallest prime divisor of |G| and p2 does not divide |G|, then G has at most one subgroup of index p (HW)
10/05/2018Thm. 7.3
Week 8Topics
10/08/2018Test 1 returned and reviewed
Prop: if φ:GH is an isomorphism, then so is φ1HG.
Prop: “isomorphic to” is an equivalence relation
Thm. 6.1 Cayley's theorem
Aut(G), Inn(G)
10/09/2018Thm 6.4 Aut(G) is a group and Inn(G) is a subgroup of Aut(G)
Example: Inn(D4)K4
Prop: Let G=<a> cyclic and H a group
1. A homom φ:GH is uniquely determined by φ(a).
2. If G has order n and uH has order d where d|n, then there is (unique) homomorphism φ:GH s.t. φ(a)=u. Moreover, φ is injective iff d=n.
3. If G has infinite order and uH, then there is (unique) homomorphism φ:GH s.t. φ(a)=u. Moreover, φ is injective iff u has infinite order.
Example: Aut(Zn)Un
10/10/2018Board presentations PS 6
Thms. 10.2 and 6.3
10/12/2018Fall break
Week 9Topics
10/15/2018Prop. Let N \leq G. TFAE
(i) gNg1N for all gG
(ii) gNg1=N for all gG
(iii) gN=Ng for all gG
(iv) the product of any two left cosets is a left coset.
Moreover, in the last one, we have (gN)(hN)=ghN
Def: normal subgroup
Examples: 1. AnSn
<R360/n>⊴Dn
Prop: if H is a subgroup of G of index 2, then H is a normal subgroup of G
2. Prop: if φ:GˉG is a homomorphism, then ker(φ) is a normal subgroup of G
3. If G is abelian, then every subgroup of G is normal
4. Z(G) is a normal subgroup of G.
5. G and {1} are normal subgroups of G.
Thm 9.2 proof using (iv) above.
Example 9.10 Generalize Z/nZZn
10/16/2018 Example 9.12
Thm 10.3 1st Isom Thm
Example φ:ZZn
Thm 9.4
Thm (N/C theorem) Let HG. NG(H)/CG(H) is isomorphic to a subgroup of Aut(H).
10/17/2018 proof of N/C theorem
Example 10.17 |G|=35
Thm 9.3
Corollary: If |G|=pq and Z(G)1 then G is abelian.
Thm 9.5 Cauchy's thm for abelian gps.
10/19/2018Thm 10.4 NG, q:GG/N is an epimorphism with ker(q)=N
Chapter 8 Direct Product
Def: G1G2
Prop: 1) G1G2 is a group.
2) If G1, G2 are abelian then so is G1G2.
Examples: (1) Z2Z3 is abelian of order 6, so it is isomorphic to Z6
(2) G{1}G{1}G
Cor: G1G2 contains subgroups isomorphic to G1 and G2 respectively.
Def: G1Gn
Thm 8.1
Week 10Topics
10/22/2018Thm 8.2 G1, G2 finite. G1G2 is cyclic iff G1 and G2 are cyclic or relatively prime orders.
10/23/2018RSA cryptography. Public vs private keys
Prop: medm(modn).
Internal direct product
Thm.: Let H,KG be such that HK=G and HK={1}. Then GHK.
Def: When H,KG are such that HK=G and HK={1}, we say that G is the internal direct product of H and K, and write G=H×K.
Example: Consider Dn with n=2m and m odd.
Thm. 9.7 and corollary
Prop: Let H,NG.
(1) If NG then HNG.
(2) If H,NG then HNG
10/24/20182nd, 3rd, 4th and 5th isomorphism theorems.
Sub(D4) and Sub(V4) as examples.
10/26/2018Thm If G is a finite abelian group of order n, and m|n then G has a subgroup of order m.
Fund. Thm. of Finite Abelian Groups
Statement and examples, n=12 and n=600
Elementary divisors form, and invariant factors form
Week 11Topics
10/29/2018Board presentations. Problems sets 7 and 8
10/30/2018Ch.24 Def: conjugate, conjugate class cl(a).
Prop: (1) “conjugate to” is an equivalence relation. The equivalence classes are the conjugacy classes.
(2) cl(a)={a}aZ(G)
Thm. 24.1 without finite assumption
Cor. 1
Thm. Class equation (2 versions)
Thm. 24.2 A non-trivial p-group has non-trivial center.
Def: Finite p-group. Metabelian group.
Cor. Let p be a prime. If |G|=p2, then G is abelian.
Cor. Let p be a prime. If |G|=p3, then G is metabelian. Moreover, |Z(G)|=p or |Z(G)|=p3.
Example: Heisenber group H has order p3, and is not abelian.
10/31/2018Thm. 24.3 Sylow's 1st Theorem
Cor. Cauchy's theorem
Cor. If |G|=pq where p<q are primes and p(q1), then G is cyclic.
Lemma 1. (1) Let HG and C={gHg1gG} the set of all conjugates of H. Then |C|=[G:NG(H)].
Definition of Sylow p-subgroup.
(2) Let H,KG. If HK=KH then HKG.
Lemma 2. Let P be a Sylow p-subgroup G. If gNG(P) and |g| is a power of p, then gP.
Lemma 3. Let |G|=pkm and pm. Let P be a Sylow p-subgroup of G, i.e. |P|=pk, and HG with |H|=pl for some lk. Then there is a conjugate of P that contains H, i.e. there is gG such that HgPg1.
11/02/2018Board presentations. Problems set 9
Proof of Lemma3
Week 12Topics
11/05/2018Sylow Theorems
Examples: (1) |G|=35    (2) |G|=455    (3) |G|=21    (4) |G|=256
11/06/2018Test 2
11/07/2018Rings. Definitions: ring, unity, ring with unity (unitary ring), commutative ring, units of a unitary ring
Examples
Prop: The units of a ring, U(R) form a multiplicative group.
11/09/2018No class.

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people/fer/401ws/fall2018/daily_topics.txt · Last modified: 2018/11/19 08:57 by fer