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Week 7 | Topics |
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10/01/2018 | Test 1 |
10/02/2018 | Lagrange's corollary 1 |
Orbit-Stabilizer theorem | |
Examples: cube, truncated icosahedron (soccer ball) | |
10/03/2018 | Corollaries 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 $p^2$ does not divide $|G|$, then $G$ has at most one subgroup of index $p$ (HW) | |
10/05/2018 | Thm. 7.3 |
Week 8 | Topics |
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10/08/2018 | Test 1 returned and reviewed |
Prop: if $\varphi:G\to H$ is an isomorphism, then so is $\varphi^{-1}H\to G$. | |
Prop: “isomorphic to” is an equivalence relation | |
Thm. 6.1 Cayley's theorem | |
$\aut(G)$, $\inn(G)$ | |
10/09/2018 | Thm 6.4 $\aut(G)$ is a group and $\inn(G)$ is a subgroup of $\aut(G)$ |
Example: $\inn(D_4) \isom K_4$ | |
Prop: Let $G = <a>$ cyclic and $H$ a group | |
1. A homom $\varphi:G\to H$ is uniquely determined by $\varphi(a)$. | |
2. If $G$ has order $n$ and $u\in H$ has order $d$ where $d|n$, then there is (unique) homomorphism $\varphi:G\to H$ s.t. $\varphi(a)=u$. Moreover, $\varphi$ is injective iff $d=n$. | |
3. If $G$ has infinite order and $u\in H$, then there is (unique) homomorphism $\varphi:G\to H$ s.t. $\varphi(a)=u$. Moreover, $\varphi$ is injective iff $u$ has infinite order. | |
Example: $\aut(\Z_n) \isom U_n$ | |
10/10/2018 | Board presentations PS 6 |
Thms. 10.2 and 6.3 | |
10/12/2018 | Fall break |
Week 9 | Topics |
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10/15/2018 | Prop. Let N \leq G. TFAE |
(i) $gNg^{-1} \subseteq N$ for all $g\in G$ | |
(ii) $gNg^{-1} = N$ for all $g\in G$ | |
(iii) $gN = Ng$ for all $g\in G$ | |
(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. $A_n \normaleq S_n$ | |
$<R_{360/n}> \normaleq D_n$ | |
Prop: if $H$ is a subgroup of $G$ of index 2, then $H$ is a normal subgroup of $G$ | |
2. Prop: if $\varphi:G\to \bar{G}$ is a homomorphism, then $ker(\varphi)$ 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 $\Z/n\Z \isom \Z_n$ | |
10/16/2018 | Example 9.12 |
Thm 10.3 1st Isom Thm | |
Example $\varphi:\Z \to \Z_n$ | |
Thm 9.4 | |
Thm (N/C theorem) Let $H \leq G$. $N_G(H) / C_G(H)$ 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 $Z(G) \neq 1$ then $G$ is abelian. | |
Thm 9.5 Cauchy's thm for abelian gps. | |
10/19/2018 | Thm 10.4 $N\normaleq G$, $q:G \to G/N$ is an epimorphism with $ker(q)=N$ |
Chapter 8 Direct Product | |
Def: $G_1 \oplus G_2$ | |
Prop: 1) $G_1 \oplus G_2$ is a group. | |
2) If $G_1$, $G_2$ are abelian then so is $G_1 \oplus G_2$. | |
Examples: (1) $\Z_2 \oplus \Z_3$ is abelian of order 6, so it is isomorphic to $\Z_6$ | |
(2) $G \oplus \{1\} \isom G \isom \{1\}\oplus G$ | |
Cor: $G_1 \oplus G_2$ contains subgroups isomorphic to $G_1$ and $G_2$ respectively. | |
Def: $G_1 \oplus \cdots \oplus G_n$ | |
Thm 8.1 |
Week 10 | Topics |
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10/22/2018 | Thm 8.2 $G_1$, $G_2$ finite. $G_1 \oplus G_2$ is cyclic iff $G_1$ and $G_2$ are cyclic or relatively prime orders. |
10/23/2018 | RSA cryptography. Public vs private keys |
Prop: $m^{ed}\equiv m \pmod n$. | |
Internal direct product | |
Thm.: Let $H,K\leq G$ be such that $HK=G$ and $H\intersection K=\{1\}$. Then $G\isom H\oplus K$. | |
Def: When $H,K\leq G$ are such that $HK=G$ and $H\intersection K=\{1\}$, we say that $G$ is the internal direct product of $H$ and $K$, and write $G=H\times K$. | |
Example: Consider $D_n$ with $n=2m$ and $m$ odd. | |
Thm. 9.7 and corollary | |
Prop: Let $H,N\leq G$. | |
(1) If $N\normaleq G$ then $HN\leq G$. | |
(2) If $H,N\normaleq G$ then $HN\normaleq G$ | |
10/24/2018 | 2nd, 3rd, 4th and 5th isomorphism theorems. |
$\sub(D_4)$ and $\sub(V_4)$ as examples. | |
10/26/2018 | Thm 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 11 | Topics |
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10/29/2018 | Board presentations. Problems sets 7 and 8 |
10/30/2018 | Ch.24 Def: conjugate, conjugate class $\cl(a)$. |
Prop: (1) “conjugate to” is an equivalence relation. The equivalence classes are the conjugacy classes. | |
(2) $\cl(a)=\{a\} \iff a\in Z(G)$ | |
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 $|G|=p^2$, then $G$ is abelian. | |
Cor. Let $p$ be a prime. If $|G|=p^3$, then $G$ is metabelian. Moreover, $|Z(G)|=p$ or $|Z(G)|=p^3$. | |
Example: Heisenber group $H$ has order $p^3$, and is not abelian. | |
10/31/2018 | Thm. 24.3 Sylow's 1st Theorem |
Cor. Cauchy's theorem | |
Cor. If $|G|=pq$ where $p<q$ are primes and $p\not\mid (q-1)$, then $G$ is cyclic. | |
Lemma 1. (1) Let $H\leq G$ and $C=\{gHg^{-1}\mid g\in G\}$ the set of all conjugates of $H$. Then $|C|=[G:N_G(H)]$. | |
Definition of Sylow $p$-subgroup. | |
(2) Let $H,K\leq G$. If $HK=KH$ then $HK\leq G$. | |
Lemma 2. Let $P$ be a Sylow $p$-subgroup $G$. If $g\in N_G(P)$ and $|g|$ is a power of $p$, then $g\in P$. | |
Lemma 3. Let $|G|=p^km$ and $p\not\mid m$. Let $P$ be a Sylow $p$-subgroup of $G$, i.e. $|P|=p^k$, and $H\leq G$ with $|H|=p^l$ for some $l\leq k$. Then there is a conjugate of $P$ that contains $H$, i.e. there is $g\in G$ such that $H\leq gPg^{-1}$. | |
11/02/2018 | Board presentations. Problems set 9 |
Proof of Lemma3 |
Week 12 | Topics |
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11/05/2018 | Sylow Theorems |
Examples: (1) $|G|=35$ (2) $|G|=455$ (3) $|G|=21$ (4) $|G|=256$ | |
11/06/2018 | Test 2 |
11/07/2018 | Rings. 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. | |
11/09/2018 | No class. |