I'm studying algebra with some friends, and i will try to write here what subjects are we studying right now.
We start with Abstract Algebra using Hungerford book, but i want to follow the MITOpenCourseWare about Algebra I (http://ocw.mit.edu/OcwWeb/Mathematics/18-701Fall-2007/Readings/index.htm).
Composition Law:A composition law $f$ select a pair of elements on a set $S$ and return an element of $S$, ie:
$f : S \times S \rightarrow S$, I will write $f(a,b)$ as $a*b$ for seek of simplicity.
If $*$ holds that $(a*b)*c = a*(b*c) \quad \forall a,b,c \in S$ then we say that is an associative law.
If $*$ holds that $a*b=b*a \quad \forall a,b \in S$ we say that is a commutative law.
SemigroupsLet $S$ be a set and $*$ be a composition law on $S$, if $*$ is associative then $(S,*)$ is a semigroup.
MonoidsLet $(S,*)$ be a semigroup if $S$ has an identity element $e$ such that:
$\forall x \in S$ $x*e = x = e*x$,
then $(S,*)$ is a Monoid.
GroupsLet $(S,*)$ be a Monoid if:
$\forall x \in S$ $\exists x^{-1} \in S$ s.t. $x*x^{-1} = e = x^{-1}*x$,
then $(S,*)$ is a Group.
if $*$ is commutative then $(S,*)$ is a commutative or Abelian group.
SubgroupsLet $(G,*)$ be a group, if a proper subset $H$ of $G$ is also a group with $*$ then $H$ is a subgroup of $G$
TheoremIf $(G,*)$ is a group and $H$ is a nonempty subset of $G$:
$(H,*)$ is a subgroup of $(G,*)$ if and only if $a*b^{-1} \in H \quad \forall a,b \in H$
RelationLet $A$ and $B$ be sets and $R$ a map from $A$ to $B$ s.t.:
$R : A \rightarrow B$
with $a\in A$ and $b=R(a) \in B$ and
$(a,b) = (a,R(a)) \in A \times B$
for seek of simplicity we write $(a,R(a)) \in A \times B$ as $aRb$.
Equivalence RelationsLet $S$ be a set and $R$ a relation in $S \times S$, if:
i) $aRa \quad \forall a \in S$ (Reflexive)
ii) $aRb \implies bRa$ (Symmetric)
iii) $aRb \wedge bRc \implies aRc$ (Transitive)
then $R$ is an Equivalence Relation on $S$. (When we speak of an equivalence relation we'll use the symbol $=$)
Partial Order RelationsA partial order relation is a relation $R$ in a set $S$ with a equivalence relation $=$ if $R$ is reflexive and transitive and:
$aRb \wedge bRa \implies a=b$ (antisymmetric)
We write a partial order relation as $\leq$. The above oreder is a non-strict relation.
A strict partial relation order $R$ in $S$ is:
i) $a\not R a \quad \forall a \in S$ (irreflexive)
ii) $aRb \wedge bRa \implies a=b$ (antisymmetric)
iii) $aRb \wedge bRc \implies aRc$ (Transitive)
We write a strict partial order relation as $<$.
Complete Order RelationsThe above partial relations don't hold $\forall a,b \in S$, if they do then the order is a complete order.
Congruence RelationLet $(G,*)$ be a Monoid, if a equivalence relation $=$ on $G$ also holds that:
$a_1=a_2 \wedge b_1 = b_2 \implies a_1*b_1 = a_2*b_2$,
then $=$ is also congruence relation.
Equivalence ClassesLet $S$ be a set with an equivalence relations $=$, the equivalence class $\bar{a}$ (or $[a]$) of an element $a \in S$ is:
$\bar{a} = \{ x \in S | x = a \}$
The class of all the equivalence classes on $S$ is denoted as $(S/=)$, and is called the quotient class of $S$ by $=$.
The union of all the equivalence classes is $S$, i.e.
$\bigcup_{a \in S} \bar{a} = A = \bigcup_{\bar{a} \in (S/=)} \bar{a}$ or $\bar{a} = \bar{b}$
If $a,b \in S$ either $\bar{a} \cap \bar{b} = \emptyset$
Theorem:Let $(G,*)$ be a Monoid and $=$ a congruence relation on $G$, then:
$((G/=), *)$ is a Monoid with
$* : (G/=) \times (G/=) \rightarrow (G/=)$ and $[a]*[b] = [a*b] \quad [a],[b] \in G/=$ (sorry for the change on notation but the bar don't expand enough in the a*b)
If $(G,*)$ is Abelian then $(G/=,*)$ is Abelian too.
Group HomomorphismLet $(G,\circ)$ and $(H,\diamond)$ be two groups and a function $f : G \rightarrow H$ such that:
$f(x \circ y ) = f(x) \diamond f(y) \quad \forall x,y \in G$,
then $f$ is a Group homomorphism.
If $f$ is injective then $f$ is a Group Monomorphism.
If $f$ is surjective then $f$ is a Group Epimorphism.
If $f$ is bijective then $f$ is a Group Isomorphism.
If a Group Homomorphism is $f: G \rightarrow G$ then $f$ is a Group Endomorphism of $G$.
If a Group Isomorphism is $f: G \rightarrow G$ then $f$ is a Group Automorphism of $G$.
KernelLet $(G,\circ)$ and $(H,\diamond)$ be two groups and a homomorphism $f : G \rightarrow H$. $e_G$ and $e_H$ are the ideintity element of each group, then, the kernel of $f$ is:
$Ker f = \{a \in G | f(a) = e_H\}$,
The kernel is a subset of $G$ that maps to the identity element on $H$ with the homomorphism $f$.
Theorem:If $f : G \rightarrow H$ is a monomorphism $\Longleftrightarrow$ $Ker f = \{e_G\}$
Theorem:If $f : G \rightarrow H$ is a isomorphism $\Longleftrightarrow$ $f^{-1} : H \rightarrow G$ is a homomorphism and $f(f^{-1}(a)) = a \quad \forall a \in H$ and $f^{-1}(f(a)) = a \quad \forall a \in G$
Cyclic groupsLet $(G,*)$ be a group and $X$ a subset of $G$.
Let $\{ H_i | i \in I \}$ be the family of all the subgroups ($H_i$) of $G$ that contains $X$ i.e. $X \subset H_i \forall i \in I$
$\bigcap_{i \in I} H_i$ is a subgroup of $G$ generated by $X$ and denoted $\langle X\rangle$
The elements of $X$ are the generators of $\langle X\rangle$.
Several different subsets on $G$ can generate the same $\langle X\rangle$, so, in general $\langle X\rangle = \langle Y \rangle$ with $X \neq Y$.
If $X$ is a finite set such that $X = \{a_1, a_2, \cdots , a_n \}$, we write $\langle a_1, a_2, \cdots, a_n \rangle$ instead of $\langle X\rangle$.
If $G = \langle a_1,a_2, \cdots, a_n \rangle$ with $a_i \in G$, then $G$ is said to be finitely generated.
If $a \in G$, then the subgroup $\langle a\rangle$ is the Cyclic Group or Cyclic Subgroup generaterd by a.
TheoremLet $(G,*)$ be a group and $X$ a nonempty subset of $G$, then the subgroup $\langle X \rangle$ generated by $X$ consists of all finite products $a_1^{n_1}a_2^{n_2}\cdots a_1t^{n_t} \quad a_i \in X; \quad n_i \in \mathbb{Z}$.
And $\forall a \in G, \langle a \rangle = \{ a^n | n \in \mathbb{Z}\}$
TheoremLet $H$ be a cycli subgroup of $(\mathbb{Z},+)$, either $H=\langle 0 \rangle$ or $H=\langle m \rangle$, with $m$ the least positive integer in $H$, then if $H \neq \langle 0 \rangle$, then $H$ is infinite.
TheoremEvery infinite cyclic group is isomorphic to $(\mathbb{Z},+)$ and every finite cyclic group of order $m$ is isomorphic to $(\mathbb{Z}_m,+)$ (¿¿¿$\langle m \rangle$???).
CosetsFirst we define that $a \equiv b (mod m) \Longleftrightarrow m | a - b \Longleftrightarrow a-b \in \langle m \rangle$.
Let $(G,*)$ be a group, and $(H,*)$ a subgroup of $(G,*)$ and $a,b \in G$.
a is
right congruent to $b$ modulo $H$, denoted as $a \equiv_r b (mod H)$ if $a*b^{-1} \in H$.
a is
left congruent to $b$ modulo $H$, denoted as $a \equiv_l b (mod H)$ if $a^{-1}*b \in H$.
If $G$ is abelian then if one of the congruencies hold then the other does:
$a*b^{-1} \in H \Longleftrightarrow (a*b^{-1})^{-1} \in H$
$(a*b^{-1})^{-1} = b*a^{-1} = a^{-1}*b$
and
$a^{-1}*b \in H \Longleftrightarrow (a^{-1}*b)^{-1} \in H$
$(a^{-1}*b)^{-1} = b^{-1}*a = a*b^{-1}$
TheoremLet $(H,*)$ be a subgroup of $(G,*)$, then
i) Right(Left) congruence modulo $H$ is an equivalence relation on $G$.
ii) The equivalence class of $a \in G$ with the right (left) congruence is the set $Ha = \{h*a | h \in H\} ($aH = \{a*h | h \in H\})
iii) $|Ha| = |H| = |aH| \quad \forall a \in G$
$Ha$ is called a right coset of $H$ in $G$.
$aH$ is called a left coset of $H$ in $G$.
Corollaryi) $G$ is the union of the right(left) cosets of $H$ on $G$.
ii) Two right(left) cosets of $H$ on $G$ are either disjoint or equal.
iii) $\forall a,b \in G, \quad Ha = Hb \Longleftrightarrow a*b^{-1} \in H$ and $aH = bH \Longleftrightarrow a^{-1}*b \in H$
iv) If $R$ is the set of distinct right(left) cosets of $H$ in $G$, and $L$ is the set of distinct left cosets of $H$ in $G$, then
DefinitionLet $(G,*)$ be a group, and $(H,*)$ a subgroup of $(G,*)$ the index of $H$ in $G$, denoted as $[G : H]$ is the cardinal number of the set of distinct right(left) cosets. of $H$ in $G$.
