Groups

Semigroup, monoid, group

A semigroup is a nonempty set $G$ with a binary operation on $G$ that is

1. associative: $a(bc)=(ab)c$

a monoid is a semigroup with a

2. identity element $e\in G$ such that $ae=ea=a$

a group is a monoid $G$ such that

3. for every $a\in G$ there is an inverse $a^{-1}\in G$ such that $a^{-1}a=a{a}^{-1}=e$.

a gruop is abelian or commutative if its binary operation is

4. commutative: $ab=ba$

The order of a group $G$ is its cardinality $|G|$. For monoids, the identity element is necessarily unique: $e=e{e}^{\prime }=e^{\prime }$.

For a group $G$:

1. $cc=c$ implies $c=e$
2. $ab=ac$ implies $b=c$; likewise $ba=ca$ implies $b=c$
3. the inverse $a^{-1}$ is unique
4. For all $a\in G$, $(a^{-1}{)}^{-1}=a$.
5. For all $a,b\in G$, $(ab{)}^{-1}=b^{-1}{a}^{-1}$
6. The equations $ax=b$ and $ya=b$ have unique solutions, namely $x=a^{-1}b$ and $y=b{a}^{-1}$.

Subgroup

Let $G$ be a group and $H$ a nonempty subset that is closed under the product in $G$. If $H$ is a group, then $H$ is a subgroup of $G$ and denoted $H<G$.

As a quick test of subgroup: A nonempty subset $H$ of a group $G$ is subgroup of $G$ if and only if $a{b}^{-1}\in H$ for all $a,b\in H$.

Symmetric Group

Let $S$ be a nonempty set and $A(S)$ be the set of all bijections $S\to S$. Under function composition the set $A(S)$ is called the group of permutations. If $S=\{1,2,\dots ,n\}$ then $A(S)$ is the symmetric group on n letters and denoted $S_n$. The order of $S_n$ is $|S_n|=n!$.

An element $\sigma \in S_n$ is typically denoted

$$\left(\begin{array}{ccccc}1& 2& 3& \cdots & n\\ i_1& i_2& i_3& \cdots & i_n\end{array}\right)$$

For two elements $\sigma =\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 1& 2& 4\end{array}\right)$ and $\tau =\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)$ of $S_4$ the product $\sigma \tau$ maps $k\mapsto \sigma (\tau (k))$. For example, $1\mapsto \sigma (\tau (1))=\sigma (4)=4$. Thus,

$$\sigma \tau =\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 1& 2& 4\end{array}\right)\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 3& 1& 2\end{array}\right).$$

Direct Product

Let $G$ and $H$ be groups. Then the direct product $G\times H$ is a group with the underlying binary operation given by

$$(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b{b}^{\prime })\text{ where }a,a^{\prime }\in G,b,b^{\prime }\in H.$$

Clearly, the identity element is $(e_G,e_H)$ and for $(a,b)$ the inverse element is $(a^{-1},b^{-1})$. Moreover, $|G\times H|=|G||H|$.

If $G$ and $H$ are abelian, then $G\times H$ is abelian and we write $G\oplus H$.

Homomorphisms

Homomorphism

Let $G$ and $H$ be semigroups. A function $f:G\to H$ is a homomorphism if $f(ab)=f(a)f(b)\text{ for all }a,b\in G.$ If $f$ is injective it is called a monomorphism. If $f$ is surjective it is called a epimorphism. If $f$ is bijective it is an isomorphism and we say $G$ and $H$ are isomorphic (denoted $G\cong H$). If $f:G\to G$ then it is called an endomorphism; if it is an isomorphism then it is called an automorphism.

A few facts:

1. If $f\in \text{Hom}(G,H)$ and $g\in \text{Hom}(H,K)$, then $gf\in \text{Hom}(G,K)$.
2. For homomorphism $f:G\to H$, $f(e_G)=e_H$
3. For homomorphism $f:G\to H$, $f(a^{-1})=f(a{)}^{-1}$

Kernel, image

Let $f:G\to H$ be a homomorphism. The kernel of $f$ is the set $\text{Ker}f=\{a\in G\mid f(a)=e_H\}$ and the image of $f$ is the set $\text{Im}f=\{f(a)\mid a\in G\}.$

One may see that a homomorphism $f:G\to H$ is a monomorphism if and only if $\text{Ker}f=\{e_G\}$.

($\Rightarrow$): Let $a\in \text{Ker}f$. Then $f(a)=e_H=f(e_G)$. Since $f$ is injective, $a=e_G$.

($\Leftarrow$): Let $f(a)=f(b)$ for $a,b\in G$. Notice that $e_H=f(a)f(b{)}^{-1}=f(a)f(b^{-1})=f(a{b}^{-1})$ implies that $a{b}^{-1}=e_G$. Thus, $b^{-1}=a^{-1}$ implying that $a=b$.

The kernel is a subgroup of $G$ and the image is a subgroup of $H$.

1. Let $a,b\in \text{Ker}f$. Then $f(a{b}^{-1})=f(a)f(b{)}^{-1}=e_H{e}_H=e_H$.
2. Let $a,b\in \text{Im}f$. Notice that $\exists a^{\prime },b^{\prime }\in G$ such that $a=f(a^{\prime })$ and $b=f(b^{\prime })$. Notice that $a{b}^{-1}=f(a^{\prime })f(b^{\prime }{)}^{-1}=f(a^{\prime }(b^{\prime }{)}^{-1}).$