Homomorphisms, Normal Subgroup & Conjugation

Homomorphisms

Group Homomorphism

Let $G$ and $H$ be groups. A homomorphism $\phi :G\to H$ is a map that satisfies $\phi (ab)=\phi (a)\phi (b)$ for all $a,b\in G$.

Proposition

If $\phi :G\to H$ is a group homomorphism, then the identity of $G$ is mapped to the identity of $H$.

Kernel & Image

Let $\phi :G\to H$ be a group homomorphism. The kernel and the image of $\phi$ are defined by $\begin{array}{rl}\ker (\phi )& =\{g\in G\mid \phi (g)=e_H\}\subset G,\\ \mathrm{im}(\phi )& =\{h\in H\mid \exists g\in G,\phi (g)=h\}\subset H.\end{array}$

Proposition

A homomorphism $\phi :G\to H$ is injective if and only if $\ker (\phi )=\{e_G\}$.

Proof If $\phi$ is injective then clearly $\ker (\phi )=e$. Conversely, suppose that $\ker (\phi )=\{e_G\}$ and let $a,b\in G$ with $\phi (a)=\phi (b)$. Then $\phi (a^{-1}b)=\phi (a{)}^{-1}\phi (b)=e_H$. Hence $a^{-1}b=e_G$ and therefore $a=b$. $\square$

Normal Subgroup

Conjugate

Let $G$ be a group. For any $a,b\in G$, we say $a$ and $b$ and conjugate if there exists $g\in G$ such that $a=gb{g}^{-1}$.

Normal Subgroup & Simple Group

Let $G$ be a group. A subgroup $N\le G$ is called normal if $gN{g}^{-1}\subset N$ for every $g\in G$. We denote a normal subgroup by $N◃G$.
A group is simple if there’s no non-trivial normal subgroups.

e.g. Let $\phi :G\to H$ be a homomorphism, then $\ker (\phi )◃G$. Fix some $x\in \ker (\phi )$, then for all $g\in G$ we have $\phi (gx{g}^{-1})=\phi (g)\phi (x)\phi (g^{-1})=\phi (g)\phi (g{)}^{-1}=e$.

Proposition

The followings are true:

1. $N\le G$ is normal iff the right coset $Ng$ coincides with the left coset $gN$ for every $g\in G$.
2. We required in the definition that $gN{g}^{-1}\subset N$ , but it follows that $gN{g}^{-1}=N$.
3. Every subgroup of an abelian group is normal.
   

Normaliser

Let $G$ be a group and $H\le G$. The normaliser is the subgroup of $G$ that
$N_G(H)=\{g\in G\mid gH{g}^{-1}=H\}\supseteq H.$
In other words, it is the largest subgroup of $G$ in which $H$ is normal.

Proposition

$N_G(H)◃G$.

Isomorphisms

Group Isomorphism

A group homomorphism $\phi :G\to H$ is called isomorphism if it has an inverse, that is a homomorphism $\psi :H\to G$ such that $\psi \circ \phi =e_G$ and $\phi \circ \psi =e_H$.

Proposition

Prop A homomorphism $\phi :G\to H$ is an isomorphism if and only if it is bijective.

Proposition

Prop Groups of prime order are cyclic.

Proposition

Prop Any two cyclic group of the same order are isomorphic.

Definition

Def Lambda-Map
For every $g\in G$ define a map by left multiplication with $g$: ${\lambda }_g:G\to G,x\mapsto gx$

Proposition

Prop ${\lambda }_g\in \text{Sym}(G)$.

Proposition

Prop ${\lambda }_{g^{-1}}$ is the inverse of ${\lambda }_g$.

Cayley’s Theorem

The map $\lambda :G\to \text{Sym}(G)$ given by $g\mapsto {\lambda }_g$ is an injective group homomorphism.

Proof Clearly $\lambda (x)\circ \lambda (y)=\lambda (xy)$. Set $\lambda (g)={\lambda }_g=\text{id}$ where $\text{id}$ is the identity map from $G$ to $G$. Then for all $x\in G$, we have $gx=x$. It follows that $g=1$. Thus $\lambda$ is injective. $\square$

Corollary

Corollary Every finite group is isomorphic to a subgroup of a symmetric group $S_n$.
Proof For any finite group $G$, the $\lambda$ map forms an isomorphism, thus $G\cong S_n$ for some $n$.

Definition

Def Automorphism
An automorphism is an isomorphism from a group $G$ to $G$. Denote $\text{Aut}(G)$ as the set of all automorphisms on $G$.

Conjugation

Conjugation

Let $G$ be a group and $g\in G$. The map${\gamma }_g:G\to G,a\mapsto ga{g}^{-1}$is called conjugation by $g$. We say that the elements $a$ and $ga{g}^{-1}$ are conjugate.

Proposition

Let $G$ be a group and $g\in G$. Then ${\gamma }_g:G\to G$ is an isomorphism.

Proof We first check that ${\gamma }_g$ is a homomorphism:${\gamma }_g(ab)=gab{g}^{-1}=ga(g^{-1}g)b{g}^{-1}=(ga{g}^{-1})(gb{g}^{-1})={\gamma }_g(a){\gamma }_g(b)$The map is injective because ${\gamma }_g(a)=e⟹ga{g}^{-1}=e⟹ga=g⟹a=e$. And is surjective since for all $a\in G$, we have ${\gamma }_g(g^{-1}ag)=a$. $\square$