MATH 415 Lecture 10

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Homorphism

Let $\left\langle G,\cdot \right\rangle$ and $\left\langle G',*\right\rangle$ be groups. The function $\phi :G\to G'$ is a homomorphism if and only if for all $x,y\in G$

$\phi (x\cdot y)=\phi (x)*\phi (y)$

Example: Integration

Let $F$ represent the group of all functions over addition.

$\phi :F\to \mathbb {R}$ where $\phi (f)=\int _{a}^{b}f(x)\,\mathrm {d} x$ is a homomorphism:

$\int _{a}^{b}(f+g)(x)\,\mathrm {d} x=\int _{a}^{b}f(x)\,\mathrm {d} x+\int _{a}^{b}g(x)\,\mathrm {d} x$

Images and Ranges

Let $\phi :X\to Y$ and $A\subseteq X$.

• $\phi \left[A\right]=\left\{\phi (a)~\mid ~a\in A\right\}$ is the image of $\phi$ over $A$
• $\phi \left[X\right]$ is the range of $\phi$.

It is also possible to find inverse images:

$\phi ^{-1}\left[B\right]=\left\{x\in X~\mid ~\phi (x)\in B\right\}$

Theorem 13.12

$\phi :G\to G'$ is a homomorphism

1. if $e$ is the identity element in $G$, then �$\phi (e)=e'$ is the identity in $G'$.
2. if $a\in G$, then $\phi (a^{-1})=\phi (a)^{-1}$.
3. if $H$ is a subgroup of $G$, then $\phi \left[H\right]$ is a subgroup in $G'$.
4. if $K'$ is a subgroup of $G'$, then $\phi ^{-1}\left[K'\right]$ is a subgroup in $G$.

Proof.

1. $\phi (a)=\phi (a\,e)=\phi (a)\,\phi (e)=\phi (a)\,e'$. By transitivity, $\phi (a)=\phi (a)\,e'$, so $\phi (e)=e'$.
2. Since $e'=\phi (e)=\phi (a\,a^{-1})$, we can say $\phi (a\,a^{-1})=\phi (a)\,\phi (a^{-1}=e'$. Therefore $\phi (a^{-1})=\phi (a)^{-1}$.
3. Let $\phi (a),\phi (b)\in \phi \left[H\right]$ for $a,b\in G$. Since $G$ is a group, then $a\,b\in G$ and thus $\phi (a\,b)=\phi (a)\,\phi (b)\in \phi \left[H\right]$.Furthermore, if $\phi (a)\in \phi \left[H\right]$, then $\phi (a^{-1})=\phi (a)^{-1}\in \phi \left[H\right]$.

quod erat demonstrandum

Kernel

Let $\phi :G\to G'$ be a homomorphism. $\left\{e'\right\}$ is the trivial subgroup of $G'$.

We define $\phi ^{-1}\left[\left\{e'\right\}\right]=\left\{x\in G~\mid ~\phi (x)=e'\right\}$ to be the kernel of $\phi$.

The kernel is a subgroup in $G$ that maps all of its members to $e'$ in $G'$. Thus the kernel is collapsing down to $e'$.

Example

Let $\phi :\mathbb {R} ^{n}\to \mathbb {R} ^{m}$, where $\phi ({\vec {x}})=A\,{\vec {x}}$, where $A$ is a $m\times n$ matrix.

The kernel of $\phi$ is the null space of $A$

Theorem 13.15

Let $\phi :G\to G'$ be a homomorphism, $H=\mathrm {kernel} (\phi )$, and let $a\in G$. Then the set $\phi ^{-1}\left[\left\{\phi (a)\right\}\right]=\left\{x\in G~\mid ~\phi (x)=\phi (a)\right\}$ is the left coset $a\,H$ and at the same time is the right coset $H\,a$ (thus implying $a\,H=H\,a$)

Proof. Consider $x\in a\,H$, thus $x=a\,h$ for $h\in H$. Because $\phi$ is a homomorphism, we have $\phi (x)=\phi (a\,h)=\phi (a)\,\phi (h)=\phi (a)\cdot e'=\phi (a)$.

$x\in A$, then $\phi (x)=\phi (a)$, $\phi (x)\,\phi (a)^{-1}=e'=\phi (x)\,\phi (a^{-1})$. Therefore $(x\,a^{-1}=h)\in (H=\mathrm {kernel} (H))$ implies $x=h\,a$, so $x\in H\,a$. a similar strategy shows that $x\in a\,H$. Thus the right and left cosets are identical.

quod erat demonstrandum

Example

Let $D$ be the group of differentiable mappings from $\mathbb {R} \to \mathbb {R}$ over addition.

Therefore $f'$ is well-defined. Consider $\phi :D\to D$ with $\phi (f)=f'$.

Observe that $\phi$ is a homomorphism because $\phi (f+g)=\phi (f)+\phi (g)=f'+g'$.

The kernel of $\phi$ is the set of all functions whose derivative is 0, thus it is the set of constant functions.

Corollary

Corollary. A group homomorphism $\phi :G\to G'$ is injective if and only if its kernel is $\left\{e\right\}$.

If $\left|\mathrm {kernel} (\phi )\right|>1$, then $\phi$ is not one-to-one since $x,y\in \mathrm {kernel} (\phi )$ for $x\neq y$. Alternatively, if $\mathrm {kernel} (\phi )=\left\{e\right\}$, then assuming $\phi (a)=\phi (x)=e$, we arrive at the contradiction that the kernel must have at least two elements.

Relation to Isomorphism

Recall that if $\phi :G\to G'$ is an isomorphism, then

1. $\phi$ is a homomorphism
2. $\phi$ is injective (hence its kernel is $\left\{e\right\}$)
3. $\phi$ is surjective

Normal Subgroups

A subgroup $H$ of a group $G$ is normal if its left and right cosets coincide. That is, if $g\,H=H\,g$ for all $g\in G$.

Notation: $H\trianglelefteq G$

Corollary. The kernel of a homomorphism $\phi$ is a normal subgroup of $G$.

Factor Groups

Given a group $G$ partitioned into cosets, we want to collapse all cosets $g\,H$ (for $g\in G$) into individual elements in another group $G'$. In general, this is not possible, but it is possible if $H$ is a normal subgroup.

Theorem 14.1

Let $\phi :G'$ be a group homomorphism and $H=\mathrm {kernel} (\phi )$. Then the cosents of $H$ form a group (called a factor group; denoted $G/H$), where $(a\,H)\,(b\,H)=(a\,b)\,H$. The identity element is $e\,H=H=\mathrm {kernel} (\phi )$.

Furthermore, the map $\mu :G/H\to \phi \left[G\right]$ defined by $\mu (a\,H)=\phi (a)$ is an isomorphism.

Both coset multiplication and $\mu$ are well-defined independent of the choices $a$ and $b$ from the cosets.

Note: It's important to note that each coset contains multiple elements, so choosing any representative still refers to the same coset.

Proof. By definition, $(a\,H)(b\,H)=(a\,b)\,H$. What about $(a_{1}\,H)(b_{1}\,H)$? If $a_{1}\,H=a\,H$, then there exists an $h_{1}\in H$ such that $a_{1}=a\,h_{1}$. Similarly, if $b_{1}\in b\,H$, then $b_{1}=b\,h_{2}$ for $h_{2}\in H$. Thus $a_{1}\,b_{1}=a\,h_{1}\,b\,h_{2}$ Since $h_{1}\,b\in H\,b$ and $H$ is a normal subgroup, we can say $H\,b=b\,H$, so $h_{1}\,b\in b\,H$ implies $h_{1}=b\,h_{3}$ for some $h_{3}\in b\,H$. Thus $a_{1}\,b_{1}=a\,b\,h_{3}\,h_{2}$. $H$ is closed, so let $h_{3}\,h_{2}=h'\in H$. Thus $(a_{1}\,b_{1})\,H=(a\,b)\,H$.

…

quod erat demonstrandum