MATH 415 Lecture 10

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Homorphism

Let $⟨G,\cdot ⟩$ and $⟨G^{\prime },*⟩$ be groups. The function $\varphi :G\to G^{\prime }$ is a homomorphism if and only if for all $x,y\in G$

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

Example: Integration

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

$\varphi :F\to \mathbb{ℝ}$ where $\varphi (f)=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$ is a homomorphism:

$\underset{a}{\overset{b}{\int }}(f+g)(x)\mathrm{d}x=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x+\underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$

Images and Ranges

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

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

It is also possible to find inverse images:

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

Theorem 13.12

$\varphi :G\to G^{\prime }$ is a homomorphism

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

Proof.

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

quod erat demonstrandum

Kernel

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

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

The kernel is a subgroup in $G$ that maps all of its members to $e^{\prime }$ in $G^{\prime }$. Thus the kernel is collapsing down to $e^{\prime }$.

Example

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

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

Theorem 13.15

Let $\varphi :G\to G^{\prime }$ be a homomorphism, $H=\mathrm{k}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{e}\mathrm{l}(\varphi )$, and let $a\in G$. Then the set ${\varphi }^{-1}\left[\{\varphi (a)\}\right]=\{x\in G\mid \varphi (x)=\varphi (a)\}$ is the left coset $aH$ and at the same time is the right coset $Ha$ (thus implying $aH=Ha$)

Proof. Consider $x\in aH$, thus $x=ah$ for $h\in H$. Because $\varphi$ is a homomorphism, we have $\varphi (x)=\varphi (ah)=\varphi (a)\varphi (h)=\varphi (a)\cdot e^{\prime }=\varphi (a)$.

$x\in A$

, then

$\varphi (x)=\varphi (a)$

,

$\varphi (x)\varphi (a{)}^{-1}=e^{\prime }=\varphi (x)\varphi (a^{-1})$

. Therefore

$(x{a}^{-1}=h)\in (H=\mathrm{k}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{e}\mathrm{l}(H))$

implies

$x=ha$

, so

$x\in Ha$

. a similar strategy shows that

$x\in aH$

. 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 $f^{\prime }$ is well-defined. Consider $\varphi :D\to D$ with $\varphi (f)=f^{\prime }$.

Observe that $\varphi$ is a homomorphism because $\varphi (f+g)=\varphi (f)+\varphi (g)=f^{\prime }+g^{\prime }$.

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

Corollary

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

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

Relation to Isomorphism

Recall that if $\varphi :G\to G^{\prime }$ is an isomorphism, then

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

Normal Subgroups

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

Notation: $H⊴G$

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

Factor Groups

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

Theorem 14.1

Let $\varphi :G^{\prime }$ be a group homomorphism and $H=\mathrm{k}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{e}\mathrm{l}(\varphi )$. Then the cosents of $H$ form a group (called a factor group; denoted $G/H$), where $(aH)(bH)=(ab)H$. The identity element is $eH=H=\mathrm{k}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{e}\mathrm{l}(\varphi )$.

Furthermore, the map $\mu :G/H\to \varphi \left[G\right]$ defined by $\mu (aH)=\varphi (a)$ 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, $(aH)(bH)=(ab)H$. What about $(a_{1}H)(b_{1}H)$? If $a_{1}H=aH$, then there exists an $h_1\in H$ such that $a_1=a{h}_1$. Similarly, if $b_1\in bH$, then $b_1=b{h}_2$ for $h_2\in H$. Thus $a_1{b}_1=a{h}_{1}b{h}_2$ Since $h_{1}b\in Hb$ and $H$ is a normal subgroup, we can say $Hb=bH$, so $h_{1}b\in bH$ implies $h_1=b{h}_3$ for some $h_3\in bH$. Thus $a_1{b}_1=ab{h}_3{h}_2$. $H$ is closed, so let $h_3{h}_2=h^{\prime }\in H$. Thus $(a_1{b}_1)H=(ab)H$.

…

quod erat demonstrandum