MATH 415 Lecture 24

« previous | Thursday, November 21, 2013 | next »

Group Rings and Group Fields

Consider group of two elements $G=\{e,a\}$ and field $F={\mathbb{\mathbb{Z}}}_2$. We construct the group ring ${\mathbb{\mathbb{Z}}}_{2}G=\{0e+0a,0e+1a,1e+0a,1e+1a\}=\{0,a,1,e+a\}$.

The addition and multiplication tables are:

$+$	$0$	$a$	$e$	$e+a$
$0$	$0$	$a$	$e$	$e+a$
$a$	$a$	$0$	$e+a$	$e$
$e$	$e$	$e+a$	$0$	$a$
$e+a$	$e+a$	$e$	$a$	$0$

$\cdot$	$0$	$a$	$e$	$e+a$
$0$	$0$	$0$	$0$	$0$
$a$	$0$	$e$	$a$	$e+a$
$e$	$0$	$a$	$e$	$e+a$
$e+a$	$0$	$e+a$	$a+a$	$0$

The Quaternions

This is a noncommutative division ring (skew field)

Discovered by Hamilton

Denoted $\mathbb{ℍ}={\mathbb{ℝ}}^4=\mathbb{ℝ}\times \mathbb{ℝ}\times \mathbb{ℝ}\times \mathbb{ℝ}$

Addition

Addition defined as $(a_1,a_2,a_3,a_4)+(b_1,b_2,b_3,b_4)=(a_1+b_1,a_2+b_2,a_3+b_3,a_4+b_4)$

"identity" elements:

• $1=(1,0,0,0)$
• $i=(0,1,0,0)$
• $j=(0,0,1,0)$
• $k=(0,0,0,1)$

For scalar multiplication (any $\alpha \in \mathbb{ℝ}$):

• $\alpha \cdot 1=(\alpha ,0,0,0)$
• $\alpha \cdot i=(0,\alpha ,0,0)$
• $\alpha \cdot j=(0,0,\alpha ,0)$
• $\alpha \cdot k=(0,0,0,\alpha )$

Thus any $(a_1,a_2,a_3,a_4)\in \mathbb{ℍ}$ can be written as the sum of components $a_1\cdot 1+a_2\cdot i+a_3\cdot j+a_4\cdot k$.

Multiplication

• for any $a\in H$, $1\cdot a=a\cdot 1=a$
• $1^2=1$ (idempotent)
• $i^2=j^2=k^2=-1$

Cyclic

• $ij=k$, $ji=-k$
• $jk=i$, $kj=-i$
• $ki=j$, $ik=-j$

Think of moving CCW around triangle:

  k

i   j

Carefully expand the following product:

$\begin{array}{rl}(a_1+a_{2}i+a_{3}j+a_{4}k)(b_1+b_{2}i+b_{3}j+b_{4}k)& =(a_1{b}_1-a_2{b}_2-a_3{b}_3-a_4{b}_4)\\ & +(a_1{b}_2+a_2{b}_1+a_3{b}_4-a_4{b}_3)i\\ & +(a_1{b}_3-a_2{b}_4+a_3{b}_1+a_4{b}_2)j\\ & +(a_1{b}_4+a_2{b}_3-a_3{b}_2+a_4{b}_1)k\end{array}$

Conjugates

If $a=a_1+a_{2}i+a_{3}j+a_{4}k$, the conjugate of $a$, denoted $\overline{a}$ is given by $\overline{a}=a_1-a_{2}i+a_{3}j-a_{4}k$

Now $a\cdot \overline{a}=a_1^2+a_2^2+a_3^2+a_4^2$

This also happens to be ${\left|a\right|}^2$, where $\left|a\right|=\sqrt{a\cdot \overline{a}}=\sqrt{a_1^2+a_2^2+a_3^2+a_4^2}$ is the "length" or "norm" of $a$.

By the way, $\left|a\right|=0$ if and only if $a=0$.

Observe $a\cdot \left(\frac{\overline{a}}{{\left|a\right|}^2}\right)=\frac{a\cdot \overline{a}}{{\left|a\right|}^2}=1$

Theorem 24.9

The quaternions $\mathbb{ℍ}$ form a strictly skew field under $+$, $\cdot$.

Theorem 24.10: Wedderburn's Theorem

Every finite division ring is a field.

Group Action on a Set

Let $X$ be a set and $G$ be a group

Definition

An action of $G$ on $X$ is a map $*:G\times X\to X$ such that

1. $e\cdot x=x$
2. $(g_1*g_2)*x=g_1*(g_2*x)$ for all $g_1,g_2\in G$ and for all $x\in X$(associativity)

The notation Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle G \lefttorightarrow X} means "$G$ acts on $X$"

Alternatively, we say that $X$ is a $G$-set.

Example

Consider $S_n$ (symmetric group of size $n$) and $X=\{1,2,\dots ,n\}$.

Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle S_n \lefttorightarrow X} : Let $\sigma \in S_n$, then $\sigma :x\mapsto \sigma (x)$

Also consider the alternating group $A_n<S_n$ consisting of even permutations. Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle A_n \lefttorightarrow X} .

• In general, if Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle G \lefttorightarrow X} and $H\le G$, then Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle H \lefttorightarrow X} as well.

If $\left|X\right|=\mathrm{\infty }$, then it happens that Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle S_X \lefttorightarrow X} (even for infinite sets!)

Theorem 16.3

Let $X$ be a $G$-set. Then for each $g\in G$, the function ${\sigma }_g:X\to X$ defined by ${\sigma }_g(x)=gx$ for $x\in X$ is a permutation of $X$.

Also, the map $\varphi :G\to S_X$ defined by $\varphi (g)={\sigma }_g$ is a homomorphism with the property that $\varphi (g)(x)=gx$

In other words, we claim that ${\sigma }_g:X\to X$ is a bijection:

1. injectivity: suppose ${\sigma }_g(x)={\sigma }_g(y)$ for distinct elements $x$ and $y$. Then $gx=gy$. Apply the inverse action to both sides, and we get $g^{-1}gx=g^{-1}gy$ implies $x=y$
2. surjectivity: ???

Properties

Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle G \lefttorightarrow X} is transitive if for all $x,y\in X$, there existts a $g\in G$ such that $gx=y$.

We define orbits of $x$ (notation: $O_x$) as

$O_x=\{y\mid y=gx\text{for some}g\in G\}$

In fact, $X$ is a disjoint union of orbits.

Membership in an orbit imposes an equivalence relation on $X$: each orbit $O_x$ represents the equivalence class $\left[x\right]$.

The action is transitive if and only if there is only one orbit.

Lazy Elements

We call an element "lazy" if $gx=x$ for all $x\in X$. Let $N$ denote the set of lazy elements. Then $N⊴G$.

In fact, $N$ is the kernel of our action ($N=\mathrm{K}\mathrm{e}\mathrm{r}(\varphi )$), in which case the kernel of a homomorphism is always a normal subgroup.

We say that the action Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle G \lefttorightarrow X} is faithful if $N=\left\{e\right\}$ (i.e. if the identity element is the only lazy element.

If this is the case, our homomorphism $\varphi$ is injective and therefore an isomorphism on its image $\varphi [G]\le S_X$.

Self-Actions

We can represent the action of a group on itself by left (or right) multiplication

Isotropy

Suppose Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle G \lefttorightarrow X} . Then for a particular element $x\in X$, we define the isotropy subgroup (or stabilizer of $x$) as

$G_x={\mathrm{S}\mathrm{t}}_G(x)=\{g\in G\mid gx=x\}$

Note: Do not confuse the isotropy subgroup with the lazy elements above!

Similarly, we can consider the elements $x\in X$ for which $gx=x$ for a particular $G$. The group $\mathrm{F}\mathrm{i}\mathrm{x}(g)=\{x\in X\mid gx=x\}\subseteq X$ of such elements contains fixed points for $g$.

Theorem 16.16

We define $Gx=\{gx\mid g\in G\}=O_x$. Observe that Failed to parse (unknown function "\lefttoright"): {\displaystyle G \lefttoright O_x} is a transitive relation (in fact it is an equivalence relation forming a single equivalence class).

Theorem.

• If $X$ is a $G$-set and $x\in X$, then $\left|Gx\right|=(G:Gx)$.
• If $\left|G\right|<\mathrm{\infty }$, then $\left|Gx\right|$ is a divisor of $\left|G\right|$

Proof. Let $H=G_x$ and let $\psi :Gx\to \{gH\mid g\in G\}$ (set of cosets of H, or $G/H$.

We will show that $\psi$ is a bijection:

injective. Let $x_1\in Gx$. Then there is $g_1\in G$ such that $g_{1}x=x_1$. Is $\psi (x_1)=g_{1}H$ well-defined? Suppose there exists $g_2\in G$ such that $g_{2}x=x_1$. By transitivity, we have

$\begin{array}{rl}{g}_{1}x& =g_{2}x\\ {g_2}^{-1}(g_{1}x=g_{2}x)\\ {g_2}^{-1}{g}_{1}x=\cancel{{g_2}^{-1}{g}_2}x\end{array}$

Therefore ${g_2}^{-1}{g}_1\in H$ and thus $g_{1}H=g_{2}H$.

surjective. For any $g_{1}H$ coset, we have $x_1=g_{1}x$ and $\psi (x_1)=g_{1}H$.

Therefore the order of the orbit $Gx$ is identical to the order of the set of cosets, which is the same as the index:

$\left|Gx\right|=\left|\{gH\mid g\in G\}\right|=(G:H)$

If

$\left|G\right|<\mathrm{\infty }$

, then

$\left|G\right|=\left|H\right|(G:H)$

.

quod erat demonstrandum

If we have two orbits $Gx$ and $Gy$, we claim that they are conjugate; that is, there exists an element $a\in G$ such that $Gx=a^{-1}Gya$:

Let $h\in Gy$. Then $\left(a^{-1}ha\right)(x)=a^{-1}h(a(x))=a^{-1}h(y)=a^{-1}(y)=x$. Hence $a^{-1}ha\in Gx$.