MATH 415 Lecture 24

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Group Rings and Group Fields

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

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

• $i\,j=k$, $j\,i=-k$
• $j\,k=i$, $k\,j=-i$
• $k\,i=j$, $i\,k=-j$

Think of moving CCW around triangle:

  k

i   j

Carefully expand the following product:

${\begin{aligned}(a_{1}+a_{2}i+a_{3}j+a_{4}k)\,(b_{1}+b_{2}i+b_{3}j+b_{4}k)&=(a_{1}\,b_{1}-a_{2}\,b_{2}-a_{3}\,b_{3}-a_{4}\,b_{4})\\&+(a_{1}\,b_{2}+a_{2}\,b_{1}+a_{3}\,b_{4}-a_{4}\,b_{3})i\\&+(a_{1}\,b_{3}-a_{2}\,b_{4}+a_{3}\,b_{1}+a_{4}\,b_{2})j\\&+(a_{1}\,b_{4}+a_{2}\,b_{3}-a_{3}\,b_{2}+a_{4}\,b_{1})k\end{aligned}}$

Conjugates

If $a=a_{1}+a_{2}\,i+a_{3}\,j+a_{4}\,k$, the conjugate of $a$, denoted ${\bar {a}}$ is given by ${\bar {a}}=a_{1}-a_{2}\,i+a_{3}\,j-a_{4}\,k$

Now $a\cdot {\bar {a}}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}$

This also happens to be $\left|a\right|^{2}$, where $\left|a\right|={\sqrt {a\cdot {\bar {a}}}}={\sqrt {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}}}$ is the "length" or "norm" of $a$.

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

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

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 $X=\left\{1,2,\ldots ,n\right\}$.

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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n \lefttorightarrow X} .

• In general, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G \lefttorightarrow X} and $H\leq G$, then Failed to parse (unknown function "\lefttorightarrow"): {\displaystyle H \lefttorightarrow X} as well.

If $\left|X\right|=\infty$, 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 $\sigma _{g}(x)=g\,x$ for $x\in X$ is a permutation of $X$.

Also, the map $\phi :G\to S_{X}$ defined by $\phi (g)=\sigma _{g}$ is a homomorphism with the property that $\phi (g)(x)=g\,x$

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 $g\,x=g\,y$. Apply the inverse action to both sides, and we get $g^{-1}\,g\,x=g^{-1}\,g\,y$ 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 $g\,x=y$.

We define orbits of $x$ (notation: $O_{x}$) as

$O_{x}=\left\{y~\mid ~y=g\,x\ {\mbox{for some}}\ g\in G\right\}$

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 $g\,x=x$ for all $x\in X$. Let $N$ denote the set of lazy elements. Then $N\trianglelefteq G$.

In fact, $N$ is the kernel of our action ($N=\mathrm {Ker} (\phi )$), 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 $\phi$ is injective and therefore an isomorphism on its image $\phi [G]\leq S_{X}$.

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

$G_{x}=\mathrm {St} _{G}(x)=\left\{g\in G~\mid ~g\,x=x\right\}$

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

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

Theorem 16.16

We define $Gx=\left\{g\,x~\mid ~g\in G\right\}=O_{x}$. Observe that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 $\left|G\right|<\infty$, then $\left|Gx\right|$ is a divisor of $\left|G\right|$

Proof. Let $H=G_{x}$ and let $\psi :Gx\to \left\{gH~\mid ~g\in G\right\}$ (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

${\begin{aligned}g_{1}\,x&=g_{2}\,x\\{g_{2}}^{-1}\,\left(g_{1}\,x=g_{2}\,x\right)\\{g_{2}}^{-1}\,g_{1}\,x={\cancel {{g_{2}}^{-1}\,g_{2}}}\,x\end{aligned}}$

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:

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

If $\left|G\right|<\infty$, 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 $Gx=a^{-1}\,Gy\,a$:

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