MATH 415 Lecture 13

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Simple Group

$G$ is called a simple group if the only normal subgroups are trivial.

That is $H\trianglelefteq G$ implies $H=\left\{e\right\}$ or $H=G$.

Example

Alternating group $A_{n}$ is an example of a simple group if $n>5$.

Maximal Group

File:MATH 415 Alternate Group Notation 1.svg

Alternate Subgroup Notation

$H\leq G$. If there are no intermediate subgroups between $H$ and $G$, then $H$ is called maximal.

That is, if $H\leq K\leq G$, either $K=G$ or $K=H$.

Maximal Normal Subgroup

Consider the case above where $H$ and $K$ are normal in $G$.

Theorem 15.18

$H$ is a maximal normal subgroup in $G$ if and only if $G/H$ is a simple group.

If we have $\gamma :G\to G/H$, and $N\triangleleft G/H$, then $\gamma ^{-1}\left(N\right)\trianglelefteq G$ for $\gamma ^{-1}:N\to N$. Furthermore $H=\gamma ^{-1}\left(e_{G/H}\right)\to \left\{e_{G/H}\right\}$.

Every finite group is either simple or has a maximal normal subgroup.

Not sure whether Maximal Normal Subgroups (or was it Simple Groups?) can be classified as finitely generated abelian groups can.

Center Subgroup

Group $G$.

$Z(G)\leq G$ is called the center, where

$e\in \mathbb {Z} (G)=\left\{x\in G~\mid ~x\,g=g\,x\forall g\in G\right\}$

If $G$ is abelian, then the entire group is the center (all elements are commutative)

In fact, this group is normal ($Z(G)\trianglelefteq G$), so we can compute its quotient $G/Z(G)$, and then take its center $Z(G/Z(G))$, etc. until we get the trivial group.

Commutator Subgroups

The commutator subgroup $[G,G]$ is the subgroup of $G$ generated by all the commutators:

$[G,G]=G'=\left\langle [a,b],\ a,b\in G\right\rangle$

Where $[a,b]=a^{-1}\,b^{-1}\,a\,b$ is called the commutator of $a$ and $b$

Moreover, it is a nontrivial normal subgroup in $G$

Theorem: $G/G'$ is an abelian group.

Generating Sets

Given a group $G$, if $A\subset G$ (subset)

$\left\langle A\right\rangle =\left\{g\in G~\mid ~g=a_{i_{1}}^{\pm 1}\,a_{i_{2}}^{\pm 1}\dots a_{i_{n}}^{\pm 1}\right\}$ for all $a_{i_{j}}\in A$.

This is a subgroup in $G$, and we call $\left\langle A\right\rangle$ the subgroup $G$ generated by $A$.

If $\left\langle A\right\rangle =G$, then $A$ is a generating set for $G$.

If $B\supset A$, then $B$ is also a generating set

In trivial case, we can take $A=G$ to be the whole generating set.

If there is a finite generating set $|A|<\infty$, then $G$ is called finitely generated.

Examples

Cylcic Groups: groups generated by a single element; of form $\left\langle a\right\rangle$

$\mathbb {Z}$ can be generated by 1, and it can also be generated by $2,3$, hence $A=\left\{2,3\right\}$ is a generating set for $\mathbb {Z}$.

$\mathbb {Z} \times \mathbb {Z}$ can be generated by $\left\{(0,1),(1,0)\right\}$.

$\left\langle \mathbb {Q} ,+\right\rangle$ is not finitely generated. However, if we take a finite subset $A\subset \mathbb {Q}$ and generate a group, that group will be cyclic. (where $\mathbb {Q}$ itself is not cyclic.

Cayley Graph

Graph consists of vertices and edges, so $\Gamma =(V,E)$

• vertices are elements of the group ($V=G$)
• (directed) edges are of form $(g,a\,g)$ (labeled by $a$) for $g\in G$ and $a\in A$, where $A$ is a generating set.

Examples

$\mathbb {Z} ,A=\left\{1\right\}$

For $\mathbb {Z} ,A=\left\{2,3\right\}$

Free Groups

Usually denoted $F_{k}$.

They have generating set $A=\left\{a_{1},\ldots ,a_{k}\right\}$, and it is called a free generating set or a basis

Suppose we take $A\cup A^{-1}=\left\{a_{1}\ldots ,a_{k},a_{1}^{-1},\ldots ,a_{k}^{-1}\right\}$.

Freely reduced words in $A\cup A^{-1}$ is set of products of all generating elements (e.g. $a_{1}\,a_{2}\,a_{1}^{-1}\,a_{3}\,a_{2}$.

• Reduced in that no word has $\dots a_{i}\,a_{i}^{-1}\dots$ inside since this can be reduced to $e$ and omitted.

We define our operation on $F_{k}$ as follows: If we have $U,V\in F_{k}$, then we just concatenate them $U\,V$ and cancel at concatenation point.

For example, $U=a_{1}\,a_{2}\,a_{5}$ and $V=a_{5}^{-1}\,a_{2}^{-1}\,a_{1}$: the product $U\,V=a_{1}\,a_{2}\,a_{5}\,a_{5}^{-1}\,a_{2}^{-1}\,a_{1}=a_{1}\,a_{1}$.

$F_{k}$ forms a group:

• Inversion: The inverse element is reversal and inversion of "letters": $(a_{1}\,a_{2}\,a_{3})^{-1}=a_{3}^{-1}\,a_{2}^{-1}\,a_{1}^{-1}$
• Identity: Empty word $e$.
• Associativity: easy.

Cayley diagram is a cool-looking snowflake / tree pattern centered at $e$ with branching degree $2\,\left|A\right|$.

In free group, there are no relations to other words; only new words can be created from each node.

Presentation of a Group by Generators and Relators

We present $F_{k}=\left\langle a_{1},\ldots ,a_{k}~\mid ~\emptyset \right\rangle$. The first part of "bra-ket" is the generating set, and the second part is the set of relations.

Consider $G=\left\langle b_{i},\ldots ,b_{k}\right\rangle$ and a map $\phi :A\to B$ where $\phi (a_{i})\to b_{i}$.

Theorem

$\phi$ naturally extends to a subjective homomorphism $\phi :F_{k}\to G$. Hence all finitely generated groups are homomorphic to a free group, and moreover $G\simeq F_{k}/\mathrm {Ker} (\phi )$.

$R\subset G$

We denote the minimal normal subgroup in $G$ containing $R$ as $\langle \langle R\rangle \rangle$

Consider $R\subset F_{k}$.

Suppose $G$ is a $k$-generated group, hence it is isomorphic to $F_{k}/N$, where $N=\langle \langle R\rangle \rangle$.

$G=\left\langle a_{1},\ldots ,a_{k}~\mid ~r=1\quad r\in R\right\rangle$

Consider $G=\langle \langle a,b~|~[a,b]=1\rangle \rangle \simeq F_{2}/\langle \langle [a,b]\rangle \rangle \simeq \mathbb {Z} ^{2}$

If $G$ has a presentation of the form $G=\left\langle a_{1}\,\ldots ,a_{k}~\mid ~r=1\quad r\in R\right\rangle$, where $R$ is finite, we say that $G$ is finitely presented.

Not every finitely generated group can be finitely presented.