MTH3003 Lecture 2

Simon Smith

I accidentally finished

As discussed briefly in the last lecture, the set of all permutations of a set is a Group, albeit not an Abelian (commutative) one, unless disjoint. To show these, we need to understand what the inverse of a cycle is…

Quick Inverse Proposition

Let $X$ be a set and suppose $\sigma \in \text{Sym}(X)$. We know that we can write $\sigma$ as a product of disjoint cycles, $\sigma =c_1{c}_2\dots c_n$, hence:

1. The inverse of any cycle $(x_1{x}_2\dots x_m)$ is $(x_m{x}_{m-1}\dots x_1)$, and
2. ${\sigma }^{-1}=c_1^{-1}{c}_2^{-1}\dots c_n^{-1}$.

This can be proven directly.

Groups

If we want to prove that the set $\text{Sym}(X)$ is a group, then we must recall the definition of a group…

A Group is a set $G$ together with an operation $\ast$ on $G$ such that each of the following holds:

1. Closure: for all $\forall g,h\in G$, $g\ast h\in G$.
2. Existence of an identity element: $\exists e_G\in G$ such that for all $g\in G$, $g\ast e_G=e_G\ast g=g$.
3. Existence of inverse elements: For all $g\in G$ there exists an inverse element $g^{-1}\in G$ such that $g\ast g^{-1}=g^{-1}\ast g=e_G$.
4. Associativity: for all $g,h,k\in G$, $g\ast (h\ast k)=(g\ast h)\ast k$.
5. Rarely, if Abelian, Commutativity: for all $g,h\in G$, $g\ast h=h\ast g$.

Usually however, we don’t write the operation $\ast$, instead $g\ast h$ would be written as $gh$. By definition, $g^0:=e_G$ for all $g\in G$.

Recall also that a subgroup $H$ of a group $G$ is just a subset of $G$ that is also a group under the operation of $G$. Also recall that if $G$ and $H$ are groups, then their direct product, or Cartesian product, $G\times H$ is also a group, where each element is operated on each other piecewise.

Given that a group $G$ follows these axioms for all $g\in G$, then that also gives the group six properties:

1. $G$ has only one identity.
2. $g$ has only one inverse.
3. $(g^{-1}{)}^{-1}=g$.
4. Let $h_1,h_2\in G$. If $g{h}_1=g{h}_2$ or $h_{1}g=h_{2}g$ then $h_1=h_2$.
5. $(h_1{h}_2{)}^{-1}=h_2^{-1}{h}_1^{-1}$.
6. $(g^n{)}^{-1}=g^{-n}$.

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Rough-Lecture Notes from University Notes

Didn’t have access to lecture notes beforehand, so did rough notes during instead of pre-lecture notes - enjoy the yapping!