Introduction to Groups Review

# Introduction to Groups Review

We will now review some of the recent material regarding our introduction to groups.

• On the Groups page we formally defined an algebraic structure known as a group. Recall that if $\cdot$ is a binary operation on $G$ then $G$ is a Group under the operation $\cdot$ denoted $(G, \cdot)$ if the following properties are satisfied:
• 1) For all $a, b, c \in G$ we have that $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ (Associativity of elements in $G$ under $\cdot$).
• 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (Existence of an identity element in $G$ for $\cdot$).
• 3) For all $a \in G$ there exists an $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (Existence of inverses elements for each element in $G$ under $\cdot$).
• Also recall that $(G, \cdot)$ is a Finite Group if $G$ is a finite set and is an Infinite Group if $G$ is an infinite set. The Order of $(G, \cdot)$ is defined to be the number of elements in $G$ and is denoted $\mid G \mid$.
• On the Subgroups and Group Extensions page we said that if $(G, \cdot)$ is a group, $H \subseteq G$, and $(H, \cdot)$ forms a group then $(H, \cdot)$ is said to be a Subgroup of $(G, \cdot)$. Furthermore, $(G, \cdot)$ is said to be a Group Extension of $(H, \cdot)$.
• We then looked at a very nice theorem which gives us a simpler criterion for determining whether a subset of a group is a subgroup. We proved that $(S, \cdot)$ is a subgroup of $(G, \cdot)$ if and only we have that both:
• $S$ is closed under $\cdot$.
• For every $a \in S$ there exists an $a^{-1} \in S$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$.
• On the Basic Theorems Regarding Groups page we looked at quite a few elementary results regarding groups which are summarized in the following table. Let $(G, \cdot)$ be a group and let $e \in G$ be the identity element.
Theorem
The identity element $e$ is unique.
For every $a \in G$, $a^{-1}$ is unique.
For every $a \in G$, $(a^{-1})^{-1} = a$.
For every $a, b \in G$, $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$.
For every $a, b \in G$, if $a \cdot b = e$ then $a = b^{-1}$ and $b = a^{-1}$.
For every $a \in G$, if $a^2 = a \cdot a = a$ then $a = e$.
• On the The Cancellation Law for Groups page we noted a very important result regarding groups called the Cancellation Law which says that if $(G, \cdot)$ is a group and $a, b, c \in G$ are such that $a \cdot b = a \cdot c$ or $b \cdot a = c \cdot a$ then $b = c$.
(1)
\begin{align} \quad a^m &= \underbrace{a \cdot a \cdot ... \cdot a}_{m \: \mathrm{many}} \\ \quad a^{-m} &= \underbrace{a^{-1} \cdot a^{-1} \cdot ... \cdot a^{-1}}_{m \: \mathrm{many}} \\ \quad a^0 &= e \end{align}
• And we proved some basic laws of exponents, such as $a^m \cdot a^n = a^{m + n}$ for all $a \in G$ and for all $m, n \in \mathbb{Z}$.
• On the The Intersection and Union of Two Subgroups page we looked at two results regarding groups. We saw that if $(G, \cdot)$ is a group and $(S, \cdot)$, $(T, \cdot)$ are two subgroup of $(G, \cdot)$ then we ALWAYS have that $(S \cap T, \cdot)$ is also a subgroup of $(G, \cdot)$.
• However we proved that $(S \cup T, \cdot)$ is a subgroup of $(G, \cdot)$ if and only if $S \subseteq T$ or $S \supseteq T$.
• We then looked at a bunch of examples of groups which are linked below: