Formalisation of groups and monoids

The test.tex file was deleted and all formalisations of groups and monoids was moved to the fitting document of the library. Some proof steps of the new formalisation were optimized for proof time

2 files changed, 101 insertions, 1 deletions

diff --git a/library/algebra/group.tex b/library/algebra/group.tex
index 48934bd..a79bd2f 100644
--- a/library/algebra/group.tex
+++ b/library/algebra/group.tex
@@ -1 +1,82 @@
-\section{Groups}
+\import{algebra/monoid.tex}
+\section{Group}
+
+\begin{struct}\label{group}
+    A group $G$ is a monoid such that
+    \begin{enumerate}
+        \item\label{group_inverse} for all $g \in \carrier[G]$ there exist $h \in \carrier[G]$ such that $\mul[G](g, h) =\neutral[G]$.
+    \end{enumerate} 
+\end{struct}   
+
+\begin{corollary}\label{group_implies_monoid}
+    Let $G$ be a group. Then $G$ is a monoid.
+\end{corollary}
+
+\begin{abbreviation}\label{cfourdot}
+    $g \cdot h = \mul(g,h)$.
+\end{abbreviation}
+
+\begin{lemma}\label{neutral_is_idempotent}
+    Let $G$ be a group. $\neutral[G]$ is a idempotent element of $G$.
+\end{lemma}
+
+\begin{lemma}\label{group_divison_right}
+    Let $G$ be a group. Let $a,b,c \in G$.
+    Then $a \cdot c = b \cdot c$ iff $a = b$.
+\end{lemma}
+\begin{proof}
+    Take $a,b,c \in G$ such that $a \cdot c = b \cdot c$.
+    There exist $c' \in G$ such that $c \cdot c' = \neutral[G]$.
+    Therefore $a \cdot c = b \cdot c$ iff $(a \cdot c) \cdot c' = (b \cdot c) \cdot c'$.
+    \begin{align*}
+        (a \cdot c) \cdot c'
+            &= a \cdot (c \cdot c')
+                \explanation{by \cref{semigroup_assoc,group_implies_monoid,monoid_implies_semigroup}}\\
+            &= a \cdot \neutral[G]
+                \explanation{by \cref{group_inverse}}\\
+            &= a
+                \explanation{by \cref{group_implies_monoid,monoid_right}}
+    \end{align*}
+    \begin{align*}
+        (b \cdot c) \cdot c'
+            &= b \cdot (c \cdot c')
+                \explanation{by \cref{semigroup_assoc,group_implies_monoid,monoid_implies_semigroup}}\\
+            &= b \cdot \neutral[G]
+                \explanation{by \cref{group_inverse}}\\
+            &= b
+                \explanation{by \cref{group_implies_monoid,monoid_right}}
+    \end{align*}
+    $(a \cdot c) \cdot c' = (b \cdot c) \cdot c'$ iff $a \cdot c = b \cdot c$ by assumption. 
+    $a = b$ iff $a \cdot c = b \cdot c$ by assumption.
+\end{proof}
+
+
+\begin{proposition}\label{leftinverse_eq_rightinverse}
+    Let $G$ be a group and assume $a \in G$.
+    Then there exist $b\in G$ 
+    such that $a \cdot b = \neutral[G]$ and $b \cdot a = \neutral[G]$.
+\end{proposition}
+\begin{proof}
+    There exist $b \in G$ such that $a \cdot b = \neutral[G]$.
+    There exist $c \in G$ such that $b \cdot c = \neutral[G]$.
+    $a \cdot b = \neutral[G]$.
+    $(a \cdot b) \cdot c = (\neutral[G]) \cdot c$.
+    $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
+    $a \cdot \neutral[G] = \neutral[G] \cdot c$.
+    $c = c \cdot \neutral[G]$.
+    $c = \neutral[G] \cdot c$.
+    $a \cdot \neutral[G] = c \cdot \neutral[G]$.
+    $a \cdot \neutral[G] = c$ by \cref{monoid_right,group_divison_right}.
+    $a = c$ by \cref{monoid_right,group_divison_right,neutral_is_idempotent}.
+    $b \cdot a = b \cdot c$.
+    $b \cdot a = \neutral[G]$.
+\end{proof}
+
+\begin{definition}\label{group_abelian}
+    $G$ is an abelian group iff $G$ is a group and for all $g,h \in G$ $\mul[G](g,h) = \mul[G](h,g)$. 
+\end{definition}
+
+
+\begin{definition}\label{group_automorphism}
+    Let $f$ be a function. $f$ is a group-automorphism iff $G$ is a group and $\dom{f}=G$ and $\ran{f}=G$. 
+\end{definition}
diff --git a/library/algebra/monoid.tex b/library/algebra/monoid.tex
new file mode 100644
index 0000000..06fcb50
--- /dev/null
+++ b/library/algebra/monoid.tex
@@ -0,0 +1,19 @@
+\import{algebra/semigroup.tex}
+\section{Monoid}
+
+\begin{struct}\label{monoid}
+    A monoid $A$ is a semigroup equipped with
+    \begin{enumerate}
+        \item $\neutral$
+    \end{enumerate}
+    such that
+    \begin{enumerate}
+        \item\label{monoid_type} $\neutral[A]\in \carrier[A]$.
+        \item\label{monoid_right} for all $a\in \carrier[A]$ we have $\mul[A](a,\neutral[A]) = a$.
+        \item\label{monoid_left} for all $a\in \carrier[A]$ we have $\mul[A](\neutral[A], a) = a$.
+    \end{enumerate}
+\end{struct}
+
+\begin{corollary}\label{monoid_implies_semigroup}
+    Let $A$ be a monoid. Then $A$ is a semigroup.
+\end{corollary}
\ No newline at end of file