\problemdata
{S00 tut1}
{intro}

\begin{problem}

A {\em group} is a set $G$ together with a binary operation $\cdot: G
\times G \rightarrow G$ that satisfies the following axioms.


Identify at least two ``normal'' sets and operations on them
  that satisfy the group axioms.

\begin{enumerate}

\item $\forall a,b,c \in G,(a \cdot b) \cdot c = a \cdot (b \cdot c)$.

\item $\exists e \in G$ (called the {\em identity}) 
such that $\forall a \in G, a \cdot e = e \cdot a = a$.

\item $\forall a \in G, \exists a^{-1} \in G$ (called the {\em inverse} 
of $a$), such that $a \cdot a^{-1} = a^{-1} \cdot a = e$.

\end{enumerate}

So addition defines a group on the integers, and multiplication a
group on the positive rationals.  

Using the three group axioms prove the following propositions.

\begin{problemparts}

\problempart Show, by contradiction, that the identity element is unique.

\solution{%
  \textbf{Proof by Contradiction.} Suppose there are two identities
  $e$ and $f$.  Then $ef=e$ since $f$ is an identity, but $ef=f$ since
  $e$ is an identity.  Thus $e=f$, a contradiction.  }

\ppart Show that for every $a \in G$, the inverse of $a$ is unique.
We can therefore write the inverse element as $a^{-1}$ with no fear of
ambiguity.

\solution{%
  \textbf{Proof by Contradiction.} Suppose there are two inverses $b$
  and $c$.  Then $ab=e$, which implies that $c(ab)=ce=c$.  But also
  $c(ab)=(ca)b=eb=b$ since $c$ is an inverse.  So $b=c$, a
  contradiction.  }

\problempart Show $\forall a \in G, (a^{-1})^{-1} = a$

\solution{%
  \textbf{Direct Proof.} Write $b=a^{-1}$.  By definition of $a^{-1}$,
  we have $ba=ab=e$.  Therefore, $a$ satisfies the ``inverse''
  definition for $b$---that is, $a=b^{-1}=(a^{-1})^{-1}$.  }

\problempart Show $\forall a,b \in G, (a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$

\solution{
We have
\[
(b^{-1}a^{-1})(ab) = (b^{-1})(a^{-1}a)b = b^{-1}eb = b^{-1}b = e.  
\]
Therefore $(b^{-1}a^{-1})$ satisfies the inverse definition for $ab$.
}


\end{problemparts}
\end{problem}


