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\begin{document}

\noindent
{\bf Describe all groups of size $3$ up to isomorphism.}

\medskip

How do you describe all groups of a certain size?
We begin by experimenting. Since we are told
that the size is $3$, we can assume that the set
of elements we will use is $G = \{a, b, c\}$.
We need to try to construct group operations on this
set. For multiplication, we need to fill in 
the following table:

\[\begin{array}{|c||c|c|c|}
\hline
* & a & b & c \\
\hline
\hline
a &  &  &  \\
\hline
b &  &   &  \\
\hline
c &  &   &  \\
\hline
\end{array}\]

\noindent
One of these elements, say $a$, will be the identity
element, so that helps us to
fill out part of the table. But we still need to fill in 
the remaining holes:

\[\begin{array}{|c||c|c|c|}
\hline
* & a & b & c \\
\hline
\hline
a & a & b & c \\
\hline
b & b &   &  \\
\hline
c & c &   &  \\
\hline
\end{array}\]

\noindent
If we try certain guesses and see what happens, we soon
realize that each row and column must contain each 
element at most once. Otherwise, multiplication 
will not be cancellative. As it turns out, there is
only one way to complete the table so that each row and column
contains every element at most once. It is:

\[\begin{array}{|c||c|c|c|}
\hline
* & a & b & c \\
\hline
\hline
a & a & b & c \\
\hline
b & b & c & a \\
\hline
c & c & a & b \\
\hline
\end{array}\]

\noindent
What about the other operations?
Are there different kinds of inverse
operations that would work with this
multiplication table? Is there a group
with this multiplication table, where the specified
identity element is not $a$ but is $b$?

It turns out that the identity element 
is uniquely determined by the multiplication,
since the only element $u$ that satisfies $u*u = u$
is the identity element.
Therefore, by scanning the ``diagonal''
of the previous multiplication table, 
we see that the only choice for the
identity element is $a$.

It turns out that the inverse operation is
uniquely determined by the multiplication
table, since if $u\in G$, then
$u^{-1}$ is the unique solution to
$u*x = 1$. If we know the multiplication
we can determine the inverse.

Thus, the other operations \underline{\it must be}:
\[ \begin{array}{lcr}
1 = a & \hspace{1in} &
{\begin{array}{|c||c|c|c|}
\hline
x & a & b & c\\
\hline
\hline
x^{-1} & a & c & b\\
\hline
\end{array}}
\end{array}\]

After doing all this work, we feel ``morally certain''
that there is only one group of size three.
We feel this way because we made no ``free choices''
while filling out the tables: everything was
forced on us once we said that $a$ = identity.
But how do we write the proof that there is only 
one group???

The obvious proof is to try to follow what we
did informally, and turn our moral certainty
into absolute certainty. This is a messy approach
since you must consider several cases to prove
that there is essentially one way to fill out the
multiplication table.
But if you study the multiplication table you will
notice something: $b^2 = c$. Therefore, all of $a, b$ and
$c$ are powers of $b$. That means that $G$ is a 
cyclic group! All we need to show is that

\begin{enumerate}
\item[(1)] every $3$-element group is cyclic, and 
\item[(2)] two cyclic groups of the same
size are isomorphic.
\end{enumerate}

How do you write this up?

\bigskip

\noindent
{\it Solution.}
We first argue that every $3$-element group is cyclic.
Assume that $G = \{1, b, c\}$ and that 1 is the 
identity. We cannot have $bc = b$, for then cancellation
yields $c = 1$ which is false.
Similarly we cannot have $bc = c$. Thus we must have
$bc = 1$, or $b^{-1} = c$. 

Now we cannot have $b^2 = b$, since $b\neq 1$, and 
we cannot have $b^2 = 1$, since this would
force $b^{-1} = b$ and we already showed that
$b^{-1} = c\neq b$. The only remaining possibility
is that $b^2 = c$. This proves
that $G = \{1, b, b^2\}$ and that $b^3 = 1$.
So $G$ is indeed cyclic.

Let's prove that two cyclic groups of the same
size are isomorphic. Assume that 
$G = \{1, b, b^2, \ldots, b^k\}$ and
$H = \{1, c, c^2, \ldots, c^k\}$ are cyclic groups
of the same size. Define $\varphi:G\rightarrow H$
by $\varphi(b^i) = c^i$. This is clearly a bijection.
To see that it is a homomorphism, it is enough
to see that multiplication is preserved:
%$\varphi(b^i*b^j)=\varphi(b^i)*\varphi(b^j)$.
%This is argued as follows:
\[\varphi(b^i*b^j)=\varphi(b^{i+j})=c^{i+j}
=c^i*c^j = \varphi(b^i)*\varphi(b^j). \hspace{.3in}\Box\]

\bigskip

Soon we will learn that every group of prime
order is cyclic. The argument just given applies
to show that there 
is only one group of order $p$ for each prime $p$.

\end{document}

