<s>[INST] Finding all groups of order $7$ up to isomorphism? I'm learning group theory but I didn't learn any concepts of building groups.
I know that there exists the identity group $\{e\}$, and the group with 2 elements: $\{e,a\}$.
If I try to create a group with 3 elements, let's say: $\{e,a,b\}$ then we would have:
$ea = a, eb = b, aa = ?$, and what about $ab$?
Am I supposed to try this for the $7$ elements?
What does the statement ``up to isomorphism" mean? In particular, how many groups of order $7$ are there up to isomorphism?
I'm really confused, and my book says nothing about it.
 [/INST] Groups of order $1$, $2$, $3$, $5$ and $7$ are unique up to isomorphism: for $1$ it's obvious, the others are prime numbers.
Let's tackle the $4$ case; one of the elements is the identity, so the group is $\{1,a,b,c\}$. Every group of even order has an element of order $2$ (a very easy case of Cauchy's theorem). Let's say it's $a$, so $a^2=1$.
Make the Cayley table for the group, where we fill in the information we already have. 
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 \\
b & b\\
c & c\\
\end{array}
$$
For the product $ab$ we have only one choice, namely $c$, because an element cannot appear twice in a row or column. Hence $ac=b$. Similarly, $ba=c$ and $ca=b$. So we have
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c \\
c & c & b \\
\end{array}
$$
Now we have two choices for $c^2$: either $c^2=1$ or $c^2=a$. Let's look at what we have, because once we decide for $c^2$ the other slots can be filled in uniquely:
$$
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & 1 & a \\
c & c & b & a & 1
\end{array}
\qquad\qquad
\begin{array}{c|cccc}
 & 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & a & 1 \\
c & c & b & 1 & a
\end{array}
$$
The structure on the left is Klein's group, the one on the right is the cyclic group. So we have two groups of order $4$.
For the groups with six elements we need something more advanced, because the Cayley table method is too long if started from nothing. Another application of Cauchy's theorem says that the group has a cyclic subgroup of order $3$. So there are three elements $1,a,b$ such that $a^2=b$, $ab=ba=1$ and $b^2=a$. Let's fill in this information for the set $\{1,a,b,c,d,e\}$. Again we can also assume $c^2=1$.
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 &   &   &   \\
b & b & 1 & a &   &   &   \\
c & c &   &   & 1 &   &   \\
d & d \\
e & e
\end{array}
$$
For $ac$ we can choose either $ac=d$ or $ac=e$; it doesn't matter which one, because at the end exchanging $d$ with $e$ will give an isomorphism; so let's choose $ac=d$ that forces $ad=e$ and $ae=c$
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c &   &   & 1 &   &   \\
d & d \\
e & e
\end{array}
$$
Let's now examine $ca$: again we have either $ca=d$ or $ca=e$, but this time we can't choose one and we need to go with two alternatives:
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c & d &   & 1 &   &   \\
d & d & e \\
e & e & c
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a &   &   &   \\
c & c & e &   & 1 &   &   \\
d & d & c \\
e & e & d
\end{array}
$$
We now have no choice for $bc$: in both cases it must be $e$; similarly, $bd=c$ and also $cb$ has a determined result (and we can also fill in some other slots):
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 &   &   \\
d & d & e & c &   &   &   \\
e & e & c & d &   &   &
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 &   &   \\
d & d & c & e &   &   &   \\
e & e & d & c
\end{array}
$$
We're almost done. Observe that interchanging $a$ and $b$ is irrelevant, because both are generators of the (normal) subgroup with three elements. So we can arbitrarily choose $cd=a$ in both cases and we can complete the tables. Indeed, if in the table on the left we choose $dc=b$, we'd have $a(dc)=ab=1$, but $(ad)c=ec=a$ against associativity. Similarly for the table on the right.
$$
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 & a & b \\
d & d & e & c & a & b & 1 \\
e & e & c & d & b & 1 & a
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
  & 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 & a & b \\
d & d & c & e & a & b & 1 \\
e & e & d & c & b & 1 & a
\end{array}
$$
So we have a maximum of two groups of order six. But we know two non isomorphic one: the cyclic group and $S_3$. So we're done.
</s>