TITLE: Prove that every group of order $4$ is abelian
QUESTION [36 upvotes]: How can I show that every group of order $4$ is abelian?
Let's denote $e,a,b,c$ as the four elements of the group.
Since $e$ is identity, we have $e*x=x*e$ for every $x$:
$$\begin{array}{|c|cccc|}
\hline
    & e & a & b & c \\\hline
  e & e & a & b & c \\
  a & a &   &   &   \\
  b & b &   &   &   \\
  c & c &   &   &   \\\hline
\end{array}$$
Now $a*a=$? We have several possibilities.
If I choose $a*a=b$ I can show $a*b=a*(a*a)=(a*a)*a=b*a$. 
But so far I have shown only that $a$ and $b$ commute, there are also other pairs. And I have only discussed the choice $a*a=b$, there are also other possibilities. Is there a simpler way to do this?

REPLY [3 votes]: There aren't as many choices as you'd think.
$(a, b, c)^4 = e$ so $(a^2, b^2, c^2)^2  = e$. So either $a^2 = e$ or $a^2 = b$ or $c$ and $b^2$ or $c^2 = e$.  Whatever.  Either way there is some non-identity element, x, is such that $x^2 = e$. 
We might as well assume that element is $a$ (without loss of generality).  So $a^2 = e$.
Now you can fill in the third row and column with:
$\begin{array}{|c|cccc|}
\hline
    & e & a & b & c \\\hline
  e & e & a & b & c \\
  a & a & e &  c & b  \\
  b & b & c  &   &   \\
  c & c & b  &   &   \\\hline
\end{array}$
There's only 4 squares left.  The only choice you have is $ b^2 = a$ or $b^2 = e$.  So basically there were only two options:
1: $\begin{array}{|c|cccc|}
\hline
    & e & a & b & c \\\hline
  e & e & a & b & c \\
  a & a & e &  c & b  \\
  b & b & c  & e  &  a \\
  c & c & b  &  a &  e \\\hline
\end{array}$
or
2: $\begin{array}{|c|cccc|}
\hline
    & e & a & b & c \\\hline
  e & e & a & b & c \\
  a & a & e &  c & b  \\
  b & b & c  &  a &  e \\
  c & c & b  &  e &  a \\\hline
\end{array}$
Note:
$\#1 \cong \mathbb Z_4$ with $b = b; b^2 = a; b^3 = c; b^4 = e$
And  $\#2 \cong \mathbb Z_2 \times \mathbb Z_2$ With $a^2 = b^2 = c^2 = e; a*b =c = b*a; b*c = a = c*b; c*a =b=a*c$. 
That's it.  Only two groups of order 4.