Group tables for a group of four elements.

Hints: we are talking about groups of order 4, which narrows the possible binary relations and elements available:

• each must contain the identity,
• each must be associative,
• each must be closed under inverses (if $a \in G,\;a^{-1} \in G$)
• (and of course, closed under the group operation).

There are essentially (up to isomorphism) only two groups of order 4:

• one of course will be the additive cyclic group $\mathbb{Z}_4$, and
• the other will be the Klein 4-group, which is indeed abelian.

If you follow the suggestions for solving the problem, you'll find, indeed, that any possible GROUP of order 4 can be shown isomorphic to one of the two groups mentioned simply by a renaming elements.

Yes: use of the Cayley table will be of great importance:

• for a group: no element can appear twice in any column,
• no element can appear twice in any row.

You'll find that the only ways of completing a table satisfying these criteria are limited, and then you show that a simple renaming of elements will reveal the group is isomorphic to $\mathbb{Z}_4$ or else the Klein-4 group.

There aren't that many. One of the elements will be the identity, let's call it $1$. (I am writing the operation as multiplication.) So the table looks like $$ \begin{matrix} 1 & a & b & c\\ a\\ b\\ c\\ \end{matrix} $$ (I'm not writing the row and column labels, as they are the same as the first column and row). Now what could $ab$ be? Not $a$, not $b$... And $ba$? Proceed with $ac$, $ca$, $bc$, $cb$, and then you are left with the squares. They might all be $1$, or...

This simplifies it:

If a group has 4 elements, we separate two cases:

Case1: all the elements have order 2, then the group is abelian, and $a^2=1\Rightarrow a=a^{-1}$ for all $a\in G$: $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ the table should be easy to make in this case: there is only one.

Case2: there is an element $a\not =1$ that has order different from 2, then by LAgrange's theorem, the order must be 4, and so $G=\langle a\rangle$, and so the group is cyclic, and therefore it's abelian. We have proved that a 4-element group is always abelian. The table shouldn't be difficult in this case either.