Group tables for a group of four elements.

I should consider group tables obtained by renaming elements as essentially the same and then show that there are only two essentially different groups of order 4.

There seems to be so many different possible group tables for all the different binary operations - which is why I'm confused. I was thinking about using Cayley's table to show their commutativity but I'm really not too sure. Any help please!

Edit: After all of your help, I completely understand how to show that there are only two different groups of order four. Thank you. The only thing which I am still unclear of is how to note down the 'other tables' before stating that they are essentially the same as one of the other tables - meaning that there are just two different ones. It's pointless work but I think it's what the question requires.

Answer: After a bit of playing around - I've realised that there are four different tables, however, 3 tables are the same as each other, just with different values (the Klein 4 Group with 3 different generators). Hence there are two tables.

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.

The order of an element must divide $4$, so it must be either $1$, $2$, or $4$. Only the identity element can have order $1$. If one element has order $4$, then it generates the whole group and you have a cyclic (hence abelian) group.

Otherwise the three non-identity elements each have order $2$.

$e=\text{the identity}$.

$a,b,c=\text{the other three}$.

$a^2=b^2=c^2=e$, and each of these is its own inverse.

So what is $ab$? It can't be $e$ since if $ab=e$ then $(ab)b^{-1} = b^{-1}=b$, and hence $a=b$. It can't be $a$ or $b$ since if $ab=b$ then $(ab)b^{-1}=bb^{-1}=e$ and so $a=e$, and similar reasoning shows it can't be $a$ (but you have to multiply on the left be the inverse).

So $ab=c$.

Similar reasoning shows $ab=ba=c$, and $ac=ca=b$ and $bc=cb=a$.

So the group is abelian.