TITLE: Group tables for a group of four elements.
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.