I was asked to find number of different binary operation which forms a groups for a set containing 4 elements in a class test.
So first what i did was select an identity element 'e'. (Let other element be a,b,c)
Necessary conditions :- ee=e , ae=ea=a , ec=ce=c , eb=be=b
case 1) aa=e , bb = e , cc=e
Now ab cannot be equal to e (that would imply a=b) or a,b (that would imply b=e or a=e).
Similarly , ab=ba=c , ac=ca=b , bc=cb=a
(Only one way from 3 elements)
Case 2) ab=ba=e and that would imply cc=e
As pointed above : bc=cb=a , ca=ac=b
(ab)c=a(bc) implies c=aa and similarly bb= c
(3 ways to do from 3 elements , selecting x such that xx=e has 3 ways)
Then i checked associativity for both case 1 and case 2 , which pretty much holded .
So the ways of what we want is = (numer of ways to select an identity element) × (number of ways of (case1) + number of ways of (case2))
= 4×(1+3)=16.
Is the method and answer correct. And as you can see it's not really a good method (especially due to need of checking of associativity).
Is there more simple way of checking associativity (i made each combination and checked it). Is there a more simpler way to do this , or for a general 'n' or can we just do it for smaller 'n' by only counting each case? Is there any recursions that we can develop?
Any thoughts , link , a way to make calcutions etc. , All are appreciated. Thanks for the help.
Your answer is correct.
The slightly more advanced way to do this for a set of size $n$ is to compute the isomorphism types of groups of order $n$, then compute $n!\sum \frac{1}{|\mathrm{Aut}(G)|}$, where the sum runs over the isomorphism types. For $n=4$, there are two isomorphism types: $C_4$, with automorphism group of order $2$, and $C_2^2$, with automorphism group of order $6$. So we get $4!(1/2+1/6)=16$.