In: Advanced Math
(You are welcome to analyze groups with 5 elements using the same technique, but you will soon know enough about groups to be able to avoid such brute-force approaches.)
If G has exactly one element say then we must have
the group
identity and so the table is fixed at
* | ![]() |
![]() |
![]() |
If G has exactly two elements, one of them must be group
identity so we must have
The multiplications
must hold and
So we must have the
group identity and so the group table is fixed at
* | ![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
If G has exactly 3 elements we must
have
Then we must have
Consider the element . This can't be
equal to
since
This means we must have and
similarly
And so our table is again fixed as:
* | ![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |
![]() |