Ifis a group of order 6, the orders of each element may only be 1, 2, 3 or 6, since the order of an element must divide the order of the group.
Ifhas an element of order 6, it is cyclic and is isomorphic to
Supposehas no element of order 6, but has an elementof order 3, thencontains the elementsThere must be some elementso by composingwithandwe obtain the setBy cancellation none of these elements can be equal so these elements constitute the group. We can partially fill out the group table as below.
|
||||||
|
|
|
||||
|
|
|
||||
|
|
|
||||
|
|
|
||||
|
|
|
can't be equalby cancellation (ifthen), and can't beor (since each ealement can only appear once in each row and column. We are left withor
Supposethen
so the order ofis 6 and the group is cyclic, but the group hase no element of order 6. The same analysis is done for the caseso that
Now consider the productin the second row, the same row asorand the same column asso can't be equal to any of them, soor
SupposeThe powers ofare
since
and the order ofis 6. As before this is a contradiction sinceis not cyclic, so
The other possibility is
We then have
The group table is then
|
||||||
This group isomorphic to the permutation groupor the group of symmetries of the equilateral triangle, also called