If
is 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.
If
has an element of order 6, it is cyclic and is isomorphic to![]()
Suppose
has no element of order 6, but has an element
of order 3, then
contains the elements
There must be some element
so by composing
with
and
we obtain the set
By cancellation none of these elements can be equal so these elements constitute the group. We can partially fill out the group table as below.
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can't be equal
by cancellation (if
then
), and can't be
or
(since each ealement can only appear once in each row and column. We are left with
or![]()
Suppose
then
![]()
![]()
![]()
![]()
so the order of
is 6 and the group is cyclic, but the group hase no element of order 6. The same analysis is done for the case
so that![]()
Now consider the product
in the second row, the same row as
or
and the same column as
so can't be equal to any of them, so
or![]()
Suppose
The powers of
are
since![]()
![]()
![]()
![]()
![]()
and the order of
is 6. As before this is a contradiction since
is not cyclic, so![]()
The other possibility is![]()
We then have
![]()
![]()
The group table is then
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This group isomorphic to the permutation group
or the group of symmetries of the equilateral triangle, also called![]()