The set of automorphisms of a group is itself a group. Since automorphisms preserve order and inverses, we can analyse the orders of each element to decide on possible automorphisms.
The orders of each element of
are given below
|
Element |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
Order |
0 |
10 |
5 |
10 |
5 |
2 |
5 |
10 |
5 |
10 |
If
is an automorphism of
then we can sort the elements into sets by the elements they can be mapped to by an automorphism. Each element in the same set has the same order.
{0}, {1, 3, 7, 9}, {2, 4, 6, 8} and {5}.
If 1 is mapped to 3 then 1+1=2 is mapped to 3+3=6, 1+1+1=3 is mapped to 3+3+3=9, 1+1+1+1=12 is mapped to 3+3+3+3=12 ≡ 2 (mod 10). Continuing in this way we obtain the mapping
![]()
Similarly, if 1 is mapped to 7,![]()
If 1 is mapped to 9,![]()
As soon as
is decided,
is fixed since
and similarly for all the other elements.
Call e the element![]()
Call a the element![]()
Call b the element![]()
Call c the element![]()
Then
sinc
and
since ![]()
The group of automorphisms of
is cyclic, with four elements, and is isomorphic to![]()