00:01
Hi, in this question to prove that pi is an automorphism, we need to show that it is both one to one and autofirst.
00:07
Let's show that pi is one to one.
00:09
So pi is one to one.
00:14
Suppose pi of a is equal to pi of b for some a, b in g.
00:26
Then we have that a power n is equal to b power n and a is equal to b power minus one into a, b.
00:37
Since n is relatively prime to the order of three, we can use the fact that a power n is equal to b power n to write that a is equal to b power k for some integer k.
01:02
Substituting this into the second equation, a is equal to b power a into the second equation.
01:08
So we get this second equation and this is first equation.
01:12
So substituting second equation, we get that a is equal to b power minus k, a b power k is equal to b power minus one into a, b, which implies that a is equal to b.
01:29
Since conjugation by b is an automorphism of g, conjugation by b is an automorphism of g.
01:47
Therefore, pi is one to one...