00:01
Okay, so before we go on with the solution of this exercise, let me make a couple of remarks.
00:07
Remark number one, well, x, which is the set of functions going from a group g to itself, such that f is a bijective, well, this guy here is a group under composition of functions.
00:31
So under composition, perfect.
00:34
Remark number two, if f going from zn to itself is a homomorphism, then f of x is defined by multiplication, that is, f of the equivalence class of x is equal to the equivalence class of rx for some r belonging to zn and depending on f only.
01:30
Perfect.
01:32
Now, after these two remarks, our exercise is very easy to solve.
01:37
Indeed, for part a, we just need to observe that the set of automorphisms of g, which is a subset of x, well, this one is a subgroup, is a subgroup because the neutral element of x, which is the identity of g, is an automorphism, so it belongs to this guy here.
02:14
And moreover, if f and g are automorphisms of g, then f of the inverse of g is an automorphism.
02:29
Perfect.
02:30
Now, part b of our exercise, well, here we are going to use remark two, that is, we know that a homomorphism from z6 to z6 is defined by multiplication.
02:47
So if f going from z6 to z6 is a homomorphism, okay, homo, homomorphism, then f of the equivalence class of x is equal to the equivalence class of rx...