 So that description of group automorphisms might have been a bit quick. So let's look at an example. So we're going to look at an example. Let's create a set. And our set contains three elements. And if I were to consider just the symmetric group on those three elements, we know that that would just be all the possible permutations. Now let's take a subset of that and that is just going to be the cyclic group on three elements. And we're going to have three of them. The first one is just going to map everything to itself. One to one, two to two, and three to three, or at least have that permutation. The second permutation is going to have one go to two and then two go to three. And then the next cyclic one is going to be k on three. One goes to three. Three goes two on, which goes to two. And there we go. We have the identity element. We have sigma and we have sigma squared. That's what we would call that because if we had a group composition of sigma, we'll get to this. So those are the three elements in my cyclic group. So my cyclic group in three elements is going to be this set of these three elements, sigma and sigma squared. And my group operation on that is just going to be the composition of these three elements. So let's have a look at the group automorphism. So I'm going to set up this mapping and it's going to map C3 to C3. And let's have a look at some possible ones to see if we indeed this is, base at least in this property of this isomorphic property. So let's have a look. Let's map F. Let's say that it maps E to E and it maps sigma to sigma squared. And it maps sigma squared to sigma. That might be one mapping and that would be an automorphism. It's a bijection because it is one to one and onto this mapping. So it is that. And what we just need now I need to see if I have this property that if I take one element and I compose it with another element if that is going to equal the same as doing this. We've seen that before. So let's just take two arbitrary ones and let's leave out the identity element. Let's look at this. So let's just look at our composition table, our Cayley's table. Under this we have E with E. We have sigma. We have sigma squared. We have sigma and we have sigma squared. We can have E, sigma, sigma squared, sigma and sigma squared. And if I have a sigma and a sigma that's going to leave me with sigma squared. And that means I leave, I've got to have an identity element. Sigma and sigma squared leaves me with identity element. Sigma squared and sigma leaves me with identity element. And sigma squared and sigma squared leaves me with sigma. So that's my Cayley's table that we have there. Let's do this properly. Sigma that leaves me there. Now let's take g1 composed with g2. So and this is then remember this is for all g1 and g2 now elements of c3 which is what we're dealing with here. So let's compose, let's take f that's my mapping. And we're going to take sigma. So that's going to be sigma and that's going to be sigma squared. So we're going to take sigma composed with sigma squared. And what does that leave us? If we compose sigma and sigma squared we get the identity element. And f maps the identity element to the identity element. Let's look see what happens on this side. I have the f of g1 we set that's sigma. And I'm going to compose that with sigma squared. And if sigma maps to sigma squared. And sigma squared maps to sigma. And if we have sigma squared and sigma we see we get the identity element. And those two are that is a beta. We have this isomorphic property. So indeed this is a lovely example of this group automorphism where I can map elements in a bijective form to some of the elements. One of the elements in the set to elements in the set. And it doesn't have to be the identity mapping. The identity mapping would just be e to e sigma to sigma squared sigma squared. Here's another one. So if I'm looking at the group of automorphisms of g. We're definitely going to have fe. And then we're going to have let's call this f1. And we're going to have f1. And indeed as we've proved before this is a group. So we can even compose those. We're still going to have closure. And we're still going to have associativity. We're going to have the identity element that is there. And we have the inverse. So yeah each one would be the inverse of itself of course. And so very beautiful example because it's an easy example. Short we only have three elements here. But I think now you can really understand what the concept is of this. Of group automorphisms.