10:21 In the fifth point, it looks like we need to prove φ(x^-1)= (φ(x))^-1. It can be done by the following steps. Let G be a group and φ be an isomorphism between G and H. First, let’s prove that the image of the identify element in G is the identify element in H, i.e φ(e) = e’. φ(g)*φ(e)= φ(g*e)= φ(g) for any element g from G. Next, φ(x^-1) φ(x) = φ(x^-1 * x) = φ(e) = e’. Thus, φ(x^-1) is also the inverse element to φ(x). Therefore, it can be written as φ(x)^-1, i.e. φ(x^-1)= (φ(x))^-1.
"sick"
lol, never expected to hear that.
I lost it, super funny
great video... helped alot!
Great sir💫
In (4), you used twice the fact that it's abelian. I think you meant cyclic.
10:21 In the fifth point, it looks like we need to prove φ(x^-1)= (φ(x))^-1.
It can be done by the following steps.
Let G be a group and φ be an isomorphism between G and H.
First, let’s prove that the image of the identify element in G is the identify element in H, i.e φ(e) = e’.
φ(g)*φ(e)= φ(g*e)= φ(g) for any element g from G.
Next, φ(x^-1) φ(x) = φ(x^-1 * x) = φ(e) = e’.
Thus, φ(x^-1) is also the inverse element to φ(x). Therefore, it can be written as φ(x)^-1, i.e. φ(x^-1)= (φ(x))^-1.