00:01
Hello, we are given that g is a finite group, h is a subgroup of g and n is a normal subgroup of g and it is given that if we have to show that if the order of h and the index of n in g, these are co -prime, this is dct, if are co -prime, show that we have to show that h is a subgroup of g, sorry h is a subgroup of n.
01:08
Okay, consider the canonical homomorphism p from g to g mod n.
01:20
Restrict this homomorphism to the subgroup h so that phi restricted to h is a group homomorphism from h to g mod n.
01:33
Let us call this, then we have that for this homomorphism by the first isomorphism theorem, the, before we go there, let us observe one more thing, the kernel, the kernel of phi restricted to h is equal to the kernel of phi of this map intersected with h, which is equal to n intersected with h.
02:19
For this map by the first isomorphism theorem, we have that image of phi restricted to h is isomorphic to h modulo the kernel of phi restricted to h.
02:53
Now, the order of this group, so this implies that the order of the image of phi restricted to h is equal to the order of h, order of h divided by the order of, the kernel is n intersection h, so it is n intersection h.
03:28
But note something, that the order of g mod n is equal to the index of n in g and the image of phi restricted to h is a subgroup of g mod n, the image is a subgroup of the codomain.
03:57
This implies that the order of image of phi restricted to h will divide the order of g mod n, which is equal to the index of n in g.
04:15
So, this order of this thing divides this, but this is nothing but the order of h divided by order of h intersection n...