Verified answer

You probably want to say that H is a proper subgroup of finite index and prove there is a proper normal subgroup of finite index. Otherwise you could just take H = G and G is a normal subgroup of finite index.

So suppose the index of H in G is n.

Then the homomorphism you have described is a homomorphism from G to S_n.

You can verify that it is, in fact, a homomorphism.

Let N be the kernel of this homomorphism.

Then N is normal in G and, by the isomorphism theorems, G/N is isomorphic to a subgroup of S_n. So G/N is finite. But the index of N in G is equal to the order of G/N. So N is a normal subgroup of finite index.

If x is in N, then under the homomorphism, x maps to the identity permutation. In other words xaH = aH for all a in G. So xH = H, so x is in H. Thus N is a subgroup of H, which shows that N is a proper subgroup.

If you write out the homomorphism a little more carefully you'll see that N is actually the intersection of all the conjugates of H.

documents by using skill of induction on the order 2n. assume your fact is incredibly for all abelian communities of order 2m with m unusual and m<n. choose g in G and permit H be the cyclic subgroup generated by using skill of g. by using skill of Lagrange the order of H divides 2n. (a) If | H | = 2n then G is cyclic and g^n has order 2. (b) Now assume H is a suited subgroup then a million < |H| < 2n. If 2 divides |H| then by using skill of the induction it has an element to reserve 2. (c) If 2 does not divide |H|, then it divides |G/H|. by way of fact G is abelian, H is (of course) properly-known, and G/H is a team. by way of fact |G/H| < 2n, then G/H as an element to reserve 2, represented by using skill of g ' H. So g '2 = g^r is in H. enable s be the order of g^r in G, then (g ')^s has order 2. In all circumstances there is an element to reserve 2. If G contained 2 factors of order 2 then they might generate a subgroup of order 4 it rather is impossible by way of fact 4 won't have the skill to divide 2n.