This one is a well-known and important result:
Let be a group and a subgroup of Prove that if then
I love these problems.
It is well known that for any subgroup of finite index in there exists a normal subgroup of which is contained in This is proved by considering the set of all left cosets of in any element induces a permutation of by Then the mapping from to the symmetric group of is a homomorphism with kernel
Consider first the case when is finite. The number of conjugates of in is equal to the index of the normalizer of in As we have – and as is proper, Hence
This proves the result for the finite case.
For infinite, consider the normal subgroup above. The quotient group is a finite group since it is isomorphic to a subgroup of the symmetric group of degree So is as it is a subgroup of Thus is a proper subgroup of finite index of the finite group and so the union of its conjugates in is proper subset of The result follows since the mapping is a bijection from the set of all conjugates of in to the set of all conjugates of in