Suppose n is an even positive integer and H is a subgroup of Zn.
Prove that either every member of H is even or exactly half of the
members of H are even.
Let E be the set of all even elements in Z/nZ, and suppose that HnE is not H (so H contains at least one odd element). Define the map h:H --> {-1,1} by mapping even elements to 1 and odd elements to -1. Endowing {-1,1} with the multiplicative group structure, this is a group homomorphism. By assumption, this map is surjective, and so by the first isomorphism theorem of groups we have that H/ker(h) is isomorphic to {-1,1}. In particular, ker(h) = HnE, and so the collection of all even elements of H has an index of two. Thus 2|HnE| = |H| by Lagrange's theorem, which is precisely what we wanted to show.
Let E be the subgroup of Zn of all even residues. (Observe that evenness is well defined here. E is a subgp.)
Must show that either |H intersect E| = |H| (all elements of H are even), or |H intersect E| = |H|/2 (half the elements of H are even).
Consider the group HE = {h+e in Zn | h in H, e in E} (Notation, in this abelian case, it could be written H+E).
What are the possible groups that HE could be? (Hint - there are only 2, and you can name them, and prove it via Lagrange).
Then apply the 2nd isomorphism theorem.