Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb{Z}_n$.
Prove that either every member of $H$ is even or exactly half of the members of $H$ are even.
Well, I know that if $0 \in H$, then all even members of $\mathbb{Z}_n$ also must belong to $H$. And no odd numbers are needed in $H$ in order for it to be a group.
$0$ must belong to $H$ because it is the group's identity (and consequently all other even numbers) but if we put an odd number there, then all other odd number will also have to be in $H$.
Is it a good solution? I would appreciate all your help.
Thanks.
Best Answer
Sketch of proof:
Consider the homomorphism $f:H\to (\{\pm 1\},.)$ which sends only even numbers to $1$. Now apply (the corollary to) Lagrange's theorem, namely, $|H|=|$ker $f||$Im $f|$.
Clearly ker $f=\{\text{even numbers}\}$. If $|$Im $f|=1$ then $H$ consists only of even numbers. If $|$Im $f|=2$ then $|$ker $f|=|H|/2$ so that exactly half the numbers are even.