Either all elements of a subgroup of $\mathbb{Z}_n$ are even or exactly half of them 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.
 A: 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.
A: Hint: Show that $K=H \cap 2 (\mathbb{Z}/n\mathbb{Z})$ (ie. the even integers of $H$) is a subgroup of $H$ of index at most two.
A: My motivation for answering this question:
Andrew your proof lacks both clarity and depth, and other proofs in the answer section are using very strong results to prove the point. So here is my simple (but long and detailed) proof avoiding strong results.
Proof:
Let us assume $\mathbb{Z}_n$ be a group under addition modulo $n$ where $n$ is positive even integer. Let $H$ be subgroup of $\mathbb{Z}_n$.
For any two elements in $\mathbb{Z}_n$ :
Odd + Odd = Even,
Odd + Even = Odd,
Even + Odd = Odd,
Even + Even = Even,
(Remember it happens only if n is even) otherwise in $\mathbb{Z}_7$, $1+2 = 3$ is odd but $1+6 = 0$ is even.
Now we show that
Part 1) if $H$ has both even and odd elements then both odds and evens are equal in number.
Part 2) if subgroup is of odd order then it can't possess any odd element.
Part 1)
Let us assume $H$ has both odd and even elements. Then $H$ can be written as:
$H$ = $\{0=a_1, a_2, a_2, ... a_r\} U \{b_1, b_2, ... b_t\}$ where $a_i$'s are even and $b_j$'s are odd.
Case 1: Let us say $r < t$ that is, we have more odds than even. Then consider following:
$b_1+b_1, b_1+b_2, b_1+b_3, ... , b_1+b_t$
Notice that all of $b_1+b_i$'s are even (as n is even) and for $i ≠ j$, $b_1+b_i ≠ b_1+b_j$ for say if they are then by cancellation law $b_i = b_j$ which is a contradiction as $H$ has $r+t$ elements. So there are atleast $t$ evens or we can say $r >= t$
Case 2: Let us say $t < r$ that is, we have evens more than odd in $H$, now consider the following:
$a_1+b_1, a_2+b_1, a_3+b_1, ... , a_r+b_1$
Notice that all of them are odd and unique (as n is even), for say $i ≠ j, a_i+b_1 = a_j+b_1$ then $a_i = a_j$ which is contradiction as $H$ has exactly $r+t$ elements. So there are atleast $r$ odd elements or we can say $t >= r$
From case 1 and case 2 we conclude that if $H$ has both even and odd elements then number of even elements have to match number of odd elements. Which can only happen if order of $H$ is even. So if subgroup is of even order, either all of them are even or exactly half of them are even.
Part 2):
We now show that if order of subgroup is odd then subgroup can't have odd member.
It is not possible for any subgroup to have all odd members as subgroup has to contain identity $0$ (in the case of operation addition modulo n), and $0$ is even, so if order of group is odd either all of them(elements of subgroup) are even or $H$ contains both even and odd member, but then from case 1 and case 2 of part 1), we conclude that $H$ has half the members even and half of them odd then order of $H$ has to be even, which is contradiction. Thus if $H$ is of odd order then it can only possess even members.
Finally proved !
