Question: 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.
I saw solution in
Either all elements of a subgroup of $\mathbb{Z}_n$ are even or exactly half of them
But I didn't understand the concept because I have not studied homomorphism yet. Also I asked the same question some time ago got one answer but the proof contained Lagrange's theorem and I also have not studied yet.
My background: Currently studying chapter 3 from the book Contemporary Abstract Algebra by Gallian.
Here is my attempt.
Proof:
If $H$ contains only even elements, then we are done.
Suppose $H$ contains both even and odd elements. Let
$O=\{x ∈H:x$ is odd $\}$ and $E = \{x ∈ H: x$ is even $\}$ s.t $H = E ∪ O$. Now we want to show both sets contain exactly same number of elements. Since $n$ is even, thus the sum of two even elements of $H$ is even and sum of two odd elements is even as well. Let $a ∈ O$ and define a map $f : O →E$ by $x ↦ x +a$. We clearly see $x ↦ x + a$ is bijective thus $|E|=|O|$ as desired. ∎
My Question:
Is my proof correct? If yes , then is it necessary to show that the map is bijective or not? And same for "sum of two evens (odds) is even" ?