How to do this without knowledge of cyclic groups? I did the following exercise:

Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb{Z}_n$ (integers mod n with addition). Prove that either every member of $H$ is even or exactly half of the members of $H$ are even.

My answer:
Since $\mathbb{Z}_n$ is cyclic so is $H$. If $k$ generates $H$ when $k$ is even then every element in $H$ is even. If $k$ is odd then exactly every other element is even which proves the claim. 
Assuming my proof is correct I was wondering how else to do this. The exercise appears before the chapter about cyclic groups. 

How to answer this question without using any knowledge of cyclic
  groups, generators, etc.?

 A: Suppose there is an element $x$ that isn't even. Let $A$ be the set of even elements in the subgroup and define $B=\{x+a:a\in A\}$. Then every element of $B$ is odd. Prove that $A$ and $B$ have the same number of elements and the subgroup is the disjoint union of $A$ and $B$.
A: Here's a sketch of an alternate alternate proof: consider the map $f: x\mapsto x+x$, with $ran(f):=A$ the set of even elements. 

For every $a, b\in A$, $f^{-1}(a)$ has the same number of elements as $f^{-1}(b)$.

Proof sketch: Fix $c+c=a$ and $d+d=b$, and consider the map $g: x\mapsto x+d-c$. It's not hard to check that $g$ is a bijection from $f^{-1}(a)$ to $f^{-1}(b)$. $\Box$
Let $\xi$ be the number of elements in $f^{-1}(a)$ for $a\in A$.

$\xi=1$ or $\xi=2$.

Proof sketch: Clearly, $\xi\ge 1$, so we just have to show that $\xi\le 2$. To do this, note that for $0\le i<n$ we have $0\le 2i<2n$; so if $a$ is even, then $i+i\equiv a$ implies $i+i=a$ or $i+i=n+a$. $\Box$

If $\xi=1$, then every element is even.

Proof: Then $x\mapsto x+x$ is injective, hence bijective. $\Box$

If $\xi=2$, then exactly half the elements are even.

Proof: Consider the equivalence relation $\approx$ on $H$ given by $a\approx b$ if $a+a=b+b$. Since $\xi=2$, the $\approx$-classes all have exactly two elements, that is, $\approx$ partitions $H$ into pairs. The number of pairs is $\vert H\vert/2$, and each pair corresponds to a unique element of $A$. $\Box$
