How to do this without knowledge of cyclic groups?

Question

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.?

Answer 1

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 $B$.

Answer 2

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 $A$. $\Box$