TITLE: Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n \mathbb Z$. Prove that either every element of $H$ is...
QUESTION [7 upvotes]: Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n\mathbb Z$. Prove that either every element of $H$ is even or exactly half of its elements are even. (Gallian, Contemporary Abstract Algebra, Exercise 24, Chapter 3.)

REPLY [9 votes]: This proof uses very elementary group theory (as per Gathdi's request) — it does not use homomorphisms or cosets, and it does not even assume the well-known result that every subgroup of a cyclic group is cyclic (a special case of it is proved here itself).
$\mathbb Z / n \mathbb Z = \{0, 1, \ldots, n - 1\}$.
Let $H$ be any subgroup of $\mathbb Z / n \mathbb Z$. If all elements of $H$ are even, we have the desired result. Otherwise, if at least one element of $H$ is odd, let $k$ be the least such odd element. Then we can prove that

$H = \langle k \rangle$, the cyclic subgroup generated by $k$, and
$H$ has even order, and half the elements of $H$ are even.

To prove the first statement, let $m$ be any element of $H$. Then using the division algorithm,
\begin{equation*}
m = qk + r, \quad 0 \le r < k
\end{equation*}
for some integers $q$ and $r$. Now, $r < k$ is an element of $H$ (since $r = m - qk$, $m, k \in H$), so it cannot be odd, due to our assumption on $k$. But if $r$ is even, then $k - r > 0$ is an odd element of $H$, which is possible only if $k - r \ge k$, so in fact, $k - r = k$ (since $r \ge 0$). Thus, $r = 0$, and $m = qk$, which proves that $H = \langle k \rangle$.
Now, the order of $H$ is the order of the element $k$, which is the smallest positive integer $s$ such that $sk = ln$, for some integer $l$ (as $n$ is the order of the group $G$). But given that $n$ is even, and $k$ is odd, it must be that $s$ is even. Thus, $H$ has even order, and since $H = \{0, k, 2k, 3k, \ldots, (s - 1)k \}$, exactly half of the elements of $H$, namely $0, 2k, 4k, \ldots, (s - 2)k$, are even.