If $H$ is a cyclic group of even order, $H$ has exactly two elements which square to $1.$ 
If $H$ is a cyclic group of even order, then $H$ has exactly two elements which square to $1.$

This was used in a answer (Pete Clark's answer) here: Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions
but I am not sure why this is true. Could someone please provide a proof to fill in some extra details?
 A: In $ℤ/2nℤ$, the equation $2x = 0$ has the solutions $x = 0$ and $x = n$ and no other solutions. Every cyclic group $H$ of even order is isomorphic to $ℤ/2nℤ$ for some $n ∈ ℕ$, and a multiplicative equation $x^2 = 1$ in $H$ then translates to $2x = 0$ in $ℤ/2nℤ$.

If you want to prove this directly in $H$: Let $h ∈ H$ be a generator of $H$ and $n = \frac{|H|}{2}$. Then the order of $h$ is $2n$, so $h^n·h^n = 1$. Next to $1·1 = 1$, this must be the only solution to $x^2 = 1$, because for all other $g ∈ G$, $g·g = h^k·h^k = h^{2k} ≠ 1$ for some $k ∈ \{1,…, n-1\}$, because the order of $h$ is the minimal positive integer $m$ such that $h^m = 1$.
A: Every subgroup of a cyclic group is cyclic.
In particular, the subgroup of elements of order dividing two is cyclic, and this clearly implies that there is at most one element of order two.
A: In additive form  $\ \Bbb C_{2n}\cong\, \Bbb Z/2n\ $ where $\ x\cdot x = 1\,$ additively is $\ x\!+\!x = 0.\,$ This has solution
$2x\equiv 0\pmod{\! 2n}\!\iff\! 2n\mid 2x\!\iff\! n\mid x\!\iff\! x\equiv 0\pmod n\!\iff\! x\equiv\color{#c00}{0,n}\pmod{\!2n}$
A: Given an integer $n$, one way to view your question is:

how many congruence classes $\xi$ in $\mathbb Z/2n\mathbb Z$ are such that $2\xi=0$?

Now, if $\xi$ is a congruence class in that quotient, we know that there is an integer $x$ in $\xi$ such that $0\leq x<2n$ and we have $2\xi=0$ in $\mathbb Z/2x\mathbb Z$ iff $2n\mid 2x$ in $\mathbb Z$, which happens exactly when $n\mid x$. Clearly, there are two possible values of $x$ satisfying this condition, namely $0$ and $n$, so the answer to the question as phrased above is two.
A: $H$ cyclic of even order means $H=\langle h\rangle$, with $h^{2n}=1$ (and $2n$ is the minimum such an integer). So the elements of $H$ are of the form $h^i,\;\;i=1,\dots,2n$.
Now the square of an element $h^i$ is thus $h^{2i}$ which is $1$ iff $i=n,2n$. Hence $H$ contains exactly two elements whose square is $1$: they are $1$ and $h^n$.
