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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?

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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.

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    $\begingroup$ "at most one element of order two." So we take this and the identity as the two elements which square to $1$? $\endgroup$ – St Vincent May 2 '15 at 21:11
  • $\begingroup$ @StVincent, it is not that «we take them»... they are! $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 0:21
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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 $h^m = 1$.

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  • $\begingroup$ :we know in every group of even order there exist element of order two.if my group is cyclic just one element of order two.if my group not cyclic we can found more than.is it true? $\endgroup$ – pink floyd Apr 30 '15 at 19:49
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    $\begingroup$ There are finite abelian groups that only do have one element of order $2$, yet are not cyclic. Take $ℤ/2ℤ × (ℤ/3ℤ)^2$. $\endgroup$ – k.stm Apr 30 '15 at 20:27
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In additive form $\ \Bbb C_{2n}\cong\, \Bbb Z/2n\ $ where $\ x\cdot x = 1\,$ additively is $\ x\!+\!x = 0.\,$ This has solution

$2x\equiv 0\pmod{\! 2n}\!\iff\! 2n\mid 2x\!\iff\! n\mid x\!\iff\! x\equiv 0\pmod n\!\iff\! x\equiv\color{#c00}{0,n}\pmod{\!2n}$

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    $\begingroup$ I've read your answer a few times the other day and a couple more just now, and I have to say I find it most incomprehensible. :-/ $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 0:22
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    $\begingroup$ @Mariano Thanks, that surely a compliment coming from you! But, seriously, if you truly don't understand something then ask for an explanation. That's what comments are for (not your usual snide remarks). $\endgroup$ – Bill Dubuque May 3 '15 at 0:24
  • $\begingroup$ Oh, I do understand —what surprises me is the cryptic way of expressing it. For one thing: you follow the phrase «this has solution» (and you do not say exactly solution of what) with a sequence of equivalences, and surely those equivalences are not the solution of the equation. &c. $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 0:57
  • $\begingroup$ My comment is neither nonsensical nor, certainly, rude (and I have you in sufficient esteem to be sure that you do not think it is). If instead of «This has solution» you had written «The congruence class of an integer $x$ is a solution of this equation $x+x=0$ in the group iff $2x\equiv0\pmod{2n}$ &c» the sentence would be much better (and I cannot imagine you do not agree) $\endgroup$ – Mariano Suárez-Álvarez May 3 '15 at 1:16
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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.

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$H$ 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 $h^n$.

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