Show that any cyclic group of even order has exactly one element of order $2$


Let $G$ be a finite group of even order.

Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"

But how to show that exactly one such element if $G$ be cyclic.

NB: Please don't use isomorphic properties.

  • 4
    $\begingroup$ Let $n$ be the order of the group.Then $G$ is of the form $\{e,a,a^2,\dots,a^{n-1}\}$ for some element $a$. Show that $a^{n/2}$ has order $2$, but not the other elements. $\endgroup$
    – David
    Jan 10, 2016 at 17:59
  • $\begingroup$ G = <a>. $b^2 = 1$ $c^2 = 1$ $b = a^m$ $c = a^n$ $a^{2m} = a^{2n} = 1$ $\endgroup$
    – fleablood
    Jan 10, 2016 at 18:02
  • 1
    $\begingroup$ Remark: More generally, a cyclic group of order $n$ has precisely one subgroup of order $k$ for every positive integer $k$ which divides $n$. $\endgroup$
    – Mike F
    Jan 10, 2016 at 18:03
  • $\begingroup$ @fleablood if $a^{2m} = a^{2n} = 1$ then what you want to express exactly. Please write in details. $\endgroup$ Jan 10, 2016 at 18:07
  • $\begingroup$ It was a comment and a hint. Not an answer. You figure it out. Hint: $a^{|G|} = 1$. So $2m = k*|G|$ and $2n = j*|G|$ but $m,n < |G|$. $\endgroup$
    – fleablood
    Jan 10, 2016 at 18:43

7 Answers 7


If $G$ is a finite cyclic group of order $n$, then $G$ has a unique subgroup of order $m$ for each $m$ dividing $n$. Now, if $n$ is even, then $G$ has a unique subgroup of order $2$, hence one element of order $2$ (as if $G$ had two distinct elements of order $2$, then we'd have two distinct subgroups of order 2).

  • $\begingroup$ I've added this to my list of "favourite proofs": I love how it cuts through the fog in such a short and succinct fashion. It also readily adapts to prove that "A cyclic group of order $n$ has exactly one element of order $m$ for each value of $m$ that divides $n$. $\endgroup$ Jul 8, 2021 at 10:37
  • 1
    $\begingroup$ Has only one subgroup of order m, not one element! $\endgroup$
    – mich95
    Jul 8, 2021 at 14:14
  • $\begingroup$ Oops - Yes ! Thanks. $\endgroup$ Jul 8, 2021 at 14:41

G is cyclic of order $m$ means that G = <$a$> = {$e, a, a^2, ... a^{m-1}$} where $a^j \ne a^k$ for all $j \ne k; j,k < m$.

This means $a^m = e$. (See postscript.)

Suppose $m = 2k$; an even number.

Then $g = a^k$ means $g \ne e$ and $g^2 = (a^k)^2 = a^{2k} = a^m =e$ so $g = a^k$ has order 2.

Let $g \ne a^k; g \in G; g \ne e \implies g = a^j; j \ne k; g =a^j \ne e$ (so $j \ge 1$).

Case 1: $0< j < k$. Then $2j < 2k = m$ so $g^2 = a^{2j} \ne e$. So $g = a^j$ does not have order 2.

Case 2: $k < j < m$. Then $m = 2k < 2j < 2m$ so $0 < 2j - m < m$ and $g^2 = a^{2j} = a^m*a^{2j - m} = a^{2j - m} \ne e$. So $g = a^j$ does not have order 2.

So $g = 2^k$ is the unique element of order 2.



$a^m = e$.

Proof: $a^m = a*a^{m - 1} \in G$ so $a^m = a^j; j < m$ so $a^{m -j} = e$ so for $a^k; k \ge m - j; a^k = a^{k - (m-j)}$ is a duplicate of group elements. So order of G is at most $m - j$. But order of G is $m$. So $j = 0$ so $a^m = e$.


Use this proposition: If $|x|=n$, then $|x^{a}|=\frac{n}{(a,n)}$, where $(a,n)$ is the greatest common divisor of $a$ and $n$. Let $G=\{1,x,x^{2},\cdots,x^{n-1}\}$. We know $x^{n/2}$ is an element of order $2$. To show uniqueness, let $x^{m}$ be an element of order $2$, where $m\in\{1,\cdots,n-1\}$. We show that $m=\frac{n}{2}$. By the proposition, $|x^{m}|=\frac{n}{(m,n)}=2$, so $(m,n)=\frac{n}{2}$. Thus, $\frac{n}{2}$ divides $m$. $m=\frac{n}{2}k$ for some $k\in \mathbb{Z}$. Since $m\in\{1,\cdots,n-1\}$, $k=1$. $m=\frac{n}{2}$. So $x^{n/2}$ is the only element of order $2$.

  • $\begingroup$ Would you please explain why you conclude $k=1$. I am unable to understand this. $\endgroup$ Jan 12, 2016 at 15:55
  • 1
    $\begingroup$ Since $k$ is an integer, $k\in\{\cdots,-1,0,1,\cdots\}$. Then $m\in\{\cdots,-\frac{n}{2},0,\frac{n}{2},n,\cdots\}$. Since we also have $m\in\{1,\cdots,n-1\}$, the only way is that $m=\frac{n}{2}$. $\endgroup$
    – Delong
    Jan 13, 2016 at 1:00

Here is another proof: it's fairly elementary, but it's also a bit long-winded, I apologize for that. But hey, sometimes the scenic route is interesting.

We start with what we know, $G = \langle x\rangle$, and that $x$ has order $m = 2k$, where $k$ is a positive integer.

We'll do the "easy part" first:

$G$ has at least one element of order $2$:

Consider $x^k$. We see that $(x^k)^2 = x^{2k} = x^m = e$. Thus $x^k$ has order dividing $2$, so either $1$, or $2$. If $x^k$ had order $1$, we would have $x^k = e$, and since $0 < k$, then $k < 2k$, contradicting the fact that $2k = m$ is the LEAST positive integer $n$ such that $x^n = e$. So $x^k$ does not have order $1$, and thus has order $2$.

Now the "harder part", here is where we will take our scenic detour:

Claim: any subgroup of $G$ is also cyclic, with generator $x^r$ for some $0 \leq r < m$. Let's call our subgroup $H$. If $H = \{e\}$, there is nothing to prove, the trivial group is clearly cyclic (the identity is a generator). So assume $H$ has some non-identity element in it. So we can write:

$H = \{e,x^{r_1},x^{r_2},\dots,x^{r_t}\}$, for some positive integers $r_1,r_2,\dots, r_t$. Choose $r = \min(r_1,r_2,\dots,r_t)$. I claim $x^r$ generates $H$. For suppose it did not. Then we have some $r_j$ such that $x^{r_j}$ is not a power of $x^r$, that is $r_j$ is not a multiple of $r$. In other words, we can write:

$r_j = qr + u$, where $0 < u < r$.

It follows that $x^u = x^{r_j - qr} = x^{r_j}[(x^r)^q]^{-1}$.

Now $x^{r_j} \in H$, and $x^r \in H$, so any power $(x^r)^q \in H$ (by closure), and thus (since $H$ is a subgroup, and contains all inverses) $[(x^r)^q]^{-1} \in H$, and thus $x^u \in H$, being the product of two elements in $H$.

But $0 < u < r$, contradicting our choice of $r$ as the least positive power of $x$ in $H$ (dramatic pause).

So...since assuming $x^r$ does not generate $H$ leads to a contradiction,...can't have that.

Now...where were we? Oh yeah, we want to show that $G$ has AT MOST one element of order $2$. Again, let's suppose it has at least $2$, we'll call them $a$, and $b$.

Now cyclic groups are abelian (I hope you already know this, or I will certainly try any reader's patience proving that, too), so in particular $ab = ba$. Hence:

$(ab)^2 = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a^2b^2 = ee = e$. So either $ab$ also has order $2$, or order $1$.

If $ab = e$, then $b = a^{-1} = a$ (since $a$ has order $2$), but we started out assuming $a$ and $b$ are two different elements of $G$. So this can't be right, it must be that $ab$ has order $2$ as well.

And now, the punch line:

It is easy to see that $\{e,a,b,ab\}$ is closed under multiplication, and possesses all inverses, and is thus a subgroup of $G$. But it's not cyclic! C'est impossible! (we just wasted 10 minutes of our lives proving it above).

And that contradiction shows more than one element of $G$ of order $2$ cannot be.


Let $G$ be a cyclic group of even order $2n$ and neutral element $e$. Then there's a $g\in G$ with order $2n$ and so $$G=\{g^k\mid 0\leq k\leq 2n-1\}.$$ Then $g^n$ is the only element with order $2$.


If you have any element $g^l\in G, 1\leq l\leq 2n-1$, with order $2$ than $(g^l)^2=g^{2l}=e$. So $2l\geq 2n$ and so $l\geq n$. Suppose $l>n$. Than there is a $1\leq k<n$ with $l=n+k$ and $g^{2l}=g^{2n}g^{2k}=g^{2k}=e$ what is a contradiction because $1\leq 2k<2n-1$.

  • $\begingroup$ How to show it is is the only element with order 2 $\endgroup$ Jan 10, 2016 at 18:21
  • $\begingroup$ I have added a prove of this. $\endgroup$
    – user302982
    Jan 10, 2016 at 18:33

Let $(G.e)$ by a finite cyclic group containing an involution $v$. By Lagrange's theorem the order of $G$ is even, and for some $m \ge 1$ we can write

$\quad |G| = 2m$

If $g$ is any generator of $G$ there is a minimal integer $k$ satisfying

$\tag 1 1 \le k \lt 2m \; \land \; \{g^1,g^2,\dots,g^k\} \cap \{e,v\} = \{v\}$

The theory found here allows us to assert that $k$ must be equal to $m$.

So by fixing the generator $g$ and 'running' the above argument against involutions $v, u \in G$, we see that $v = g^m$ and $u = g^m$. Using basic properties of the equality relation, the involutions $u$ and $v$ are equal.

We have shown that if a finite cyclic group contains an involution then its order is even and, moreover, it contains exactly one element that is an involution.


Let $g$ be a generator of $G$ and $G$ has order $n$. Suppose, $(g^m)^k=e$. Then $km=nr$ for some integer $r$. Let $d=(m,n)$ and $l=mn/(m,n)$. Note that $l$ is the least common multiple of $n$ and $m,n$ and $km=nr=lt$ for some $0\le t<d$. Now, $k=n/d$ and $o(g^k)=d$. Letting $m=2$, we get that $o(g^{m/2})$ is the only element of order $2$.

  • $\begingroup$ Can it not be solved in way of solving method I have used? $\endgroup$ Jan 10, 2016 at 18:05
  • $\begingroup$ Well you let $a=g^m$, and $b=g^k$, and assume that both have order $2$. Now, you need to show that $a=b$, or equivalently, that $m=k$. $\endgroup$ Jan 10, 2016 at 18:10
  • $\begingroup$ Yes, nice, but how to show $m=k$ then? $\endgroup$ Jan 10, 2016 at 18:13
  • $\begingroup$ Well I think the easiest way to do this, is the way I did it in my answer, which is to show that for any $1\le m\le n$, that $o(g^m)=n/d$. $\endgroup$ Jan 10, 2016 at 18:15
  • $\begingroup$ @TimRaczkowski why do you start by supposing $(g^m)^k=e$? is your assumption that $g^m$ and $g^k$ are distinct elements of $G$ with order 2? and if that is your assumption, does that tell you that $(g^m)^k=e$? $\endgroup$
    – Mjoseph
    May 5, 2019 at 0:30

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