Suppose $G$ is an abelian finite group, and the number of order-2 elements in $G$ is denoted by $N$.

I have found that $N= 2^n-1$ for some $n$ that satisfy $2^n| \ |G|$. I write my proof. Would you tell me if this proof is correct? Moreover, Would you tell me can we say more about the number $N$?

My Proof

I: The subset of all elements with order 2 union $\{e \}$ is a subgroup of $G$ because for all $x \ne y: (xy)^2=x^2y^2=e$, and all elements are self inverse. Therefore, $N+1$ divides |G| by the Lagrange theorem.

II: If $N=1$ (i.e. there is only one element of order 2 in $G$), namely $x$, everything will be fine. However, if we have $x$ and $y$ as elements of order 2 in $G$, then $xy$ has order 2 and $N=3$. If there exists another element of order 2 in $G$, namely $z$, then $xz,\ yz,\ xyz$ have order 2 and $N=7$. If there exists another element of order 2 in $G$, namely $w$, then $xw,\ yw,\ zw,\ xyw,\ xzw,\ yzw, \ xyzw$ have order 2 and $N=15$. By induction, $N=$$n\choose{1}$$+\cdots +$${n}\choose{n}$ $= \ 2^n-1$ for some $n$.

With I and II, $N= 2^n-1$ for some $n$ that satisfy $2^n| \ |G|$.

Obviously, if |G| is odd, $N=0$. Or if $|G|=36$, $N=1$ or $N=3$.

Is what I wrote correct?

Can we be more specefic about the number of elements of order 2 in $G$?

  • $\begingroup$ It's not clear what you are asking. Please state, clearly, the assumptions you are making and what it is you want to deduce from them. $\endgroup$
    – lulu
    May 26 at 21:24
  • $\begingroup$ Is it better now? $\endgroup$
    – khashayar
    May 26 at 21:37
  • 1
    $\begingroup$ What you have written is correct in spirit, but it isn't an induction on the number of elements of order 2. You are actually doing induction on the size of a minimal generating set of the subgroup generated by the elements of order 2 (which as you rightly point out just comprises those elements together with $e$). $\endgroup$
    – Rob Arthan
    May 26 at 23:07

3 Answers 3


Yes, your argument is fine.

Note that the given subgroup of $N+1$ elements naturally carries a $\Bbb Z/2\Bbb Z$ vector space structure, so indeed its cardinality must be $2^n$.

Since such a vector space exists for all $n\in\Bbb N$, these ($2^n-1$) are exactly the possible numbers for $N$.


By the fundamental theorem of finite abelian groups, $G$ can be decomposed:

$G\cong\mathbb{Z_{2^{n_1}}}\times\mathbb{Z_{2^{n_2}}}\times...\times\mathbb{Z_{2^{n_k}}}\times H$

Where $H$ is a group of odd order. A cyclic group of even order has exactly one element of order $2$. So let's say $a_i$ is the element of order $2$ of the group $\mathbb{Z_{2^{n_i}}}$. Then clearly an element $g\in G$ satisfies $2g=0$ if and only if it has the form $g=(\epsilon_1a_1, \epsilon_2a_2,...,\epsilon_ka_k, 0)$ where $\epsilon_i\in\{0,1\}$. So the number of such elements is $2^k$. Thus the number of elements of order $2$ is $2^k-1$. So what you wrote is correct, and $k$ is the number of $2$-groups in the unique decomposition of $G$. (and for each $k$ there exists such a group where $N=2^k-1$, we can't say more)


One can be very explicit using the following fact : a finite abelian group is isomorphic to $\bigoplus_{i=1}^n \frac{\mathbb{Z}}{n_i \mathbb{Z}}$, so you only have to count the number of $n_i$'s which are even. If we call this number $m$, then the number of order 2 elements should indeed be $2^m-1$.


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