# Number of elements with order $2$ in a finite abelian group

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 $$G$$?

• 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.
– lulu
May 26 at 21:24
• Is it better now? May 26 at 21:37
• 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 $e$). May 26 at 23:07

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 $$2^m-1$$.