TITLE: If an abelian group has more than 3 elements of order 2 then it must have at least 7 elements with order 2.
QUESTION [9 upvotes]: This is an exercise in Gallian, Contemporary Abstract Algebra.  Intuitively, I cannot believe the statement is true.  It is essentially saying there is no abelian group with 4, 5 or 6 elements of order 2 regardless of how large the group is.  I have tried to suppose a group had exactly 4 (or 5 or 6) elements of order 2 in hopes of reaching a contradiction.  I know that if a group is cyclic then it has exactly 1 element of order 2.  I know that an element of order 2 is its own inverse and it is not the identity.  I have observed that indeed in $C_2 \times C_2$ there are 3 elements of order 2 and in $C_2 \times C_2 \times C_2$ there are 7 elements of order 2.  However, none of this is convincing me that the statement is true.

REPLY [6 votes]: The elements of order $2$ together with the identity element in an abelian group form a subgroup. That subgroup is a vector space over the field $\mathbb{F}_2$ with two elements (under the action induced by the usual action that makes any abelian group into a $\mathbb{Z}$-module). A finite vector space over $\mathbb{F}_2$ has finite dimension $d$, say, and hence contains $2^d$ elements. All but one of these elements have order $2$. So the number of elements of order $2$ in an abelian group is either infinite or it is $2^d-1$ for some $d \in \mathbb{N}$.